12
GENERAL PHYSICS 1
QUARTER 2
LEARNING ACTIVITY SHEETS
Week 1 - 4
Republic of the Philippines
Department of Education
COPYRIGHT PAGE
Learning Activity Sheet in EARTH SCIENCE
(Grade 12)
Copyright © 2020
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Table of Contents
Compentency
Calculate the moment of inertia about a given
axis of single-object and multiple-object
systems
Calculate magnitude and direction of torque
using the definition of torque as a cross
product
Describe rotational quantities using vectors
Determine whether a system is in static
equilibrium or not
Apply the rotational kinematic relations for
systems with constant angular accelerations
Determine angular momentum of different
systems
Apply the torque-angular momentum relation
Solve static equilibrium problems in contexts
but not limited to see-saws, cable-hinge-strutsystem, leaning ladders, and weighing a heavy
suitcase using a small bathroom scale
Use Newton’s law of gravitation to infer
gravitational force, weight, and acceleration
due to gravity
Discuss the physical significance of
gravitational field
Apply the concept of gravitational potential
energy in physics problems
Calculate quantities related to planetary or
satellite
motion
For circular orbits, relate Kepler’s third law of
planetary motion to Newton’s law of gravitation
and centripetal acceleration
Relate the amplitude, frequency, angular
frequency, period, displacement, velocity, and
acceleration of oscillating systems
Recognize the necessary conditions for an
object to undergo simple harmonic motion
Calculate the period and the frequency of
spring mass, simple pendulum, and physical
pendulum
Differentiate underdamped, overdamped, and
critically damped motion
Define mechanical wave, longitudinal wave,
transverse wave, periodic wave, and
sinusoidal wave
Code
Page
number
STEM_GP12RED-IIa-1
1 – 14
STEM_GP12RED-IIa-3
STEM_GP12RED-IIa-4
15 – 22
23 – 32
STEM_GP12RED-IIa-5
33- 47
STEM_GP12RED-IIa-6
48 – 57
STEM_GP12RED-IIa-9
STEM_GP12RED-IIa10
58 – 67
STEM_GP12RED-IIa-8
78 – 92
STEM_GP12G-IIb-16
93 – 105
STEM_GP12Red-IIb-18
106 – 117
STEM_GP12Red-IIb-19
118 – 130
STEM_GP12Red-IIb-19
131 – 145
STEM_GP12G-IIc-22
146 – 155
STEM_GP12PM-IIc-24
156 – 165
STEM_GP12PM-IIc-25
166 – 183
STEM_GP12PM-IIc-27
184 – 204
STEM_GP12PM-IId-28
205 – 219
STEM_GP12PM-IId-31
220 – 232
68 – 77
From a given sinusoidal wave function infer
the speed, wavelength, frequency, period,
direction, and wave number
Apply the inverse-square relation between the
intensity of waves and the distance from the
source
STEM_GP12PM-IId-32
233 – 245
STEM_GP12MWS-IIe34
246 – 259
GENERAL PHYSICS 1
Name: ________________________________
Date: ______________
Grade : ________________________________
Score: _____________
LEARNING ACTIVITY SHEETS
Moment of Inertia
Background Information for The Learners
How was your experience when it was your
first time to ride in a Ferris wheel? The moment
it starts rotating about its center, you feel as if
you want to stop it from rotating, isn’t it? But do
you know how much effort must be given to the
Ferris wheel to stop it from rotating? To answer
that question, first you need to understand
completely the moment of inertia.
Concept of Moment of Inertia
Newton’s Law of Inertia says that an object at rest tends to stay at rest, and
an object in motion tends to stay in uniform motion unless acted upon by an
unbalanced force. This tendency of the object to keep whatever it is doing and resist
any change in its state of motion is called inertia.
Just like how an object continues to be in its state of rest or in its state of uniform
motion, an object rotating about its axis tends to remain rotating about the same axis
unless hindered by any external force. This property of the object to resist any change
in its rotational state of motion is called moment of inertia. Moment of inertia is also
known as rotational inertia since it appears in objects with rotational motion. Also, it
gives us the idea of how difficult to make an object rotate and to stop an object from
rotating about its axis.
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Calculating Moment of Inertia
In translational motion, inertia depends on the mass of the object. But in
rotational motion, moment of inertia depends on how mass is distributed around an
axis of rotation and it varies depending on the chosen rotation axis.
For a single object or point-like object, moment of inertia can be generally
expressed as:
𝑰 = 𝒎𝒓𝟐
where: I = moment of inertia
m = mass of the object
r = perpendicular distance of the object from the axis of rotation
Consider a single object rotating about a fixed axis in Figure 1. Axis of rotation
is an imaginary straight line in which all parts of the object rotates. It is always
perpendicular to the rotation of the object.
For example, the object in Figure 1 is 0.1 kg. It is attached
to a 0.5-m string and is rotated about a fixed axis. What is
the moment of inertia of the object?
Solution:
I = mr2 = (0.1 kg) (0.5 m)2 = 0.025 kg·m2
So in rotating a 0.1 kg object moment of inertia is 0.025
kg·m2.
For a multiple-object system, where mass is not focused at a single point
and it consists of few particles, we can calculate its moment of inertia about the given
axis of rotation by adding up all the moments of inertia of all the particles present in
the system. In symbols:
𝐼 = ∑ 𝑚𝑟 2 = (𝑚1 𝑟 21 ) + (𝑚2 𝑟 2 2 ) + (𝑚3 𝑟 2 3 ) + ⋯
For example, three 0.1-kg balls are attached to a string and rotated about an
axis. Balls 1, 2, and 3 are 0.5 m, 0.3 m, and 0.1 m, respectively, away from the axis
of rotation. Calculate the moment of inertia of the system.
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Solution:
𝐼 = ∑ 𝑚𝑟 2
I = (mr2)1 + (mr2)2 + (mr2)3
= (0.1 kg) (0.5 m)2 + (0.1 kg) (0.3 m)2 + (0.1 kg) (0.1 m)2
= 0.035 kg·m2
Thus, the system’s moment of inertia is 0.035 kg·m2.
But most of the time, the object consists of a great number of particles. Using
integration in this case would be practical than using summation.
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The illustration below gives the moments of inertia for various objects as a
result of integration:
Image via https://openstax.org/books/university-physics-volume1/pages/10-4-moment-of-inertia-and-rotational-kinetic-energyby used
under CC BY 4.0/ Modified from the original
Learning Competency
Calculate the moment of inertia about a given axis of a single-object and multipleobject systems. (STEM_GP12REDIIa-1)
ACTIVITY 1: A Moment to Explore Rotational Inertia
Directions: Analyze the situations and then answer the questions.
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Situation A: A long pole is rotated around three different rotation axes: central core
axis, midpoint axis, and one end axis as shown in figure 2. The pole is easiest to
rotate about its central core axis, and it is hardest to rotate around its one end axis.
Analysis:
1. Which axis of rotation the pole obtains the greatest moment of inertia?
________________________________________________________
2. In which axis of rotation, the pole had the smallest moment of inertia?
________________________________________________________
3. How do the axes of rotation affect the rotation of the pole? (Hint: Relate it to
the moment of inertia.)
_____________________________________________________________
_____________________________________________________________
______________________________________________
Situation B: Two sticks as shown in figure 3 are being held to
stand on the floor with a little inclination. When the sticks are
released, the stick without an added weight on its top end rotates
to the floor faster.
Analysis:
1. Why do the two sticks rotate to the floor at a different rate or speed? (Hint:
Use the moment of inertia.)
_____________________________________________________________
_____________________________________________________________
______________________________________________
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ACTIVITY 2: A Moment to Complete Rotational Inertia
Directions: Complete the table. Indicate the rank of the objects’ moment of inertia in
a descending order.
Distance, m
Object
No.
Mass, kg
(Object to
Moment of Inertia,
Rotation Axis)
kg·m2
1
36
1
2
9
2
3
4
3
Rank
Question:
1. What are the factors that affect the moment of inertia of an object?
________________________________________________________
ACTIVITY 3: A Moment to Match Rotational Inertias
Directions: Match the word problem in column A with its answer in column B. Write
the letter of the answer in the blank provided before the item.
A
_____ 1. A mass of 10 kg, which may be a
B
A. 3.16 m
point-like object, is attached to a
rope of length 1.5 m and is being
B. 1.44 kg·m2
rotated. What is the moment of
inertia of the object?
C. 10 m
_____ 2. How far is the object from its axis of
rotation if it is 4 kg and has a
D. 22.5 kg·m2
moment of inertia 40 kg·m2?
_____ 3. Three balls are attached to a cable
E. 0.04 kg
and are being rotated. Ball A is 0.5
kg and is 1.0 m away from the axis
of rotation. Ball B is 1.0 kg and
placed 0.8 m away from the axis.
Ball C, which is 0.5 m away from
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the axis, is 1.2 kg. Calculate the
total moment of inertia of the balls.
_____ 4. The moment of inertia of the ball is
0.01 kg·m2 and is rotating around a
0.5-m string. What is the mass of
the ball?
ACTIVITY 4: A Moment to Level Up in Rotational Inertia
Directions: Read and understand the situation given below. Solve for the moment of
inertia of the system of objects and show your solution.
Situation: Figure 4 shows an object consisting of two
point-like objects of mass m connected by
a
rod of length L and mass 2m. What is the
moment of inertia of the object about an
axis through its center and perpendicular
to
the rod?
Solution:
Reflection
1. I learned that _________________________________________________
__________________________________________________________________
__________________________________________________________________
______________________________________________________
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2. I enjoyed most on ______________________________________________
__________________________________________________________________
__________________________________________________________________
______________________________________________________
3. I want to learn more on __________________________________________
__________________________________________________________________
__________________________________________________________________
______________________________________________________
References
Halliday, David, Resnick, Robert, & Walker, Jearl. Fundamentals of Physics. 6th ed.
New York: John Wiley & Sons Inc, 2001.
Hewitt, Paul G. Conceptual Physics. 10th ed. United States of America: Pearson
Addison-Wesley, 2006.
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Moore, Thomas A. Six Ideas that Shaped Physics, Unit C: Conservative Laws
Constrain Interactions. 2nd ed. New York: Mc Graw Hill, 2003.
Santos, Gil Nonato C. General Physics 1. 1st ed. Quezon City, Philippines: Rex Book
Store, 2019.
ANSWER KEY
ACTIVITY 1
Situation A
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1. The pole obtains the greatest moment of inertia when the axis of rotation is on
its one end.
2. The pole has the smallest moment of inertia when it is rotated about its central
core.
3. The closer the distribution of mass to the rotation axis, the lower its moment
of inertia, hence the easier it is to rotate. As a result, it is much easier to rotate
a pole about its central core than about its midpoint or one end.
Situation B
1. Because the two sticks have different moment of inertia. The stick with added
weight on its top end rotates slower because it has higher moment of inertia
than the other stick. So, it has greater ability to resist rotation than the other
stick.
ACTIVITY 2
Object
Moment of Inertia,
No.
kg·m2
1
36
2
36
3
36
Rank
all tie
1. Moment of inertia is proportional to the mass of the object and the square of
the object’s perpendicular distance from the axis of rotation.
ACTIVITY 3
1. D
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2. A
3. B
4. E
ACTIVITY 4
𝐿 2
𝐿 2
1
𝐼 = 𝑚 ( ) + 𝑚 ( ) + (2𝑚)(𝐿)2
2
2
12
𝑚𝐿2 𝑚𝐿2 2𝑚𝐿2
𝐼=
+
+
4
4
12
1
1
𝐼 = 𝑚𝐿2 + 𝑚𝐿2
2
6
2
𝐼 = 𝑚𝐿2
3
Prepared by:
Techie Gammad-Vera Cruz
Amulung National High School
GENERAL PHYSICS 1
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Name: ________________________________
Date: ______________
Grade: _________________________________
Score: _____________
LEARNING ACTIVITY SHEETS
Torque
Background Information for The Learners
We push or pull a door on its knob whenever we want to open or close it. We
use a wrench to tighten or loosen bolts. We also discover from experience that the
amount of force applied is not enough to rotate the object – where and how the force
is applied also matters. Try to open a door by pushing it towards its hinges. You
would notice that the door will not open well because it will not create a rotational
motion. Why? The answer lies in the concept of torque.
Concept of Torque
Torque originates from the Latin word torquere, which means to twist. It is the
rotational equivalent of force, thus also known as moment or moment of force. Just
like how force is needed to alter the object's state of linear motion, torque is
necessary to change the object's state of rotation. In vector form, it is defined as:
𝜏 =𝑟×𝐹
where 𝜏 is the torque (pronounced as tau)
F is the force acting on the object
r is the object’s lever arm or moment arm (the position vector of the
point where the force is applied relative to the axis of rotation)
When two vectors are multiplied through cross-product (A x B), the resulting
quantity is a vector. Since force, F, and lever arm, r, are both vectors, the cross
product, torque, is a vector quantity that has both magnitude and direction.
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Magnitude and Direction of Torque
The magnitude of torque is defined as follows:
𝜏 = 𝑟 × 𝐹 = |𝑟|𝐹⊥
𝜏 = |𝑟||𝐹| sin 𝜃
where r is the lever arm
F is the applied force
θ is the angle between the applied force and lever arm.
Figure 1 shows the two components of F, the
parallel (𝐹∥ ) and the perpendicular (𝐹⊥ ). Since 𝐹∥ acts
along the line of the lever arm, r, it cannot cause rotation.
Only the 𝐹⊥ does cause rotation of the object, and it is
equal to |𝐹| sin 𝜃.
The direction of the torque is always
perpendicular to both 𝐹⊥ and r as defined by the righthand rule: If you point your index finger in the direction
of the lever arm, r, and your middle finger in the
direction of the perpendicular component of force, 𝐹⊥,
then your thumb points in the direction of torque, 𝜏. See
figure 2.
Take note of the following symbols when dealing
with three-dimensional directions:
Symbol
Direction
Out of the plane or page
Into the plane or page
Hint
As if you are looking at the head of an
arrow as it moves towards you.
As if you are looking at the tail of an
arrow as it moves away from you.
In figure 1, torque 𝜏, is directed out of the page. Consider another example: A
5.0-N force is applied to one end of the lever that has a length of 2.0 meters. The
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force is applied directly perpendicular to the lever, as
shown in the diagram. What is the magnitude and
direction of the torque acting on the lever?
Solution:
For the magnitude:
𝜏 =𝑟 ×𝐹
𝜏 = |𝑟||𝐹| sin 𝜃
𝜏 = (2.0 𝑚)(5.0 𝑁) sin 90
𝜏 = 10 𝑁𝑚
For the direction:
The lever arm is pointing to the right; the force is upward; hence, the direction
of the torque is out of the page.
Note that the SI unit for torque is newton-meter (N·m).
Learning Competency
Calculate magnitude and direction of torque using the definition of torque as a crossproduct. (STEM_GP12REDIIa-3)
ACTIVITY 1: Let’s Investigate Torque
Directions: Read the statements/questions carefully. Choose the letter of the correct
answer.
1. Torque is a _______ and its direction can be determined using _______.
A. scalar; right-hand rule
C. vector; right-hand rule
B. scalar; left-hand rule
D. vector; left-hand rule
2. The magnitude of the cross product of vectors A and B is
A. 𝐴𝑥𝐵 = |𝐴||𝐵| sin 𝜃
C. 𝐴𝑥𝐵 = |𝐴||𝐵| cos 𝜃
B. 𝐴𝑥𝐵 = 𝐴𝐵 sin 𝜃
D. 𝐴𝑥𝐵 = 𝐴𝐵 cos 𝜃
For items 3-5, consider figure 3. One end
of
the stick is attached to a plank of wood.
3. The stick rotates when a force is
applied on its ______ end and is
directed along the ______.
A. fixed; z axis
C. free; x axis
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B. free; z axis
D. free; y axis
4. When torque is zero, the stick will not rotate. Which conditions result to a zero
torque?
I. The force is applied on the fixed end of the stick.
II. The force is applied along z axis at the middle of the stick.
III. The force is applied along z axis on the free end of the stick.
IV. The force is applied along x axis on the free end of the stick.
A. I only
C. III and IV
B. I and II
D. I and IV
5. The stick is pushed near its free end along -z axis. What is the direction of the
torque?
A. along +x axis
C. along +y axis
B. along -x axis
D. along -y axis
ACTIVITY 2: Let’s Twist our Hands
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Directions: Use the right-hand rule to determine the direction of the torque. Indicate
or draw the direction in the diagrams or figures.
ACTIVITY 3: Let’s Appraise Torque
Directions: Solve the following problems and show your solution.
1. The length of a bicycle pedal arm is 0.152 m, and a downward force of 111 N
is applied to the pedal by the rider’s foot. What is the magnitude of the torque
about the pedal arm pivot point when the arm makes an angle of (a) 30°, (b)
90°, and (c) 180° with the vertical?
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2. A force of 60 N is applied to the end of a wrench
12 centimeters long. How much torque is
produced? What is the direction of the torque?
3. The figure shows a stick that can
pivot about the dot marked O.
Rank the three forces (A, B, C)
according to the magnitude of the
torque they produce, greatest
first.
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Reflection
1. I learned that _________________________________________________
__________________________________________________________________
__________________________________________________________________
______________________________________________________
2. I enjoyed most on ______________________________________________
__________________________________________________________________
__________________________________________________________________
______________________________________________________
3. I want to learn more on __________________________________________
__________________________________________________________________
__________________________________________________________________
______________________________________________________
References
Halliday, David, Resnick, Robert, and Walker, Jearl. Fundamentals of Physics. 6th
ed. New York: John Wiley and Sons Inc, 2001.
Hewitt, Paul G. Conceptual Physics. 10th ed. United States of America: Pearson
Addison-Wesley, 2006.
Moore, Thomas A. Six Ideas that Shaped Physics, Unit C: Conservative Laws
Constrain Interactions. 2nd ed. New York: Mc Graw Hill, 2003.
Santos, Gil Nonato C. General Physics 1. 1st ed. Quezon City, Philippines: Rex Book
Store, 2019.
Serway, Raymond A. and Jewette, John W. Jr. Physics for Scientists and Engineers
with Modern Physics. 6th ed. Singapore: Thomson Learning Asia, 2004.
“Torque
Worksheet”.
Accessed
June
15,
2020.
https://www.npsd.k12.nj.us/cms/lib04/NJ01001216/Centricity/Domain/474/To
rque%20WORKSHEETS%202014.pdf.
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ANSWER KEY
ACTIVITY 1
1. C
4. D
2. A
5. C
3. B
ACTIVITY 2
1. Out of the page
6. Out of the page
2. Into the page
7. Out of the page
3. Into the page
8. Leftward
4. Into the page
9. Upward
5. No torque
10. Rightward
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ACTIVITY 3
1. (a) 8.44 N·m
(b) 16.87 N·m
(c) 0
2. 7.2 N·m, out of the page
3. 1st – 𝜏𝐵 = 700 𝑁𝑚
2nd – 𝜏𝐴 = 585 𝑁𝑚
3rd – 𝜏𝐶 = 0
Prepared by:
Techie Gammad-Vera Cruz
Amulung National High School
GENERAL PHYSICS 1
Name: ________________________________
Date: ______________
Grade: _________________________________
Score: _____________
LEARNING ACTIVITY SHEETS
Rotational Quantities
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Background Information for The Learners
When people are asked which horse moves faster
on a merry-go-round, some will answer that the horse
near the outside rail moves faster, while others will say
that the two horses move at the same speed. This
conflict of answers depends on the kind of motion used.
Those who chose the horse near outside the rail used
translation, while those who say that both moves at the
same speed used rotation. Translation is the motion along a straight line, while
rotation is the motion requiring an object to rotate about its fixed axis. The table
below shows the equivalence of translational and rotational motions.
Table 1: Translational quantities and their equivalence in rotational motion.
Translation
Rotation
Quantity
Symbol
Symbol
𝑥 𝑜𝑟 𝑦
𝜃
Angular Position
∆𝑥 𝑜𝑟 ∆𝑦
∆𝜃
Angular Displacement
Velocity
𝑣
𝜔
Angular Velocity
Acceleration
𝑎
𝛼
Angular Acceleration
Mass or Inertia
𝑚
𝐼
Moment of Inertia
Force
𝐹
𝜏
Torque
Linear Momentum
𝑝
𝐿
Angular Momentum
𝐹𝑑
𝜏𝜃
Work
1⁄ 𝑚𝑣 2
2
1⁄ 𝐼𝜔2
2
Fv
𝜏𝜔
Position
Displacement
Work
Kinetic Energy
Power
Quantity
Rotational Kinetic Energy
Power
Basic Rotational Quantities
The angular position is the angle through which a point revolves around a
center or through which line has been rotated about a specified axis. Its value is
positive when the rotation is counterclockwise and negative when the rotation is
clockwise (see figure 1). It is defined by:
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𝜃=
𝑠
𝑟
where θ is the angular position (θ is read as theta)
s is the length of arc along a circle
r is the radius of the circle
The SI unit for angular position is radian. But take note that one revolution in
a circle equals 2π radians or 360°.
The angular displacement is the change in the angular position of the
rotating object. In symbols:
∆𝜃 = 𝜃2 − 𝜃1
where Δθ is angular displacement (Δ is read as delta meaning change)
θ2 is final angular position
θ1 is initial angular position
If the initial angular position is the zero angular position, then angular
displacement is equal to angular position. Angular displacement is also measured by
radians. It is positive for counterclockwise rotation and negative for clockwise
rotation.
The angular velocity is the rate of change in angular position.
Mathematically, it is described as:
𝜔=
∆𝜃 𝜃2 − 𝜃1
=
∆𝑡
𝑡2 − 𝑡1
where ω is angular velocity (ω is read as omega)
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Δθ is change in angular position
Δt is change in time
The SI unit for angular velocity is radians/second (rad/s). But then we also
encounter other unit – rpm, meaning revolutions per minute.
The direction of angular velocity is
defined by right-hand rule: Curl your right
hand about the rotating object. Your fingers
are
pointing in the direction of rotation, and your
extended thumb points in the direction of
angular velocity (see figure 2). Similarly, it is
positive for counterclockwise rotation and
negative for clockwise rotation.
The angular acceleration is the change in angular velocity per unit time. Its
direction is the same with angular velocity if and only if the rotation increases in
speed. But when the rotation is slowing down, its direction is opposite of the angular
velocity’s direction. It is measured in radians per squared seconds (rad/s2). In
symbols, it is defined as:
𝛼=
∆𝜔 𝜔2 − 𝜔1
=
∆𝑡
𝑡2 − 𝑡1
where α is the angular acceleration (α is read as alpha)
Δω is change in angular velocity
Δt is change in time
These basic quantities have both magnitude and directions, then they are
vectors. However, a vector in pure rotation defines only the axis of rotation and not a
direction in which the object moves. Hence, we can describe these rotational
quantities as either positive or negative.
Learning Competency
Describe rotational quantities using vectors. (STEM_GP12REDIIa-4)
ACTIVITY 1: Quantity Search
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Directions: Find and encircle the ten quantities that are found both in translational
and rotational motions. These quantities are hidden in any directions in
the grid.
ACTIVITY 2: Rotational Motion Puzzle
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NOTE: Practice personal hygiene protocols at all times
Directions: Read the clues to complete the crossword. All words are related to
rotational motion.
ACROSS
DOWN
2. clockwise rotation
1. used to denote angular acceleration
3. revolutions per minute
5. both magnitude and direction
4. the directions of α and ω when rotation is
6. point in the direction of rotation
speeding up
8. SI unit for angular displacement
7. rule used to identify the direction of ω
9. symbol of angular velocity
10. the directions of α and ω when rotation
11. Greek letter indicating change in a
is slowing down
quantity
12. motion of wheels, planets, gears, and
13.
motors
displacement
14. used to symbolize angular position
15. points in the direction of angular
rotation
for
positive
angular
velocity
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ACTIVITY 3: Analyzing Rotational Motion
Directions: Determine the magnitude and direction of the rotational quantities asked
in the following problems. Show your solution.
1. As viewed from the north pole, the earth rotates about its axis
counterclockwise once in approximately 24 hours. What is the angular
displacement of the earth for 1 hour in radians, degrees, and revolutions?
2. What is the angular velocity of (a) the second hand, (b) the minute hand and
(c) the hour hand of a smoothly running analog watch? Answer in radians per
second and in rpm.
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3. What is the angular acceleration of the wheel of the bicycle travelling forward
when it reaches 60 rpm in 2 s? Answer in radians/seconds2.
Reflection
1. I learned that _________________________________________________
__________________________________________________________________
__________________________________________________________________
______________________________________________________
2. I enjoyed most on ______________________________________________
__________________________________________________________________
__________________________________________________________________
______________________________________________________
3. I want to learn more on __________________________________________
__________________________________________________________________
__________________________________________________________________
______________________________________________________
27
NOTE: Practice personal hygiene protocols at all times
References
Halliday, David, Resnick, Robert, and Walker, Jearl. Fundamentals of Physics. 6th
ed. New York: John Wiley and Sons Inc, 2001.
Hewitt, Paul G. Conceptual Physics. 10th ed. United States of America: Pearson
Addison-Wesley, 2006.
Moore, Thomas A. Six Ideas that Shaped Physics, Unit C: Conservative Laws
Constrain Interactions. 2nd ed. New York: Mc Graw Hill, 2003.
“Rotational
Quantities
and
Torque”.
Accessed
June
17,
2020.
http://pono.ucsd.edu/~adam/teaching/phys1a2015/worksheets/worksheet51.pdf.
Santos, Gil Nonato C. General Physics 1. 1st ed. Quezon City, Philippines: Rex Book
Store, 2019.
Serway, Raymond A. and Jewette, John W. Jr. Physics for Scientists and Engineers
with Modern Physics. 6th ed. Singapore: Thomson Learning Asia, 2004.
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ANSWER KEY
ACTIVITY 1
11. Position
16. Force
12. Displacement
17. Momentum
13. Velocity
18. Kinetic Energy
14. Acceleration
19. Work
15. Inertia
20. Power
ACTIVITY 2
1. alpha
9. omega
2. negative
10. opposite
3. rpm
11. delta
4. same
12. rotation
5. vector
13. counterclockwise
6. fingers
14. theta
7. righthand
15. thumb
8. radian
ACTIVITY 3
𝜋
1
1. ∆𝜃 = + 12 𝑟𝑎𝑑𝑖𝑎𝑛 = +0.262 𝑟𝑎𝑑𝑖𝑎𝑛 = +15° = + 24 𝑟𝑒𝑣𝑜𝑙𝑢𝑡𝑖𝑜𝑛
2. a. 𝜔 = −
𝜔=−
𝜋𝑟𝑎𝑑
30 𝑠
= −1.05 𝑥 10−1
𝜋𝑟𝑎𝑑 1 𝑟𝑒𝑣𝑜𝑙𝑢𝑡𝑖𝑜𝑛
30 𝑠
(
2𝜋𝑟𝑎𝑑
𝜋𝑟𝑎𝑑
1 ℎ𝑟
𝑠
60 𝑠
) (1 𝑚𝑖𝑛) = −
b. 𝜔 = − 1800 𝑠 = −1.75 𝑥 10−3
𝜋𝑟𝑎𝑑
𝑟𝑎𝑑
𝑟𝑎𝑑
𝑠
60 𝑟𝑒𝑣𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠
60 𝑚𝑖𝑛
= −1 𝑟𝑝𝑚
= −1.66 𝑥 10−2 𝑟𝑝𝑚
c. 𝜔 = − ( 6 ℎ𝑟 ) 3600 𝑠 = −1.45 𝑥 10−4
𝑟𝑎𝑑
𝑠
= −1.38 𝑥 10−3 𝑟𝑝𝑚
3. 𝛼 = +3.14 𝑟𝑎𝑑/𝑠
When a bicycle moves forward, its wheel is rotating counterclockwise. So, its
angular velocity is positive. Since the bicycle starts from zero to 60 rpm
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NOTE: Practice personal hygiene protocols at all times
(increase in rotation), then angular acceleration direction is the same with
the angular velocity’s direction.
Prepared by:
Techie Gamma-Vera Cruz
Amulung National High School
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GENERAL PHYSICS 1
Name: ________________________Grade Level: _______________
Date: __________________________Score: ____________________
LEARNING ACTIVITY SHEET
DETERMINE WHETHER A SYTEM IS IN STATIC EQUILIBRIUM OR NOT
Background Information for the Learners (BIL)
When you apply with a pencil, you will find it is impossible to balance the
pencil on its point. On the other hand, it is comparatively easy to make the pencil
stand upright on its flat end.
An object at rest may be in one of the three states of equilibrium. You can
distinguish between the different kinds of equilibrium by considering the illustrations
of an ice cream cone placed on a level table (see Figure 1).
A cone standing on its base will return to its original position after a little
disturbance; hence, it is in stable equilibrium on its base (Figure 1.A). On the other,
a cone placed on its tip said to be unstable equilibrium and can be easily toppled
down when slightly disturbed (Figure 1.B). A cone lying on its side stays in its position
without tending either to move further or to return to where it was before. A cone on
its side is said to be in neutral equilibrium where it can be rolled from one side to
another (Figure 1.C). The illustrations show that the equilibrium condition is affected
by the position of the object’s center of gravity. An object is in stable equilibrium if its
center of gravity is at the lowest possible position.
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Figure 1. Three States of Equilibrium: stable equilibrium (A), unstable equilibrium
(B) and neutral equilibrium (C)
Conditions for Equilibrium
First Condition
The first condition of equilibrium is that the net force in all directions must be
zero.
For an object to be in equilibrium, it must be experiencing no acceleration.
This means that both the net force and the net torque on the object must be zero.
Here we will discuss the first condition, that of zero net force.
In the form of an equation, this first condition is:
Fnet = 0 or ∑F = ma = 0
In order to achieve this conditon, the forces acting along each axis of motion
must sum to zero. For example, the net external forces along the typical x– and yaxes are zero. This is written as:
net Fx=0 and net Fy=0
The condition Fnet=0 must be true for both static equilibrium, where the
object’s velocity is zero, and dynamic equilibrium, where the object is moving at a
constant velocity.
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Below, the motionless person is in static equilibrium. The forces acting on him
add up to zero. Both forces are vertical in this case.
Figure 2. Person in Static Equilibrium: This motionless person in static
equilibrium.
Below, the car is in dynamic equilibrium because it is moving at constant
velocity. There are horizontal and vertical forces, but the net external force in any
direction is zero. The applied force between the tires and the road is balanced by air
friction, and the weight of the car is supported by the normal forces, here shown to
be equal for all four tires.
Figure 3. A Car in Dynamic Equilibrium: This car is in dynamic equilibrium
because it is moving at constant velocity.
Consider the following cases of bodies in equilibrium.
Cases 1: A box on a table
The forces acting on the box are its weight (W), acting downward, and the
normal force (FN) that the table exerts upward on the box. The box is resting on the
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table with zero acceleration. Thus, the sum of all forces acting on the box must be
zero.
∑F = FN + (-W) = 0
FN – W = 0
FN = W
Case 2: A chandelier hanging from a vertical
rope
The forces acting on the chandelier re the
weight (W), acting downward, and the tension (T) in the
rope, acting upward.
∑F = T + (-W) = 0
T–W=0
T=W
Case 3: A swing is pushed until the rope makes an angle θ with the vertical
The forces acting on the swing are the combined weight of the swing and the
boy (W), acting downward, the force (F) exerted on the swing, acting to the left, and
the tension on the rope (T) that can be resolved into its vertical (Ty) and horizontal
(Tx) components.
∑Fx = F + (-Tx) = 0
F – Tx = 0
F – T sin θ = 0
F = T sin θ
∑Fy = Ty + (-W) = 0
Ty – W = 0
T cos θ = W
T cos θ = W
Case 4: Resting in a hammock
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The forces acting on the hammock are the weight (W) on the hammock and
the Tensions ( T1 and T2 ) on the ropes that can be resolved into their vertical and
horizontal components.
∑Fx = T2x + (-T1x) = 0
T2x - T1x = 0
T2 cos β – T1 cos α = 0
T2 cos β = T1 cos α
∑Fy = T1y + T2y + (-W) = 0
T1y + T2y – W = 0
T1 sin α + T2 sin β – W = 0
T1 sin α + T2 sin β = W
Case 5: A boy on a slide
The forces acting on the boy are the
frictional force (Ff ), acting upward and
parallel to the slide, the normal force (FN), acting toward and perpendicular to the
slide, and the weight (W) of the boy that can be resolved into its components which
are parallel (W II) and (W
┴
) to the slide.
∑Fx = Ff + (-W II) = 0
µ FN – W sin θ = 0
µ FN = W sin θ
∑Fy = FN + (-W ┴ ) = 0
FN – W cos θ = 0
FN = W cos θ
µ W cos θ = W sin θ
µ=
W sin θ
W cos θ
µ = tan θ
Second Condition
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The second condition of static equilibrium says that the net torque acting on
the object must be zero.
A child’s seesaw, shown in, is an example of static equilibrium. An object in static
equilibrium is one that has no acceleration in any direction. While there might be
motion, such motion is constant. If a given object is in static equilibrium, both the net
force and the net
torque
on the object must
be
zero. Let’s break
this
down:
Figure 4. Two children on a seesaw: The system is in static equilibrium,
showing no acceleration in any direction.
The Concept of Torque
Consider the familiar seesaw you played during your childhood. Suppose a
50-kg child (W1) is placed on the right side of a seesaw and a 30-kg child (W2) is
placed on the left side as shown in Figure 5.
Figure 5. A Child’s game of seesaw demonstrates torque.
The weights of the two children exert downward forces, while the support in
the middle of the seesaw exerts an upward force which is equal to the weight of the
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NOTE: Practice personal hygiene protocols at all times
two children. Even, though the body is in transitional equilibrium, the body is still
capable of rotating. The 50-kg child on the right end moves downward, while the 30kg child on the left end moves upward; this means that the seesaw rotates in a
clockwise direction.
Torque is the quantity that measures how effectively a force (F) causes
acceleration. A torque is produced when a force is applied with leverage. It is defined
as the product of the force and the lever arm. The lever arm is the perpendicular
distance (l) from the axis of rotation to the line along which the force acts. The
magnitude of the torque (τ) can be calculated by:
torque = force x lever arm
τ = Fl
The Second Condition
A torque (a vector quantity) that tends to produce a counter clockwise rotation
is considered positive and a torque that tends to produce clockwise rotation is
negative (see Figure 6). Thus, the condition for an object to be in rotational
equilibrium is that the sum of the torques acting on the object about any point must
be zero. This means that the sum of all the clockwise
torques (τc) must be equal to the sum of all the counter
clockwise torques (τu).
∑τ = 0
∑τ = ∑τc + (-∑τu) = 0
∑τc - ∑τu = 0
∑τc = ∑τu
Figure 6. Torques make objects rotate.
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Let us consider the following cases of bodies in equilibrium:
Case 1: Bamboo pole carried at each end
In a singkil dance, two men are carrying a princess on a bamboo pole that is
5.0 m long and weighs 200 N. If the princess weighs 450 N and sits 1.5 m from one
end, how much weight must each man support? We assume that the diameter of the
bamboo pole is uniform and the weight of the pole is located at the center.
Using the first condition for equilibrium,
∑Fy = 0
∑Fy = F1 + F2 – WB – WP = 0
where, WB is the weight of the bamboo pole WP
is the weight of the princess
F1 + F2 – WB – WP = 0
F1 + F2 = WB + WP
= 200 N + 450 N
F1 + F2 = 650 N
We must specify the axis about which the torques will be computed. Let us
consider that the axis passes through point A, where man 1 is holding the pole with
force. Using the second condition for equilibrium, we can solve for F2.
∑τc = ∑τu
WBlB + WPlP = F2l2
(200 N)(2.5 m) + (450 N)(3.5 m) = F2 (1.5 m)
2075 𝑁. 𝑚
𝐹2 (5.0 𝑚)
=
5.0 𝑚
5.0 𝑚
415 N = F2
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Solving for F1,
F1 + F2 = 650 N
F1 + 450 N = 650 N
F1 = 650 N – 450 N
F1 = 235 N
Case 2: A man on a ladder
A ladder 7.5 m long is leaning against a
smooth (frictionless) wall at a point 7.0 m above
the ground with its base 2.0 m from the wall.
The ladder weighs 200 N and an 800-N painter
stands two-thirds of the way up the ladder. (a)
What is the normal force? (b) What frictional
force must act on the bottom of the ladder to
prevent it from slipping for the painter to be
safe?
Using the first condition for equilibrium, we have,
a.
∑Fy = 0
∑Fx = 0
∑Fy = FN + (-WP) + (-WL) = 0
FN – WP – WL = 0
FN = 800 N + 200 N
FN = 1000 N
b.
∑Fx = F + (-Ff) = 0
F – Ff = 0
F = Ff
Suppose the axis of rotation is the base of the ladder. Using the second
condition for equilibrium, we have
∑τ = 0
∑τc = ∑τu
Fflf = WLlL + Wlll
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F(7.0 m) = (200 N) (1.0 m) + (800 N)(1.33 m)
F=
200𝑁.𝑚+1064 𝑁.𝑚
7.0 𝑚
F = 181 N.
Learning Competency:
Determine whether a system is in static equilibrium or not. (STEM_GP12RED-IIa-5)
ACTIVITY #1: COMPLETE THE KEY CONCEPTS
Directions: Fill in the blanks with the correct word/s that complete/s the key
concept in each item.
1. An object is in stable equilibrium if it is at the ___________ possible position.
2. An object with a _________base is more stable than one with __________
base.
3. The stability of an object depends on the location of the _________________,
__________________, and amount of mass.
4. _____________ is the product of force and the lever arm.
5. Net torque always produces ______________________.
6. The lever arm is the _______________ distance from the reference point to
the direction or line of action of the force.
7. There are two conditions for a body to be in rotational equilibrium:
a. ______________ equilibrium is when the vector sum of all forces acting
on it must be zero.
b. ______________ equilibrium is when the sum of all torques about any
point must be zero.
ACTIVITY #2: CRITICAL THINKING
Directions: Analyze and structure a comprehensive reasoning to answer each
situation below. (5 points each)
1. Why does a man with a large belly or a woman in her last trimester of
pregnancy tend to lean backward when walking?
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
____________________________________________________
2. Which man carries the heavier load? Why?
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_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_______________
ACTIVITY #3: PROBLEM SOLVING
Directions: Solve the following problems and show your complete solution. (5 points
each)
1. If a person can apply a maximum force of 200N, what is the maximum length
of a wrench needed to apply 90 N.m torque to the bolts on a motorcycle
engine?
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2. A 500-N diver stands at the end of a 4.0-m
diving board. The board is attracted by two
supports 1.5 m apart as shown below. Find
the tension in each of the two supports if the
diving board weighs 150 N.
ACTIVITY #4: APPLICATION
Directions: Apply what you have learned. Aside from the given examples, construct
at least 2 systems or situations under static equilibrium. Make your samples
comprehensive and accurate.
Reflection
1. I learned that _________________________________________________
____________________________________________________________
_______________________________________________________
2. I enjoyed most on _____________________________________________
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NOTE: Practice personal hygiene protocols at all times
____________________________________________________________
__________________________________________________
3. I want to learn more on _________________________________________
____________________________________________________________
__________________________________________________
References:
Padua, Alicia L. et. al, States of Equilibrium, Practical and Explorational
Physics:
Modular Approach, 2003, pp. 98-107.
“Conditions of Equilibrium”.
https://courses.lumenlearning.com/boundless-physics/chapter/conditions-forequilibrium/#:~:text=An%20object%20in%20static%20equilibrium,no%20acc
eleration%20in%20any%20direction.
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Answer Key:
Activity #1: Complete the key Concepts
1.
2.
3.
4.
5.
6.
7.
lowest
wider , narrower
center of gravity , area of the base
Torque
angular acceleration
perpendicular
a. Translational , b. Rotational
Activity #2: Critical Thinking
1. Carrying any load in front of your stomach shifts your centers of gravity
forward. By leaning backward, you can keep your center of gravity above your
supporting feet to maintain your stability.
2. Peter carries the heavier load. He has the shorter leverage.
Activity #3: Problem Solving
1. τ = (F) (l)
τ
l=𝐹=
90 𝑁 .𝑚
200 𝑁
= 0.45 m
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2. Consider the point where F2 is applied as the axis of rotation.
∑τ = 0
∑τc = ∑τu
W1l1 + W2l2 = F1l1
(150 N)(5.0 m) + (500 N)(2.5 m) = F1(1.5 m)
75 N . m + 1250 N . m = F1(1.5 m)
1325 N . m = F1(1.5 m)
F1 = 883.33 N
We can find F2 by using the first condition for equilibrium.
∑Fy = 0
∑Fy = F2 + (-F1) + (-W1) + (-W2) = 0
F2 - F1 - W1 - W2 = 0
F2 = F1 + W1 + W2
= 883.33 N + 150 N + 500 N
F2 = 1533.33 N
Activity #4: Application
Answers may vary
Prepared by:
JOLLY MAR D. CASTANEDA
Baggao National Agricultural School Sta. Margarita Annex
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GENERAL PHYSICS 1
Name: _______________________Grade Level: _________________
Date: ________________________Score: ______________________
LEARNING ACTIVITY SHEET
APPLY THE ROTATIONAL KINEMATIC RELATIONS FOR SYSTEMS
WITH CONSTANT ANGULAR ACCELERATIONS
Background Information for the Learners (BIL)
A body in motion may be travelling either in a straight line or along a curve.
By definition, a body in pure translational motion moves such that a line drawn
between any two of its internal points remains parallel to itself after displacement. An
example is when you carry a glass full of water from one place to another. It is held
in a way such that the contents do not spill.
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NOTE: Practice personal hygiene protocols at all times
On the other hand, a body may also have rotational motion such that a line
between any two points does not remain parallel to itself. And this is explained by the
kinematics and dynamics of rotational motion.
Angles in Radians
In trigonometry, you may have encountered angle measures not only in
degrees but also in radians. In science, angles are often measured in radians (rad).
In the figure below, when the arc length is equal to the radius r, the angle θ swept by
r is equal to 1 rad. In general, any angle θ measured in radians is defined as =
𝑠
𝑟
.
Figure 1: Measuring angles in radians
Since the radian is the ratio of an arc length to the length of the radius, the
units cancel and the radian becomes a pure number. The value 3600 equals 2π
rad, or one complete revolution. This means that one revolution is equivalent to
6.28 rad.
Figure 2 shows a circle marked with both radians and degrees. Any angle in
degrees can be converted into angle in radians by multiplying it by
2𝜋
360°
, or its lowest
𝜋
term 180° . In symbols, this is written as :
𝜃 (𝑟𝑎𝑑) =
𝜋
∙ 𝜃 (𝑑𝑒𝑔𝑟𝑒𝑒𝑠)
180°
Let us have this as an example. Convert 1100 into radians.
𝜋
∙ 110° = 𝟏. 𝟗𝟐 𝒓𝒂𝒅
180°
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Figure 2: Radians and Degrees
Angular Displacement
In translational kinematics, the position of the body is defined as the
displacement x from a certain reference point. In rotational kinematics, the position
of a point on a rotating body is defined by the angular displacement θ from some
reference line that connects this point to the axis of rotation.
Study Figure 3. The body has rotated through the angular displacement θ if
the point which was originally at P1 is now at the point P2. This angular displacement
is a vector that is perpendicular to the plane of the motion. The magnitude of this
angular displacement is the angle θ itself. The direction depends whether it is a
positive or a negative quantity. If it is positive, the rotation of the body is counter
clockwise and the angular displacement vector points upward. If it is negative, the
rotation is clockwise and the vector points downward.
Just an angle in radians is defined by the ratio of the arc length to the radius,
the angular displacement is equal to the
change in the arc length, ∆𝑠, divided by
the distance from the axis of rotation, r. It
is given as,
∆𝜃 =
∆𝑠
𝑟
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which means that,
∆𝑠 = 𝑟∆𝜃.
Figure 3: Angular displacement in a rotating body
Sample Problem 1:
A boy rides on a merry-go-round at a distance of 1.25 m from the center. If the
boy moves through an arc length of 2.25 m, through what angular displacement does
he move?
Given:
r = 1.25 m
∆𝑠 = 2.25 m
Find :
∆𝜃
Solution:
∆𝜃 =
∆𝑠
𝑟
=
2.25 𝑚
1.25 𝑚
= 𝟏. 𝟖 𝒓𝒂𝒅
Angular Velocity
Angular velocity is similarly defined as the linear velocity. It is denoted by the
lowercase of the Greek letter omega (𝜔) and is defined as the ration of the angular
displacement ∆𝜃 to the time interval ∆𝑡, the time it takes an object to undergo that
displacement. It describes how quickly the rotation takes place.
In symbols, the average angular velocity is given as:
𝜔𝑎𝑣𝑒 =
∆𝜃 ∆𝑠 1
=
∙
∆𝑡 ∆𝑡 𝑟
In the limit that the time interval approaches zero,
∆𝑠
∆𝑡
becomes the
instantaneous velocity, . Angular velocity is expressed in radians per second (rad/s).
In some instances, angular velocities are expressed in revolutions per unit time such
as revolutions per second (rps) and revolutions per minute (rpm).
1 𝑟𝑒𝑣 = 2𝜋 𝑟𝑎𝑑
Linear velocity of a point on the rotating body and angular velocity of the body
are linked by the equation, 𝑠 = 𝑟𝜃 divided by t. That is,
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NOTE: Practice personal hygiene protocols at all times
𝑠 𝑟𝜃
=
𝑡
𝑡
𝑠
but we know that, 𝑡 = 𝑣 and
𝜃
𝑡
= 𝜔 . And so, 𝒗 = 𝒓𝝎.
This equation implies that, for a body rotating at an angular velocity 𝜔, the
farther the distance 𝑟 that the body is form the axis of rotation, the greater is its linear
or tangential velocity.
Sample Problem 2:
A child in a barber shop spins on a stool. If the child turns counter clockwise
through 8.0 𝜋 rad during an 8.0 s interval, what is the average angular velocity of the
child’s rotation?
Given:
∆𝜃 = 8.0𝜋 𝑟𝑎𝑑
∆𝑡 = 8.0 𝑠
Find:
𝜔𝑎𝑣𝑒
Solution:
𝜔𝑎𝑣𝑒 =
∆𝜃
∆𝑡
=
8.0𝜋 𝑟𝑎𝑑
8.0 𝑠
= 𝟑. 𝟏𝟒 𝒓𝒂𝒅/𝒔
Angular Acceleration
Angular acceleration occurs when angular velocity changes with time. We will
use the symbol alpha, α , to denote angular acceleration. The average angular
acceleration is given by this relationship,
𝛼𝑎𝑣𝑒 =
𝜔𝑓 − 𝜔𝑖
𝑡𝑓 − 𝑡𝑖
𝛼𝑎𝑣𝑒 =
∆𝜔
∆𝑡
where ∆𝜔 is the change in angular velocity and ∆𝑡 is the change in time.
This quantity is expressed in the unit radians per second squared (rad/s2).
There is a connection between instantaneous tangential acceleration and angular
acceleration. The tangential acceleration associated with motion of a point moving in
a circular path of radius r is related to the instantaneous angular acceleration through:
𝒂𝒕 = 𝜶𝒓.
Sample Problem 3:
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A figure skater begins spinning counter clockwise at an angular speed of
5.0𝜋 𝑟𝑎𝑑/𝑠. She slowly pulls her arm inward and finally spins at 9.0𝜋 𝑟𝑎𝑑/𝑠 for 3.0 s.
What is her average angular acceleration during this time interval?
Given:
𝜔𝑓 = 9.0𝜋 𝑟𝑎𝑑/𝑠
𝜔𝑖 = 5.0𝜋
𝑟𝑎𝑑
𝑠
∆𝑡 = 3.0 𝑠
Find:
𝛼𝑎𝑣𝑒
Solution:
𝛼𝑎𝑣𝑒 =
𝜔𝑓 − 𝜔𝑖
𝑡𝑓 − 𝑡𝑖
𝑟𝑎𝑑
𝑟𝑎𝑑
9.0𝜋 𝑠 − 5.0𝜋 𝑠
=
3.0 𝑠
𝜶𝒂𝒗𝒆 = 𝟒. 𝟐 𝒓𝒂𝒅/𝒔𝟐
Rotational Kinematic Equations
∆𝜽
∆𝒕
𝝎𝒇 − 𝝎𝒊
=
𝒕
𝝎𝒂𝒗𝒆 =
𝜶𝒂𝒗𝒆
𝝎𝒇 = 𝝎𝒊 + 𝜶𝒕
𝜶𝒕𝟐
𝜽 = 𝝎𝒊 +
𝟐
𝝎𝒇𝟐 − 𝝎𝒊𝟐
𝜽=
𝟐𝜶
Sample Problem 4:
A fish swimming behind a luxury cruise liner gets caught in a whirlpool created
by the ship’s propeller. If the fish has an angular velocity of 1.5 rad/s and the water
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in the whirlpool accelerates at 3.5 rad/s2, what will be the instantaneous angular
velocity of the fish at the end of 4.0 seconds?
Given:
𝑟𝑎𝑑
𝜔𝑖 = 1.5
𝛼 = 3.5
𝑠
𝑟𝑎𝑑
𝑠2
𝑡 = 4.0 𝑠
Find:
𝜔𝑓
Solution:
𝜔𝑓 = 𝜔𝑖 + 𝛼𝑡
= 1.5
𝑟𝑎𝑑
𝑠
𝝎𝒇 = 𝟏𝟓. 𝟓
3.5𝑟𝑎𝑑
+(
𝑠2
) (4.0𝑠)
𝒓𝒂𝒅
𝒔
Learning Competency:
Apply the rotational kinematic relations for systems with constant angular
accelerations. (STEM_GP12RED-IIa-6)
Activity #1: TRUE OR FALSE
Directions: Write TRUE if the statement is correct, and FALSE if otherwise.
____________ 1. The angle turned through by a body about a given axis is called
angular displacement.
____________ 2. Angular acceleration is the change in the angular displacement of
rotating body about the axis of rotation with time.
____________ 3. Angular velocity is the change in the angular speed of a tangential
motion.
____________ 4. The farther the distance 𝑟 that the body is form the axis of rotation,
the greater is its linear or tangential velocity.
____________ 5. The value 3600 equals 2π rad, or one complete revolution.
Activity #2: PROBLEM SOLVING
Directions: Solve the following problems. Show your complete solutions.
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1. Convert the following into indicated angle measure unit:
a. 360 to rad
b. 1250 to rad
c. 2 𝜋/5 to degrees
d. 𝜋/7 to degrees
2. A wheel of radius 14.0 cm starts from rest and turns through 2.0 revolutions
in 3.0s.
a. What is its average velocity?
b. What is the tangential velocity of a point on the rim of the wheel?
3. A rifle is a long gun barrel has been grooved or “rifled” on the inside with spiral
channels. Bullets fired from a rifled barrel spin. This gives them greater
stability in flight and thus greater accuracy when fired. Since 1964, the
standard infantry weapon in the US Army has been the 0.22 caliber M16 rifle.
Due to rifting, a bullet fired from an M16 rotates two and a half times on its
journey from the breech to the muzzle. Given a barrel length of 510 mm and
a muzzle velocity of 950 m/s. Determine the following.
a. the average translational acceleration
b. the average angular acceleration (in radians per second squared)
c. the final angular velocity (in rotations per second)
4. A fish swimming behind a luxury cruise liner gets caught in a whirlpool created
by the ship’s propeller. If the fish has an angular velocity of 2.5 rad/s and the
water in the whirlpool accelerates at 4.5 rad/s2, what will be the instantaneous
angular velocity of the fish at the end of 5.0 seconds?
Activity #3: CONSTRUCTIVE REASONING
Try riding on a freely rotating child’s merry-go-round. What happens to the
rotational velocity when you move inward or outward from the axis of the merry-goround? How can this be explained?
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
____________________
Reflection
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NOTE: Practice personal hygiene protocols at all times
1.I learned that ____________________________________________________
____________________________________________________________
_______________________________________________________
2.I enjoyed most on _________________________________________________
____________________________________________________________
__________________________________________________
3.I want to learn more on _____________________________________________
____________________________________________________________
__________________________________________________
References
54
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Padua, Alicia L. et. al, Rotational Kinematics , Practical and Explorational
Physics: Modular Approach, 2003, pp. 149-153.
The Physics Hypertxtbook.
https://physics.info/rotational-kinematics/practice.shtml
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Answer Keys:
Activity #1: TRUE or FALSE
1.
2.
3.
4.
5.
TRUE
FALSE
FALSE
TRUE
TRUE
Activity #2: Problem Solving
1. a.
𝜋
5
𝑟𝑎𝑑 , b.
25𝜋
𝑟𝑎𝑑 , c. 720, d. 25.710
36
𝑟𝑎𝑑
2. a. 𝜔𝑎𝑣𝑒 = 4.19
𝑠
b. 𝑣 = 0.59 𝑚/𝑠
3. a. a = 8.8 x 105 m/s2
b. t = 0.001074 s , 𝛼 = 2.7 𝑥 107 𝑟𝑎𝑑/𝑠 2
c. 𝜔 = 29,000
4. 𝜔 = 25
𝑟𝑎𝑑
𝑠
0𝑟 𝜔 = 4,700
𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛𝑠
𝑠𝑒𝑐𝑜𝑛𝑑
𝑟𝑎𝑑
𝑠
Activity #3: Constructive Reasoning
Answers may vary.
Prepared by:
JOLLY MAR D. CASTANEDA
Baggao National Agricultural School Sta. Margarita Annex
56
NOTE: Practice personal hygiene protocols at all times
GENERAL PHYSICS 1
Name: ______________________ Grade Level: _________________
Date: ________________________Score: ______________________
LEARNING ACTIVITY SHEET
DETERMINE ANGULAR MOMENTUM AT DIFFERENT SYSTEMS
Background Information for the Learners (BIL)
Rotational Inertia
An object rotating about an axis tends to continue rotating about that axis
unless an unbalanced external torque tries to stop it. This is because objects tend to
resist any change in their state of motion. This resistance is physically embodied in
the inertial mass, or simply mass. The resistance of an object to changes in its
rotational motion is called
rotational
inertia
which is also termed as
moment of inertia.
Figure 1. Dumbbell B is easier to rotate in spite of its large masses because these are near
its axis of rotation; hence, the dumbbell’s moment of inertia is smaller. The opposite can be
said of dumbbell A.
If force is needed to change the linear state of motion of an object, torque is
required to change the rotational state of motion of an object. And so, if there is no
net torque, a rotating object continues to rotate at a constant velocity.
Rotational inertia depends on the distribution of the mass. A small mass which
is at a greater distance from the axis of rotation has a greater moment of inertia than
a large mass which is near the axis of rotation.
The moment of inertia I, gives a measurement of the body to a change in its
rotational motion. The larger the moment of inertia of a body, the more difficult it is to
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NOTE: Practice personal hygiene protocols at all times
put that body into rotational motion or, the larger the moment of inertia of a body, the
more difficult it is to stop its rotational motion.
For the very special case of the moment of inertia of a single mass m, rotating
about an axis, a distance r from m, we have
I = mr2
It is important to remember that when moment of inertia is asked for, it is a
must to specify about what axis the rotation will take place. Because r is different for
each axis and, since I differs as r2, I is also different for each axis.
The unit for the moment of inertia is kg ∙ m2 and has no special name.
Calculus is usually used to sole for the moment of inertia. However, for
simplicity, you can use Table 1, which shows how values of the moment of inertia for
some reason uniform symmetrical bodies about different axes can be determined.
Sample Problem1:
Find the moment of inertia of a solid cylinder of mass 3.0 kg and radius 0.50
m, which is free to rotate about an axis through its center.
Given:
m = 3.0 kg
r = 0.50 m
Find:
I=?
Solution:
𝐼 = 2 𝑚𝑟 2
1
=
=
𝟏
𝟐
𝟏
𝟐
(𝟑. 𝟎 𝒌𝒈)(𝟎. 𝟓𝟎 𝒎) 2
(𝟑. 𝟎 𝒌𝒈)(𝟎. 𝟐𝟓 𝒎𝟐 )
𝑰 = 𝟎. 𝟑𝟖 𝒌𝒈 ∙ 𝒎𝟐
You have seen how Newton’s first law of motion is similar to rotational
motion. Newton’s three laws many be stated in terms of rotational motion.
The first law for rotational motion:
A body in motion at a constant angular velocity will continue in motion
at that same angular velocity, unless acted upon by some unbalanced external
torque.
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A very good example to illustrate this is the earth’s rotation. The earth
continues to rotate at an angular velocity of 7.27 x 10-5 rad/s since there is no external
torque acting on it.
The second law for rotational motion:
When an unbalanced external torque acts on a body with moment of
inertia 𝐼, it gives that body an angular acceleration α, which is directly
proportional to the torque 𝜏 and inversely proportional to the moment of inertia.
In symbols, this is given as
𝝉 = 𝑰𝜶
The third law for rotational motion:
If body A and body B have the same axis of rotation, and if body A
exerts a torque on body B, then body B exerts an equal but opposite torque
on body A.
Table 1. Moments of Inertia of selected bodies with Mass m
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Angular Momentum
If the rotational equivalent of force is torque, which is the moment of the force,
the rotational equivalent of linear momentum (p) is angular momentum (L), which is
the moment of momentum. Like linear momentum, angular momentum is also a
vector quantity; it has magnitude and direction. It is defined as the product of the
moment of inertia (I) of a rotating body and its angular velocity (𝝎). In equation form,
this is given as
L = I𝝎
The unit of angular momentum is kg ∙m2/s.
If an object is small compared with the radial distance to its axis of rotation,
the angular momentum is equal to the magnitude of its linear momentum mv,
multiplied by the radial distance r. In equation form,
L = mvr
The velocity is always perpendicular to radial distance. Otherwise L = r x p = mr x v.
The angular momentum of an object, such
as a stone swinging from a long string, or a planet
circling the sun, can be determined using the
equation L = mvr. This shows that the angular
momentum is directly proportional to the linear
momentum and the radius.
Figure
2.
Demonstrating
angular
momentum
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Sample Problem 2:
What is the angular momentum of a 250 g stone being whirled by a slingshot
at a tangential velocity of 6 m/s, if the length of the slingshot is 30 cm?
Given:
m = 250 g = 0.25 kg
v = 6.0 m/s
r = 30 cm = 0.30 m
Find:
L=?
Solution:
𝐿 = 𝑚𝑣𝑟 = (0.25 𝑘𝑔) (6
𝑚
) (0.30 𝑚 = 𝟎. 𝟒𝟓 𝒌𝒈 . 𝒎𝟐 /𝒔
𝑠
The table below shows the concept of momentum for linear and rotational
situtions.
Table 2. Kinematic Equations for Linear and angular Momentum.
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If no net force acts on a system, we know that the linear momentum of this
system is conserved. Angular momentum is also conserved for systems in rotation.
The Law of Conservation of Angular Momentum states that in the absence of an
unbalanced external torque, the angular momentum of a system remains constant.
Learning Competency:
Determine angular momentum at different systems. STEM_GP12RED-IIa-9
Activity 1: CRITICAL THINKING
Directions: Answer the following questions comprehensively. (3 points each)
1. Why do you bend your legs when you run?
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_______________
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2. If you walk along the top of a fence, why would extending your arms out help
you to balance?
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_______________
3. When is angular momentum conserved?
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_______________
4. By how much will the rate of spin of skater increase, if she pulls her arms in
to reduce her moment of inertia to half?
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_______________
5. Will there be a change in a gymnast’s angular momentum if he changes his
body configuration during a somersault?
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_______________
Activity 2: PROBLEM SOLVING (5 points each)
Directions: Solve the following problems. Show your complete and accurate
solutions.
1. Lara, a 50.0 kg gymnast, swings her 1.5 m long body around a bar by her
outstretched arms. What is her moment of inertia?
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2. Lito is spinning a basketball with a radius of 12 cm on the tip of his finger.
Determine the mass of the ball if its moment of inertia is 5.568 x 10 -3 kg.m2.
3. What is the angular momentum of a 300 g stone being whirled by a slingshot
at a tangential velocity of 9 m/s, if the length of the slingshot is 40 cm?
REFLECTION
1. I learned that ________________________________________________
____________________________________________________________
_______________________________________________________
2. I enjoyed most on _____________________________________________
____________________________________________________________
__________________________________________________
3. I want to learn more on _________________________________________
____________________________________________________________
__________________________________________________
64
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References
Padua, Alicia L. et. al, 2003, Rotational Kinematics , Practical and
Explorational
Physics: Modular Approach, pp. 149-153.
The Physics Hypertxtbook taken from https://physics.info/angularmomentum/practice.shtml
65
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Answer Key
Activity 1: Critical Thinking
(Answers may vary)
Activity 2: Problem Solving
1. 37.5 kg . m2
2. O.58 kg
3. 1.8 kg . m2/s
Prepared by:
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JOLLY MAR D. CASTANEDA
Baggao National Agricultural School Sta. Margarita Annex
GENERAL PHYSICS 1
Name: ____________________________
Date: _____________________________
Grade Level: _________
Score: ______________
LEARNING ACTIVITY SHEET
TORQUE-ANGULAR MOMENTUM RELATIONSHIP
BACKGROUND INFORMATION FOR LEARNERS
TORQUE
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Torque is the turning or twisting effectiveness of a force. Figure 1, helps to
explain the idea of torque. When you push on a door with a force F, as in part a, the
door opens more quickly when the force is larger. Other things being equal, a larger
force generates a larger torque. However, the door does not open as quickly if you
apply the same force at a point closer to the hinge, as in part b, because the force
now produces less torque. Furthermore, if your push is directed nearly at the hinge,
as in part c, you will have a hard time opening the door at all, because the torque is
nearly zero. In summary, the torque depends on the magnitude of the force, on
the point where the force is applied relative to the axis of rotation (the hinge in
Figure 1), and on the direction of the force. For simplicity, we deal with situations
in which the force lies in a plane that is perpendicular to the axis of rotation. In Figure
2, for instance, the axis is perpendicular to the page and the force lies in the plane of
the paper. The drawing shows the line of action and the lever arm of the force, two
concepts that are important in the definition of torque. The line of action is an
extended line drawn collinear with the force. The lever arm is the distance l between
the line of action and the axis of rotation, measured on a line that is perpendicular to
both. The torque is represented by the symbol τ (Greek letter tau), and its magnitude
is defined as the magnitude of the force times the lever arm:
Figure 1. With a force of a given
magnitude, a door is easier to open
by (a) pushing at the outer edge than
by (b) pushing closer to the axis of
rotation (the hinge). (c) Pushing into
the
hinge makes it difficult to open the door.
Figure 2. In this top view, the hinges of a door appear as a black dot (•) and define
the axis of rotation. The line of action and lever arm l are illustrated for a force applied
to the door (a) perpendicularly and (b) at an angle. (c) The lever arm is zero because
the line of action passes through the axis of rotation.
Torque Formula
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Torque = F x L
= FLsinƟ
The S.I unit of torque is Nm.
Direction: right hand rule.
Torque is calculated with respect to (about) a point. Changing the point can change
the torque’s magnitude and direction.
Sample Problem no 1:
In Figure 2 a force (magnitude = 55 N) is applied to a door. However, the lever arms
are different in the three parts of the drawing: (a) l = 0.80 m, (b) l = 0.60 m and (c) l
= 0 m.
Find the torque in each case.
First, we are going to write all the given and identify the unknown.
Given: F= 55 N
l= a= 0.80 m
b= 0.60 m
c= 0 m
τ= ?
Second, perform the needed operation in each of the following problem.
a.
τ= F x l
b.
τ= F x l
c.
τ= F x l
τ= 55 N x 0.80 m
τ= 55 N x 0.60 m
τ= 55 N x 0 m
τ= 44 Nm
τ= 33 Nm
τ= 0 Nm
In parts a and b the torques are positive, since the forces tend to produce a
counterclockwise rotation of the door. In part c, the line of action of passes through
the axis of rotation (the hinge). Hence, the lever arm is zero, and the torque is zero.
Angular Momentum
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Why does Earth keep on spinning? What started it spinning to begin with?
And how does an ice skater manage to spin faster and faster simply by pulling her
arms in? Why does she not have to exert a torque to spin faster? Questions like
these have answers based in angular momentum, the rotational analog to linear
momentum. Linear momentum p of an object is defined as the product of its mass m
and linear velocity v; that is, p = mv. For rotational motion the analogous concept is
called the angular momentum L.
The mathematical form of angular
momentum is analogous to that of linear momentum, with the mass m and the linear
velocity v being replaced with their rotational counterparts, the moment of inertia I
and the angular velocity ω.
L= I ω
SI Unit of Angular Momentum: kg • m2/s
Linear momentum is an important concept in physics because the total linear
momentum of a system is conserved when the sum of the average external forces
acting on the system is zero. Then, the final total linear momentum Pf and the initial
total linear momentum P0 are the same: Pf = P0. In the case of angular momentum,
a similar line of reasoning indicates that when the sum of the average external
torques is zero, the final and initial angular momenta are the same: Lf = L0, which is
the principle of conservation of angular momentum.
When you push a merry-go-round, spin a bike wheel, or open a door, you
exert a torque. If the torque you exert is greater than opposing torques, then
the rotation accelerates, and angular momentum increases. The greater the net
torque, the more rapid the increase in L. The relationship between torque and
angular momentum is:
ΔL
Net τ
=
Δt
This expression is exactly analogous to the relationship between force and
linear momentum, F = Δp / Δt. The equation net τ = ΔL / Δt is very fundamental and
broadly applicable. It is, in fact, the rotational form of Newton’s second law.
Calculating the Torque Putting Angular Momentum
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The image on the right shows a Lazy
Susan food tray being rotated by a person in
quest of sustenance. Suppose the person exerts
a 2.50 N force perpendicular to the lazy Susan’s
0.260-m radius for 0.150 s. What is the final
angular momentum of the lazy Susan if it starts
from rest, assuming friction is negligible?
Given:
F = 2.50 N
l = 0.260 m
t = 0.150 s
L=?
Solving net τ = ΔL / Δt for ΔL gives ΔL = (net τ) Δt.
Because the force is perpendicular to r, we see that net τ = rF , so that
L = r F • Δt
= (0.260 m)(2.50 N)(0.150 s)
= 9.75×10−2 kg ⋅ m2 / s.
Learning Competency:
Apply the torque-angular momentum relation (STEM_GP12REDIIa-10)
ACTIVITY 1 – CROSSWORD PUZZLE
Directions: Find and circle the following words below. These terms are related with
torque and angular momentum. They are hidden in the grid and can be found either
horizontally, vertically or diagonally.
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A
S
K
S
X
T
F
O
D
R
K
Q
A
U
K
E
X
C
I
U
O
P
X
C
A
N
A
S
D
E
TORQUE
PUSH
D
F
G
H
J
K
L
Q
D
E
R
Y
U
M
S
A
T
S
H
S
F
A
M
I
L
Y
E
U
I
S
D
F
Z
Q
P
U
S
H
E
R
C
E
L
Q
D
L
K
L
Q
H
O
G
X
C
D
H
J
K
E
A
S
D
T
I
H
I
U
U
R
Y
A
V
D
E
R
E
Y
H
O
P
T
H
J
K
E
F
G
H
W Z
I
K
L
I
S
D
I
R
E
C
T
I
O
N
G
A
N
G
L
E
A
L
Q
A
D
T
G
K
L
G
A
S
D
R
S
D
U
R
E
E
A
S
A
D
E
R
M
E
R
W H
W R
F
Z
M
F
G
H
F
T
I
O
D
T
W
G
U
L
A
R
M
O
M
E
N
T
U
M
R
E
Y
M
A
E
R
I
U
K
L
P
A
R
E
Y
D
F
G
H
O
P
A
S
C
F
ANGULAR MOMENTUM HINGE MAGNITUDE TAU
ANGLE
FORCE LEVER ARM DIRECTION
ACTIVITY 2 – TRUE OR FALSE
Directions: Write the word TRUE if the statement is correct and FALSE if the
statement is incorrect. Write your answer on the space provided before the number.
1. Angular momentum is conserved if the torque is not equal to zero.
2. Mass is one of the variable when dealing with torque.
3. Linear velocity and angular velocity are the same but used separately
in physics.
4. Changing the point doesn’t change the torques magnitude and
direction.
5. Torque is a vector quantity.
6. Angular momentum is a scalar quantity.
7. The lever arm is zero if the line of action passes through the axis of
rotation
8. Axis of rotation is sometimes referred to as the lever arm.
9. Friction will also stop something that is rotating.
10. Torque has both magnitude and direction.
ACTIVITY 3 – MULTIPLE CHOICES
Directions: Encircle the letter of your answer on the choices below the statement.
1. “All objects tend to keep on spinning”. What does the statement implies?
a. Objects in rotational motion, like moving objects along straight line will
keep on moving not unless acted upon by an outside force.
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2.
3.
4.
5.
b. That is only true for some objects under rotational motion.
c. The object will eventually stop soon.
d. Only outside application of force will make it stop.
What do we call the tendency of a spinning object to continue to spin?
a. Torque
b. Conservation of angular momentum
c. Conservation of Torque
d. Newton’s Second Law of spinning bodies
What is the symbol of Torque?
a. T
b. t
c. τ
d. t
What happens to angular momentum when the sum of all external torque
acting on a system of particles is zero?
a. The angular momentum is also zero
b. The angular momentum increases
c. The angular momentum decreases
d. The angular momentum remains constant
What happens when the line of action passes through the axis of rotation?
a. Lever arm is zero and so with the torque
b. Lever arm and torque is constant
c. There will be a negative rotation
d. Lever arm and torque doesn’t change
ACTIVITY 4 – PROBLEM SOLVING
Directions: Answer the following set of problem. The scoring is being provided for
you before the questions.
Given
Solution
Final Answer w/ unit
1 pt
2 pts.
2 pts.
1. In a public playground, one cloudy afternoon. Reyma and her friends wanted
to try the merry - go – round. Supposed she let her friends try it and she will
be the one to spin it for them. If she exerts 50 N force perpendicular to the 1.2
m radius for 5 s. What is the final angular momentum of the lazy Susan if it
starts from rest, assuming friction is negligible?
2. Calculate the torque of a door upon applying 5 N of force to open it and has a
lever arm of 2.3 m?
Reflection
1. I learned that ________________________________________________
____________________________________________________________
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_______________________________________________________
2. I enjoyed most on ____________________________________________
____________________________________________________________
__________________________________________________
3. I want to learn more on ________________________________________
____________________________________________________________
__________________________________________________
References
College Physics. Houston, TX: OpenStax College, 2012.
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Johnson, Kenneth W., and John D. Cutnell. Physics. Hoboken, NJ: Wiley, 2012.
Index of /~roldan/classes. Accessed November 9, 2020.
https://physics.ucf.edu/~roldan/classes.
ANSWER KEY
ACTIVITY 1 – CROSSWORD PUZZLE
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A
K
X
F
D
K
A
K
X
I
O
X
A
A
D
S
S
T
O
R
Q
U
E
C
U
P
C
N
S
E
D
M
E
R
X
I
O
K
G
K
A
F
G
R
R
F
S
U
C
C
U
P
L
A
L
S
Z
U
E
E
G
A
I
E
D
U
T
I
N
G
A
M
L
Y
Y
H
T
S
L
H
R
H
S
G
A
D
F
A
M
D
J
S
D
Q
J
Y
J
D
L
S
E
G
R
A
F
K
H
F
D
K
A
K
I
E
D
R
H
M
E
G
L
S
Z
L
E
V
E
R
A
R
M
F
O
R
H
Q
F
Q
K
A
D
F
E
L
S
E
T
M
I
O
D
A
P
L
S
E
G
C
Q
D
R
I
E
U
P
E
M
U
Q
D
R
H
T
A
U
W
O
N
K
A
R
I
S
H
T
E
W
I
D
R
H
D
T
L
S
Y
L
H
O
I
Y
Z
O
T
E
W
T
U
P
C
U
Y
E
G
H
H
I
N
G
E
R
W
M
A
F
ACTIVITY 2 – TRUE OR FALSE
1.
2.
3.
4.
5.
FALSE
FALSE
TRUE
FALSE
TRUE
6. FALSE
7. TRUE
8. FALSE
9. TRUE
10. TRUE
ACTIVITY 3 – MULTIPLE CHOICES
1.
2.
3.
4.
5.
a
b
c
d
a
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ACTIVITY 3 - PROBLEM SOLVING
1. 300 kg ⋅ m2 / s
2. 11.5 Nm
Prepared by:
CHARLES DAQUIOAG
Sanchez Mira School of Arts and Trades
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GENERAL PHYSICS 1
Name: ____________________________Grade Level: _________
Date: _____________________________Score:______________
LEARNING ACTIVITY SHEET
Static Equilibrium
Background Information for the Learners (BIL)
Static equilibrium means stationary or at rest. Statics is the study of forces
and momentum while equilibrium is a state of rest or balance due to the equal action
of opposing forces. Or an equal balance between any powers.
In static equilibrium, in every one kilogram is equivalent to 9.8 Newton. When
all the forces that act upon an object are balanced, then the object is said to be in a
state of equilibrium. The forces are considered to be balanced if the rightward forces
are balanced by the leftward forces and the upward forces are balanced by the
downward forces. This however does not necessarily mean that all the forces are
equal to each other. Consider the two objects pictured in the force diagram shown
below. Note that the two objects are at equilibrium because the forces that act upon
them are balanced; however, the individual forces are not equal to each other. The
50 N force is not equal to the 30 N force.
https://www.physicsclassroom.com/Class/vectors/u3l3c1.gif
If an object is at equilibrium, then the forces are balanced. Balanced is the key
word that is used to describe equilibrium situations. Thus, the net force is zero and
the acceleration is 0 m/s2 Objects at equilibrium must have an acceleration of 0 m/s/s.
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This extends from Newton's first law of motion. But having an acceleration of 0 m/s2
does not mean the object is at rest. An object at equilibrium is either
•
•
at rest and staying at rest, or
in motion and continuing in motion with the same speed and direction.
If an object is at rest and is in a state of equilibrium, then we would say that the
object is at static equilibrium. Static" means stationary or at rest. A common physics
lab is to hang an object by two or more strings and to measure the forces that are
exerted at angles upon the object to support its weight. The state of the object is
analyzed in terms of the forces acting upon the object. The object is a point on a
string upon which three forces were acting. See diagram at right. If the object is at
equilibrium, then the net force acting upon the object should be 0 Newton. Thus, if
all the forces are added together as vectors, then the resultant force (the vector sum)
should be 0 Newton. (Recall the resultant of adding all the individual forces head-totail.) Thus, an accurately drawn vector addition diagram can be constructed to
determine the resultant. Sample data for such are shown below.
Magnitude
Direction
Force A
3.4 N
161 deg.
Force B
9.2 N
70 deg.
Force C
9.8 N
270 deg
For most students, the resultant is 0 Newton (or at least very close to 0 N). It’s
expected - since the object is at equilibrium, the net force (vector sum of all the
forces) should be 0 N.
https://www.physicsclassroom.com/Class/vectors/u3l3c5.gif
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Another way of determining the net force (vector sum of all the forces) involves
its horizontal and vertical components. Once the components are known, they can
be compared to see if the vertical forces are balanced and if the horizontal forces are
balanced. The diagram below shows vectors A, B, and C and their respective
components. For vectors A and B, the vertical components can be determined using
the sine of the angle and the horizontal components can be analyzed using the
cosine of the angle. The magnitude and direction of each component for the sample
data are shown in the table below the diagram.
https://www.physicsclassroom.com/Class/vectors/u3l3c6.gif
Forces
A
B
C
Horizontal Component
Vertical Component
(PSYW)
(PSYW)
Ax = 3.4 N*cos (161®)
Ay = 3.4 N*sin (161®)
Ax = 3.2 N, Left
Ay = 1.1 N, up
Bx = 9.2 N *cos (70®)
By = 9.2 N *sin (70®)
Bx = 3.1 N, right
By = 8.6 N, up
Cx = 0 N
Cy = 9.8 N, down
The table above shows that the forces are nearly balanced. An analysis of the
horizontal components shows that the leftward component of A nearly balances the
rightward component of B. An analysis of the vertical components show that the sum
of the upward components of A + B nearly balance the downward component of C.
The vector sum of all the forces is (nearly) equal to 0 Newton. But what about the 0.1
N difference between rightward and leftward forces and the 0.2 N difference between
the upward and downward forces? Why do the components of force only nearly
balance? The sample data used in this analysis are the result of measured data from
an actual experimental setup. The difference between the actual results and the
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expected results is due to the error incurred when measuring force A and force B.
We have to conclude that this low margin of experimental error reflects an experiment
with excellent results. We could say it's "close enough for government work."
Analyzing a Hanging Sign
The above analysis of the forces acting upon an object
in equilibrium is commonly used to analyze situations
involving objects at static equilibrium. The most
common application involves the analysis of the forces
acting upon a sign that is at rest. For example, consider
the picture at the right that hangs on a wall. The picture
is in a state of equilibrium, and thus all the forces acting
upon the picture must be balanced. That is, all
horizontal components must add to 0 Newton and all https://www.physicsclassroom
vertical components must add to 0 Newton. The .com/Class/vectors/u3l3c7.gif
leftward pull of cable A must balance the rightward
pull of cable B and the sum of the upward pull of cable A and cable B must balance
the weight of the sign. Suppose the tension in both of the cables is measured to be
50 N and that the angle that each cable makes with the horizontal is known to be 30
degrees. What is the weight of the sign? This question can be answered by
conducting a force analysis using trigonometric functions. The weight of the sign is
equal to the sum of the upward components of the tension in the two cables. Thus,
a trigonometric function can be used to determine this vertical component. A diagram
and accompanying work is shown below.
https://www.physicsclassroom.com/Class/vectors/u3l3c8.gif
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Since each cable pulls upwards with a force of 25 N, the total upward pull of
the sign is 50 N. Therefore, the force of gravity (also known as weight) is 50 N, down.
The sign weighs 50 N.
In the above problem, the tension in the cable and the angle that the cable
makes with the horizontal are used to determine the weight of the sign. The idea is
that the tension, the angle, and the weight are related. If the any two of these three
are known, then the third quantity can be determined using trigonometric functions.
Consider the symmetrical hanging of a sign as shown at
the right. If the sign is known to have a mass of 5 kg and
if the angle between the two cables is 100 degrees, then
the tension in the cable can be determined. Assuming
that the sign is at equilibrium (a good assumption if it is
remaining at rest), the two cables must supply enough https://www.physicscla
upward force to balance the downward force of gravity. ssroom.com/Class/vect
The force of gravity (also known as weight) is 49 N (Fgrav
ors/u3l3c9.gif
= m*g), so each of the two cables must pull upwards
with 24.5 N of force. Since the angle between the cables is 100 degrees, then each
cable must make a 50-degree angle with the vertical and a 40-degree angle with the
horizontal. A sketch of this situation (see diagram below) reveals that the tension in
the cable can be found using the sine function. The triangle below illustrates these
relationships.
https://www.physicsclassroom.com/Class/vectors/u3l3c10.gif
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Thinking Conceptually
There is an important principle that emanates from some of the trigonometric
calculations performed above. The principle is that as the angle with the horizontal
increases, the amount of tensional force required to hold the sign at equilibrium
decreases. To illustrate this, consider a 10-Newton picture held by three different
wire orientations as shown in the diagrams below. In each case, two wires are used
to support the picture; each wire must support one-half of the sign's weight (5 N). The
angle that the wires make with the horizontal is varied from 60 degrees to 15 degrees.
Use this information and the diagram below to determine the tension in the wire for
each orientation.
https://www.physicsclassroom.com/Class/vectors/u3l3c11.gif
At 60 degrees, the tension is 5.8 N. (5 N / sin 60 degrees).
At 45 degrees, the tension is 7.1 N. (5 N / sin 45 degrees).
At 15 degrees, the tension is 19.3 N (5 N / sin 15 degrees).
In conclusion, equilibrium is the state of an object in which all the forces acting
upon it are balanced. In such cases, the net force is 0 Newton. Knowing the forces
acting upon an object, trigonometric functions can be utilized to determine the
horizontal and vertical components of each force. If at equilibrium, then all the vertical
components must balance and all the horizontal components must balance.
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Let’s try this sample using the trigonometric function!
After its most recent delivery, the infamous stork announces the good news.
If the sign has a mass of 10 kg (98 N), then what is the tensional force in each cable?
Use trigonometric functions and a sketch to assist in the solution
https://www.physicsclassroom.com/
Class/vectors/u3l3c16.gif
Solution!
The tension 56.6 Newtons.
Since the mass is 10.0 kg, the weight is 98.0 N. Each cable must pull
upwards with 49.0 N of force. Thus,
sine 60 (degrees) = (49.0 N) / (Ftens).
Proper use of algebra leads to the equation
Ftens = (49.0 N) / [ sine 60 (degrees) ] = 56.6 N.
Learning Competency:
Solve static equilibrium problems in context but not limited to see-saws, cable-hingestrut-system, leaning ladders, weighing a heavy suitcase using a small bathroom
scale (STEM_GP12RED-IIa-8)
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ACTIVITY 1: Show you solutions in solving the following questions
1. The following picture is hanging on a wall. Use trigonometric functions to
determine the weight of the picture.
https://www.physicsclassroom.com/Cla
ss/vectors/u3l3c12.gif
2. The image below advertized the most important truth to be found inside. The
sign is supported by a diagonal cable and a rigid horizontal bar. If the sign has a
mass of 50 kg (490 N), then determine the tension in the diagonal cable that
supports its weight.
3. The following sign can be found in Glenview. The sign has a mass of 50 kg (490
N). Determine the tension in the cables.
https://www.physicsclassroom.com
/Class/vectors/u3l3c14.gif
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https://www.physicsclassroom.com/
Class/vectors/u3l3c13.gif
ACTIVITY 2: MULTIPLE CHOICE
Directions: Choose the best answer to the following questions
1. Statics is the study of?
a. Forces and moments
b. Acceleration
c. Inertia
d. Weight and torque
2. Static equilibrium only defines an object at rest.
a. True
b. False
3. For an object in static equilibrium......
a. Acceleration is zero
b. Velocity is zero
c. No motion at all exist
d. No external force acting on it
4. Comparing the mass of the right side and the left side which greater? Why?
https://images-na.ssl-imagesamazon.com/images/l/31lHqT
skfrL._AC_SY400_.jpg
a. Left, because of its length
b. Right, because of its width
c. Left, because of its width
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d. None of the above
5. A croquet mallet balances when suspended from its center of mass, as shown
in the left part of the figure. If you cut the mallet into two pieces at its center of
mass, as shown in the right part if the figure, how do the masses of the two
pieces compare?
a. The piece with the head of the mallet has the greater mass
b. The piece with the head of the mallet has the smaller mass
c. The masses are equal
ACTIVITY 3: Answer the following questions briefly
1. Why is static equilibrium important?
2. What is an example of static equilibrium?
3. Why do we need to study static equilibrium? What is its connection to our daily
life?
4. What is an example of equilibrium in everyday life?
ACTIVITY 4: Solve for the unknown!
1. weight=?
Ftens = 45 N
Ftens = 45 N
45 N
25®
2.
Fy
25®
25®
Ftens =?
Ftens
mass= 30kg
55®
3. Ftens= ?
Fy
55®
55®
Ftens
Fy
NOTE: Practice personal hygiene protocols at all times
87
mass = 15kg
35®
Reflection
1. I learned that ________________________________________________
____________________________________________________________
_______________________________________________________
2. I enjoyed most on _____________________________________________
____________________________________________________________
__________________________________________________
3. I want to learn more on _________________________________________
____________________________________________________________
__________________________________________________
88
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References
https://www.physicsclassroom.com/class/vectors/Lesson-3/Equilibrium-and
-
Statics
https://www.dictionary.com/browse/equilibrium
https://quizizz.com/admin/quiz
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ANSWER KEY
Activity No. 1
1. The weight of the sign is 42.4 N.
The tension is 30.0 N and the angle is 45 degrees. Thus,
sine (45 degrees) = (Fvert) / (30.0 N).
The proper use of algebra leads to the equation:
Fvert = (30.0 N) • sine (45 degrees) = 21.2 N
Each cable pulls upward with 21.2 N of force. Thus, the sign must weigh
twice this - 42.4 N.
2. The tension is 980 Newtons.
Since the mass is 50 kg, the weight is 490 N. Since there is only one "upwardpulling" cable, it must supply all the upward force. This cable pulls upwards
with approximately 490 N of force. Thus,
sine (30 degrees) = (490 N ) / (Ftens).
Proper use of algebra leads to the equation
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Ftens = (490 N) / [ sine 30 (degrees) ] = 980 N.
3. The tension is 346 Newtons.
Since the mass is 50.0 kg, the weight is 490 N. Each cable must pull
upwards with 245 N of force.
Thus, sine (45 degrees) = (245 N ) / (Ftens).
Proper use of algebra leads to the equation
Ftens = (245 N) / [sine (45 degrees)] = 346 N.
ACTIVITY NO 2.
1.) A
2.) B
3.) A
4.)A
5.)A
ACTIVITY NO. 3
1. Static equilibrium is important because an object in translational equilibrium is
not travelling from one place to another, and an object in rotational equilibrium
is not rotating around an axis… Static equilibrium is a valuable tool: for
example, if two forces are acting on an object that is in static equilibrium, that
means they add up to zero
2. Static equilibrium means the resultant force is zero and the object is not
moving.
Examples: An object (e.g book) lying still on the surface (e.g table). No
resultant moment about a pivot, so clockwise moment equals anticlockwise
moment and there is no resultant force and no motion either.
3. Answer may vary
4. There are many examples of chemical equilibrium all around you. One
example is a bottle of fizzy cooldrink. In the bottle there is carbon dioxide
dissolved in the liquid. There is also gas in the space between the liquid and
the cap.
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ACTIVITY NO. 4
𝐹𝑦
1. Degree = 𝐹𝑡𝑒𝑛𝑠
𝐹𝑦
Sin 25 = 45 𝑁
Fy= Sin25 * 45N
Fy(weight) = 19.02 N
Fy
2. Ftens = 𝑑𝑒𝑔𝑟𝑒𝑒
Convert 30kg to weight is equal to 294
Each cable must pull 147 N
147 𝑁
Ftens= sin 55
Ftens= 179.45 N
Fy
3. Ftens = 𝑑𝑒𝑔𝑟𝑒𝑒
Convert 15kg to weight is equal to 147
Since there is only on upward pulling cable, the weight is equal to 147
147 𝑁
Ftens= sin 35
Ftens= 256.29 N
Prepared by:
GLENDA M. MADRIAGA
BUKIG NATIONAL AGRICULTURAL AND TECHNICAL SCHOOL
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GENERAL PHYSICS 1
Name: ____________________________Grade Level: _________
Date: _____________________________Score:______________
LEARNING ACTIVITY SHEET
Newton’s Law of Gravitational
Background Information for the Learners (BIL)
Isaac Newton compared the acceleration of the moon to the acceleration of
objects on earth. Believing that gravitational forces were responsible for each,
Newton was able to draw an important conclusion about the dependence of gravity
upon distance. This comparison led him to conclude that the force of gravitational
attraction between the Earth and other object is inversely proportional to the distance
separating the earth’s center from the object’s center. But distance is not the only
variable affecting the magnitude of a gravitational force. Consider Newton’s famous
equation
Fnet=m*a
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Newton knew that the force that caused the apple’s acceleration (gravity) must
be dependent upon the mass of the apple. And since the force acting to cause the
apple’s downward acceleration also causes the earth’s upward acceleration
(Newton’s third law), that force is also dependent pon the mass of the earth. So for
Newton, the force of gravity acting between the earth and any object is directly
proportional to the mass of the earth, directly proportional to the mass of the object,
and inversely proportional to the square of the distance that separates the center of
the earth and the object.
The universal gravitation equation
Newton’s law of universal gravitation is about the universality of gravity.
Newton’s place in the Gravity Hall of Fame is not due to his discovery of gravity, but
rather due to his discovery that gravitation is universal. All objects attract each other
with a force of gravitational attraction. Gravity is universal. This force of gravitational
attraction is directly proportional upon the masses of both objects and inversely
proportional to the square of the distance that separates their centers. Newton’s
conclusion about the magnitude of gravitational force is summarized symbolically as
𝒎𝟏∗𝒎𝟐
Fgravα =
𝒅𝟐
WhereFgravrepresents the force of gravity between two objects
αmeans “proportional to”
m1 represents the mass of object 1
m2 represents the mass of object 2
d represents the distance separating the object’s center
Since the gravitational force is directly proportional to the mass of both
interacting objects, more massive objects will attract each other with a greater
gravitational force. So as the mass of either object increase, the force of gravitational
attraction between them also increases. If the mass of one of the objects is doubled,
then the force of gravity between them is tripled. If the mass of both of the objects is
doubled, then the force of gravity between them is quadrupled, and so on
Since gravitational force is inversely proportional to the square of the
separation distance between the two interacting objects, more separation distance
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will result in weaker gravitational force. So as two objects are separated from each
other, the force of gravitational attraction between them also decreases. If the
separation distance between two objects is doubled (increased by a factor of 2), then
the force of gravitational attraction is decreased by a factor 4 (2 raised to the second
power). If the separation distance between any two objects is tripled (increased by a
factor of 3), then the force of gravitational attraction is decreased by a factor of 9 (3
raised to the second power).
Thinking proportionally about Newton’s equation
The proportionalities expressed by Newton’s universal law of gravitation are
represented graphically by the following illustration. Observe how the force of gravity
is directly proportional to the product of the two masses and inversely proportional to
the square of the distance of separation.
Another means of representing the proportionalities is to express the
relationships in
the
equation using a
proportionality.
form
of
an
constant
This
equation
is
shown below.
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The constant of proportionality (G) in the above equation is known as the
universal gravitation constant. The precise value of G was determined experimentally
by Henry Cavendish in the century after Newton’s death. The value of G is found to
be
G=6.673 x 10 -11 N m2/kg2
The units on G may seem rather odd;
Nonetheless they are sensible. When the units on G are substituted into the
equation above and multiplied by m1 x m2 units and divided by d2 units, the result
will be Newtons – the unit of force.
Using Newton’s gravitation equation to solve problems
Knowing the value of G allows us to calculate the force of gravitational
attraction between any two objects of known mass and known separation distance.
As a first example, consider the following problem.
Sample Problem #1
Determine the force of gravitational attraction between the earth (m=5.98 x 10 24 kg)
and a 70- kg physics student if the student is standing at sea level, a distance of 6.38
x
106m
from
the
earth’s
center.
*The solution of the problem involves substituting known values of G (6.673 x 10 -11N
m2/kg2, m1 (5.98 x 1024 kg), m2 (70 kg) and d (6.38 x 106m) into the universal
gravitation equation and solving for Fgrav. The solution is as follows:
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𝑚2
).(5.98𝑥 1024 𝑘𝑔).(70𝑘𝑔)
𝑘𝑔2
(6.38 𝑥 106 𝑚)2
(6.673 𝑥 10−11 𝑁
Fgav =
Fgav = 686 N
Two general concepts can be made about the results of the two sample
calculations above. First, observe that the force of gravity acting upon the students
(a.k.a. the student’s weight) is less on an airplane at 40 000 feet than at sea level.
This illustrates the inverse relationship between separation distance and the force
of gravity (or in this case, the weight of the student). The student weighs less at the
higher altitude. However, a mere change of 40000 feet further from the center of the
earth is virtually negligible. This altitude altered the student’s weight by 2 N that is
much less than 1% of the original weight. A distance of 40 000 feet (from the earth’s
surface to a high altitude airplane) is not very far when compared to a distance of
6.38 x 106m (equivalent to nearly 20 000 000 feet from the center of the earth). This
alternation of distance is like a drop in a bucket when compared to the large radius
of the earth. As shown in the diagram below, distance of separation becomes much
more influential when a significant variation is made
Force of Gravitational towards Earth for a 70-kg Physics Student at various
location
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The second conceptual to be made about the above sample calculations is
that the use of Newton’s universal gravitation equation to calculate the force of
gravity (or weight) yields the same result as when calculating it using the equation.
Fgrav = mxg = (70 kg) x (9.8 m/s2) = 686 N
Both equations accomplish the same result because the value of g is
equivalent to the ration of (G x Mearth)/(Rearth)2.
The universality of gravity
Gravitational interactions do not simply exist between the earth and other
objects; and not simply between the sun and other planets. Gravitational interactions
exist between all objects with an intensity that is directly proportional to the product
of their masses. So as you sit in your seat in the physics classroom, you are
gravitationally attracted to your partner, to the desk you are working at, and even to
your physics book. Newton’s revolutionary idea was that gravity is universal- All
objects attract in proportion to the product of their masses. Gravity is universal. Of
course, most gravitational forces are so minimal to be noticed. Gravitational forces
are only recognizable as the masses of objects become large.
Learning Competency:
Use Newton’s Law of Gravitation to infer gravitational force, weight, and acceleration
due to gravity STEM_GP12G-llb-16
ACTIVITY 1: Finding my force!!
Directions: Use Newton’s Universal Gravitation equation to calculate the force of
gravity between the following familiar objects.
a.
Mass of Object 1
Mass of Object 2
Separation
Force of
(kg)
(kg)
Distance (m)
Gravity (N)
Football Player
Earth
6.38 x 106 m
100kg
5.98 x 1024 kg
(on surface)
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b.
c.
d.
e.
f.
g.
h.
Ballerina
Earth
6.38 x 106 m
40kg
5.98 x 1024 kg
(on surface)
Physics Student
Earth
6.60 x 106 m
70 kg
5.98 x 1024 kg
(low-height orbit)
Physics Student
Physics Student
1m
70 kg
70 kg
Physics Student
Physics Student
70 kg
70 kg
Physics Student
Physics Student
70 kg
80 kg
Physics Student
Moon
1.71 x 106 m
70 kg
7.34 x 1022 kg
(on surface)
Physics Student
Jupiter
6.98 x 107 m
70 kg
1.901 x 1027 kg
(on surface)
0.2m
1m
.
ACTIVITY 2: Choose the correct answer.
Directions: Choose the best answer to the following questions
1. Which is needed to determine the amount of gravitational force between two
objects?
a. Distance and mass’
b. Weigh and time
c. Area and weight
2. The gravitational force exerted by an object depends on its
a. Volume
b. Mass
c. Weight
3. The SI units of force are measured in
a. Grams
b. Newtons
c. Pounds
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d. Kilograms
4. The force of gravity (mass x gravity) is also known as .......
a. Mass
b. Weight
c. Distance
d. Acceleration
5. Which of the following statements refers to gravitational force?
https://justdoscience.weebly .com/uploads/2/5/2/9/2529400/6274505.png481
a. It makes objects at rest start moving
b. It makes objects that are moving stop
c. The force of attraction between two objects
d. It pulls you into space
6. It is said to be that Earth’s gravity has a value of 9.8 m/s 2. Earth’s gravity is
considered as a/an?
a. Force
b. Weight
c. Acceleration
d. Mass
7. How is the gravitational force between two objects related to their mass?
a. They are directly proportional
b. They are inversely proportional
c. They do not affect each other
d. They are equal
8. Which of the following equations refers to Newton’s Law on Gravitation?
a. F=Gm1m2/d
b. F=Gm1m2/d2
c. F=Gm1m2/2d
d. F=m1m2/2d
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9. What states that every object in the universe attracts every other object”?
a. Law of Universal Gravitation
b. Newton’s First Law of Motion
c. Newton’s Third law
d. Inertia and Gravity
10. If these teams are pulling with the same amount of force what will happen?
https://www.123rf.com/photo_84007819_group-of-children-playing-tug-of-war.html
a. The left team will win
b. They will not move at all
c. The right team will win
d. Both will fall down
ACTIVITY 3: Help me!!!!Let’s find force?
Directions: Show your complete solution in solving the following problems using
the law of gravitation equation.
1. Two spherical objects have masses of 200kg and 500kg. Their centers are
separated by a d istance of 25m. Find the gravitational attraction between
them.
2. Two spherical objects have masses of 3.1 x 10 5 kg and 6.5 x 103 kg. The
gravitational attraction between them is 65 N. How far apart are their centers?
3. Two spherical objects have masses of 8000kg and 1500kg. Their centers are
separated by a distance of 1.5m. Find the gravitational attraction between
them.
4. What is the force of attraction between two people, one of mass 80kg and the
other 100 kg if they are 0.5 m apart?
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ACTIVITY 4: Think about it!
Directions: Answer the following questions.
1. Newton’s law of universal gravitation had a huge impact on how people
thought about the universe. Why do you think it was so important?
2. What equation did Newton use to represent the force of gravity between two
objects?
3. A. From your answer in question no. 2. What does each letter in the equation
stand for? B. Which letter stand for a value that never change?
Reflection
1. I learned that ____________________________________________
_______________________________________________________
_______________________________________________________
2. I enjoyed most on ________________________________________
_______________________________________________________
_______________________________________________________
3. I want to learn more on _____________________________________
________________________________________________________
____________________________________________________
102
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References
https://www.physicsclassroom.com/class/circles/Lesson3/Newton-s-Law-of-Universal_Gravitation
https://quizizz.com/admin/quiz/58e2593815f716c1e479a44/ne
wtons-law-of-universal-gravitation
https://www.ck12.org/c/physics/newtons-univeral-law-of-gravity/lesson/user:cndhz25lckbuzxzjlmsxmi5tby51cw../Newtons-Law-of-Gravity--MS-
ESS1-2/?referrer=concept_details
103
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ANSWER KEY
Activity no. 1
1. 980 N
2. 392 N
3. 641 N
4. 3.27 x 10-7 N
5. 8.17 x 10-6 N
6. 4.67 x 10-9 N
7. 117 N
8. 1823 N
Activity no. 2
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1. A
2. B
3. B
4. D
5. C
6. C
7. A
8. B
9. A
10. B
Activity no. 3
1. 𝐹 =
𝐺𝑚1 𝑚2
𝑑2
(6.67 𝑥 10−11
=
𝑁𝑚2
)(200𝑘𝑔)(500𝑘𝑔)
𝑘𝑔2
(25 𝑚)2
= 1.1x 10-8 N
2
−11 𝑁𝑚 )(3.1 𝑥 105 𝑘𝑔)(6.5 𝑥 103 𝑘𝑔)
𝑘𝑔2
(2500𝑚)2
𝐺𝑚1 𝑚2 √(6.67 𝑥 10
2. d= √
3. 𝐹 =
4. 𝐹 =
𝐹
𝐺𝑚1 𝑚2
𝑑2
𝐺𝑚1 𝑚2
𝑑2
(6.67 𝑥 10−11
=
𝑁𝑚2
)(8000𝑘𝑔)(1500𝑘𝑔)
𝑘𝑔2
(1.5 𝑚)2
(6.67 𝑥 10−11
=
𝑁𝑚2
)(80𝑘𝑔)(100𝑘𝑔)
𝑘𝑔2
(0.5 𝑚)2
= 𝟏. 𝟒 𝒙 𝟏𝟎−𝟏𝟏 𝒎𝒆𝒕𝒆𝒓𝒔
= 3.6x 10-4 N
= 2.14x 10-6 N
Activity no. 4
1. Newton’s law of gravitational was the first scientific law that applied to the
entire universe. It explains the motion of objects not only on Earth but in outer
space as well.
2. 𝐹 =
𝐺𝑚1 𝑚2
𝑑2
3. A.Fgravrepresents the force of gravity between two objects
αmeans “proportional to”
m1represents the mass of object 1
m2 represents the mass of object 2
d represents the distance separating the object’s center
B. gravity (6.673 x 10-11N m2/kg2)
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Prepared by:
GLENDA M. MADRIAGA
BUKIG NATIONAL AGRICULTURAL AND TECHNICAL SCHOOL
GENERAL PHYSICS1
Name: __________________________
Grade Level: _________
Date: ___________________________
Score: ______________
LEARNING ACTIVITY SHEET
Gravitational Field
Background Information for the Learners (BIL)
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Gravitational field like any other force field is responsible for the force
on a body. Gravitational fields originate from all the massive bodies and result
in the attractive pull known as the gravitational force of the body. More studies
are going on in the field of physics to fully understand this force and these
fields.
Gravitational Field, Gravitation, or gravity is a natural phenomenon by
which physical bodies attract with a force proportional to their masses.
The gravitational interaction between two bodies can be expressed by a
central force which is proportional to the mass of bodies and inversely
proportional to the square of the distance that separates them.
F = G m₁. m₂
r²
where:
F= central force
G= universal gravitational constant = 6.67x10ˉ¹¹ N.m²/kg²
m = mass of the objects
r = distance between the two masses
The gravitational field is a physical property that is communicated to the
space by a mass m. This field is characterized by conservative vector fields
and it can be represented with a lines of force.
The gravitational field strength at any point in space is defined as the force
per unit mass (on a small test mass) at that point.
g = F/m (in N/kg)
Gravitational field around a point mass
If we have two masses m and m distance r apart
1
2
F= Gm₁m₂
r²
Looking at the force on m due to m , F = gm
1
2
1
2
F = Gm₁m₂/r = gm
1
g (field due to m₂) = Gm₂/r²
If we have two masses m₁ and m₂ distance r apart
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F = Gm₁.m₂ /r²
m₁
Looking at the force on m₁ due to m₂, F = gm
m₂
1
F = Gm₁m₂/r² = gm₁
2
g (field due to m ) = Gm /r
2
2
For any planet;
g = Gm₂/rp²
Don’t forget that for non-point mass, r is the distance to the center of mass
Fields as the gravitational fields that are defined at each point of space by a
vector quantity are called vector fields. These fields can be represented by lines of
force. A line of force has the characteristic of being tangent at all its points to the
direction of the field at that point and its meaning is the same as that of the field.
Gravitational field is a vector, and any calculations regarding fields
(especially
involving addition of fields from more than one mass) must use vector addition.
(i)
Field here due to both masses
m₁
m₂
(ii)
Field here due to both
masses
Field due to m₁
m₁
m₂
(iii)
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Field here due to m₂
Field due to m₁
Resultant field
m₁
m₂
Superposition principle
In the case of a field which is created by several bodies we use the
superposition principle to know the aggregate field at a given point. The
principle of superposition tells us that the gravitational field created by a body
at a point is independent from gravitational fields which are created by other
bodies. We will operate by finding out the field created by each body at the
point in question and we will add all of them (vector sum) for the total field.
Escape velocity is the minimum speed that a body should be thrown
to escape from the gravitational pull of the Earth or other celestial body. This
means that the body or projectile will not fall on Earth or starting astro leaving
at rest on a sufficiently large (in principle infinite) distance from Earth or the
star.
This speed explains why some planets have atmospheres and others
not. According to the kinetic theory of gases, the gas molecules move at a
speed:
where; v = velocity
m = mass of the molecule
T = temperature in Kelvin
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K = Boltzmann constant
Newton’s Law of Universal Gravitation
(G = universal gravitational constant) = 6.67x10ˉ¹¹ N.m²/kg²
Force between two masses; F = Gm₁ m₂
r²
Gravitational Field Strength; g = _F_
m
Magnitude of gravitational field strength in a radial field; g = GM
r²
The minus sign means that the gravitational field is directed in the
opposite direction to the unit vector that it points the direction from the Earth
to the point in question.
Gravitational field patterns
A gravitational field can be represented by lines and arrows on a
diagram, in a similar way to magnetic field lines.
The closer the lines
are together, the
stronger the force
This is an
felt.
example of a
Note, gravity is
radial field
always attractive
Field around a uniform spherical mass
The figure below the direction that a mass would accelerate if placed
in the field and help us to imagine the field.
Around a spherical mass the field lines are closer together
nearer the surface, so the field strength is larger.
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https://spark.iop.org/collections/gravitational-fields#gre
Field close to the earth’s surface
Field Lines near the Earth are almost parallel.
The field is uniform. Wherever you are near the surface of the earth
you are pulled down with the same Force/Kilogram.
Uniform
https://spark.iop.org/collections/gravitational-fields#gre
Field Strength is a vector, so two values of g can be added together
111
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https://www.slideshare.net/simonandisa/gravitational-force-and-fields
Learning Competency
Discuss the physical significance of gravitational field (STEM_GP12Red-IIb18).
Learning Activity 1 -Word Search
Directions: Find the words that related to gravitational field in the grid. They can be
horizontal, vertical, diagonal, and backwards.
L
Q
F
F
I
E
L
D
L
I
N
E
S
L
I
N
E
S
O
F
F
O
R
C
E
E
F
R
K
A
S
Q
T
R
I
X
H
P
R
O
S
R
E
S
U
L
T
A
N
T
G
K
Q
L
U
V
A
E
A
S
K
U
G
F
O
S
M
W
N
M
T
G
V
B
N
I
T
Y
S
D
L
E
I
F
R
O
T
C
E
V
I
D
L
E
V
B
F
R
S
Y
X
W
F
H
V
T
A
W
P
D
O
A
S
M
B
S
T
A
R
N
L
A
C
I
R
E
H
P
S
C
G
J
H
I
T
S
R
E
O
S
L
A
W
Learning Activity 2- Problems on Field close to the earth’s surface
Directions: Read carefully the problems below and solve for the unknown
quantities. Show all your solutions.
1. What is the weight of a 25.0 kg object near the surface of the earth?
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2. What is the mass of an object if it has a weight of 80.0 N near the
earth’s surface?
3. The Earth orbits the Sun at a distance of 1.46x1010 m from center to
center. What is the strength of the Sun’s gravitational field at this
distance?
4. What is the acceleration due to gravity on the surface of the sun?
5. What is the mass of an object if it has a weight of 127 N near the
earth’s surface?
Learning Activity 3 - Problems of Gravitational field on point masses
Directions: Solve the problem sets
1. On the surface of Venus, which has a mass of 4.869×1024 kg, an
object has a weight of 213 N and a mass of 24 kg. What is the radius
of Venus? G = 6.674×10–11 N·m2/kg2.
2. What is the gravitational field (in N/kg) 1.400×105 km above the
surface of the Sun? Radius of the Sun = 6.960×105 km, mass of the
Sun = 1.989×1030 kg, G = 6.674×10–11 N·m2/kg2.
3. Two spherical balls are placed so their centers are 74 m apart. The
gravitational attraction between them is 2.362×10–7 N. If the mass of
the smaller ball is 3800 kg, find the mass of the other ball. G =
6.674×10–11 N·m2/kg2.
Learning Activity 4 – Gravitational field as a vector
Directions: Solve the problem sets
1. Three masses are located in the vertices of an equilateral triangle.
Calculate the magnitude and direction of the gravitational force on
the mass m₁. Given: m₁ = 38 kg, m₂ = 340 kg, m₃ = 340 kg, r = 38 m.
G = 6.674×10–11 N·m²/kg².
2. Three masses are located in the corners of an
equilateral triangle. Find the magnitude and direction of the
gravitational field at the center of the triangle.
Given: m₁ = 22 kg, m₂ = 30 kg, m₃ = 30 kg,
r = 12 cm. G = 6.674×10–11 N·m²/kg².
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3. Four masses are at the vertices of a square. Find the magnitude of the
gravitational force on the mass m₁. Given: m₁ = 6 kg, m₂ = 80 kg,
m₃ = 80 kg, m₄ = 80 kg r = 24 m. G = 6.674×10–11N·m²/kg².
4. Four masses are located in the corners of a square. Calculate the
magnitude and direction of the gravitational field at the center of the
square. Given: m₁ = 85 kg, m₂ = 3 kg, m₃ = 3 kg, m₄ = 85 kg r = 8 m,
G = 6.674×10–11 N·m²/kg².
5.
Two masses m₁ = 350 kg and m₂ = 350 kg are at a distance of 22 m
from each other. Find the magnitude and direction of the gravitational
field at point A. Data: h = 9 m, G = 6.674×10–11 N·m²/kg².
Learning Activity 5 – Ciphers Text Analysis
Directions: Decode the secret message in the cryptogram based on your
reading in this Learning activity Sheet.
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1. ss amayb ecaps otdet acinu mmocs itaht ytrep orpla cisyh pasid
leifl anoit ativa rgehT
2. ssam tniop adnuo radle iflan oitat ivarG
3. ss amlac irehp smrof inuad nuora dleiF
4. .tl efecr ofeht regno rtseh t,reh tegot erase nileh treso lcehT
5. .lell arapt somla eraht raEeh traen seniL dleiF
Reflection
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1. I learned that ________________________________________________
____________________________________________________________
_______________________________________________________
2. I enjoyed most on _____________________________________________
____________________________________________________________
__________________________________________________
3. I want to learn more on _________________________________________
____________________________________________________________
__________________________________________________
References
116
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https://study.com/academy/answer/discuss-the-physical-significance-of-thegravitational-field.html
https://www.slideshare.net/simonandisa/gravitational-force-andfields?from_action=save
http://www.holytrinityacademy.ca/documents/general/Lesson11%20Gravitati
onal%20Fields%20Worksheet.pdf
http://www.vaxasoftware.com/doc_eduen/fis/x_gravit_point_masses.pdf
https://www.rpi.edu/dept/phys/Courses/Astro_F97/Class03/orbiter.html
https://www.physicsclassroom.com/Physics-Interactives/Circular-andSatellite-Motion/Gravitational-Fields/Gravitational-Fields-Exercise
https://physics.gurumuda.net/gravitational-field-problems-and-solutions.htm
https://www.boxentriq.com/code-breaking/text-analysis
ANSWER KEY
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Learning Activity 1 - Word Search
1.
2.
3.
4.
5.
Gravitation
Field Lines
Uniform
Resultant
Spherical
6. Mass
7. Vector Fields
8. Weight
9. Force
10. Lines of Force
Learning Activity 2 - Problems on Field close to the earth’s
surface
1. 245 N
2. 8.16 kg
3. 0.62 msˉ²
4. 274 msˉ²
5. 13.0 kg
Learning Activity 3 - Problems of Gravitational field on point
masses
1. 6051 km.
2. 190 N/kg
3. 5100 kg.
Learning Activity 4 - Problems of Gravitational field as a vector
1.
2.
3.
4.
5.
1.034×10–9 N, downward.
1.112×10–7 N/kg, downward
1.065×10–10 N.
2.419×10–10 N/kg, to the left.
1.465×10–10 N/kg, downward.
Learning Activity 5 - Ciphers Text Analysis
1. The gravitational field is a physical property that is
communicated to the space by a mass.
2. Field around a uniform spherical mass
3. Field around a uniform spherical mass
4. The closer the lines are together, the stronger the force felt.
5. Field Lines near the Earth are almost parallel.
Prepared by:
LEONOR C. NATIVIDAD
Baggao National High School
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GENERAL PHYSICS 1
Name: ____________________________
Grade Level: _________
Date: _____________________________
Score: ______________
LEARNING ACTIVITY SHEET
Gravitational Potential Energy
Background Information for the Learners (BIL)
Gravitational Potential Energy
Gravitational Potential Energy (GPE) is the energy that an object has due to
its position relative to the Earth’s surface. For instance, we are in a system (Earth’s
atmosphere) where each body exerts force on each other. If there is a change in
position with respect to the earth’s surface, then there would be a gain in potential
energy.
The gravitational force that acts on every kg of mass near the Earth’s surface
is represented as g with a value of 10N/kg or 10m/s² so you can think of g in two
ways.
1. A gravitational force of 10N acts on every kg of mass near the
Earth’s surface.
2.
A free-falling object near the Earth’s surface will accelerate at
10m/s²
But you may ask, where did the acceleration, 10ms-2 come from? Well
you have learnt that 1N =1kg.m/s². So, if g = 10N/kg then in place of N
we would write 10 kgms-2/kg.
10N; = 10kgms-2
kg
kg
-2
g = 10 ms
g=
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Notice that kg cancels out and you are left with 10ms-2. If greater
accuracy is required in a calculation, then use g = 9.8N/kg or 9.8ms-2.
Calculating gravitational potential energy
If you decide to run up the steps of a building, the force of gravity will
act on you, thus, there is force between you and the surface of the earth. As
you make your way up the steps you are doing work by moving yourself from
the ground floor up the steps. As you move up, the force of gravity will act on
you so you will carry your own weight up the steps. This results in work
being done so you will gain gravitational potential energy
An object of mass (m) at a vertical height (h) above the ground
has a gravitational potential energy (mgh)
Work done = change in gravitational potential energy (GPE)
= Force x distance
= weight x height
= mass x acceleration due to gravity x height
= mgh
GPE = mgh
Example 1
If you weigh 60kg and ran up the building steps covering a distance of
30 meters then the GPE is calculated as follows:
GPE = mgh
= 60kg x 10m/s2 x 30m
= 18 000kg.m²/s²
= 18 000 J
OR
GPE = 60kg x 9.8m/s2 x 30m
= 17 640 kg.m²/s²
= 17 640J
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If you use g = 10m/s2 then the answer in example 1 is 18 000J. If you
use g = 9.8ms-2 then the answer is 17640J.
For every calculations dealing with GPE, use g = 10ms-2. But some
questions will require you to use g = 9.8ms-2 for more accuracy in
calculations.
Example 2
An object has a mass of 6kg. Calculate its GPE
a)
4m above the ground and
b)
8m above the ground
c)
At what height above the ground will its GPE be 360J?
Solution
a)
GPE = mgh
= 6kg x 10ms-2 x 4m
= 240J
b)
GPE = mgh
= 6kg x 10ms-2 x 8m
= 480J
c. GPE = mgh
h = GPE
mg
= 6m
Example 3
If you lift a 3kg object from an initial height of 5m to a height of 8m and
place it at the top of a shelf, you are doing work on it, since you are
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applying a force that is in the direction of its displacement (both vertical).
In doing work on it, you are also changing its GPE.
Calculate the change in GPE of the above scenario. (use g = 10ms-2)
(i)
At initial height of 5m, the GPE is:
GPE = mgh
= 3kg x 10ms-2 x 5m
= 150J
(ii)
At final height of 8m, the GPE is:
GPE = mgh
= 3kg x 10ms-2 x 8m
= 240J
(iii)
Therefore, the change in GPE is 240J, 150J, 90J
A simpler way to calculate the change in GPE above is by taking the
difference in height and then substitute the difference in the formula
mgh to find the change in GPE.
(i)
Difference in height (is also stated as change in height) is 8m,
5m, 3m. Change in height is represented by delta h
(ii)
Therefore, GPE is:
GPE = mgh
= 3kg x 10m/s2 x (3m)
= 90 kg.m²/²
= 90J
Energy is a scalar quantity
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Mass A and B have the same magnitude. A was moved up the slope
with less force but the distance moved was greater. Mass B was lifted
vertically from the ground. Same amount of work was done in each
case so both masses have the same GPE. So to calculate the
gravitational potential energy of A and B you need to know the vertical
height only but not the direction taken. Therefore, energy is a scalar
quantity because direction is not considered.
A
A
B
BB
h
A
A
Figure 6
B
BB
Illustration of energy being a scalar quantity
(Flexible Open and Distance Education Papua New Guinea)
Example 4
A 35kg beer keg is rolled up a 5m long plank, which makes a
30° inclination to the ground. What is the GPE of the keg at the top?
Solution
A 30° incline plane with a
hypotenuse of 5m has a vertical
height given by: 5.0
2.5m.
sin 30° =
GPE = mgh
= 35kg x10ms-2 x 2.5m
= 875J
(Flexible Open and Distance Education
Papua New Guinea)
123
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Learning Competency:
Apply the concept of gravitational potential energy in physics problems
(STEM_GP12Red-IIb-19)
Learning Activity 1 - Explore and Discover
Directions: Solve for the Gravitational Potential Energy of Ball A, B and C.
Now, look at the illustration below and study the calculation of their gravitational
potential energies. Ball A and B have the same mass (3kg). Ball A and C have the
same height (4m).
A
2kg
3kg
B
C
3kg
4m
4m
2m
Figure 5 Ball at different heights above ground level
(Flexible Open and Distance Education Papua New Guinea)
124
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Mass (kg)
Weight (N)
5
2
8
20
5000
0.2
67
Height (m)
Gravitational Potential
energy (J)
2
6
5
0.6
2
10
44
Learning Activity 2 – Keep Moving
Directions: Complete the table
1. Calculate the weight for the objects in the table below.
2. Assuming that the object are on Earth, where acceleration due to gravity is
10N/kg, calculate the gravitational potential energy that they had.
3. Re-calculate the weight for the same objects, if they were on Mercury (where
the acceleration due to gravity is 4N/kg)
Mass (kg)
Weight (N)
Height (m)
5
2
8
20
5000
0.2
67
Gravitational Potential
energy (J)
2
6
5
0.6
2
10
44
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Learning Activity 3 – Apply your Skills
Directions: Read and answer the following questions accordingly in the space
provided.
Assume g= 9.8 m/s² near the surface of the Earth.
1)
Climbing a vertical rope is difficult. You have to lift your full body weight with
your arms. If your mass is 60 kg and you climb 2.0 m, by how much do you
increase your gravitational potential energy?
2)
A block of bricks is raised vertically to a bricklayer at the top of a wall using a
pulley system. If the block of bricks has a mass of 24 kg, what is its weight?
It is raised 3.0 m. Calculate its increase in gravitational potential energy
when it reaches the top of the wall.
3)
Travelling in a mountainous area, a bus of mass 3 tons reaches the edge of
a steep valley. There is a 1 km vertical drop to reach the valley below, but 20
km of road to get there. What gravitational potential energy? Will the bus
lose in making its descent to the valley bottom?
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4)
Assuming the bus in question 3 does not change its cruising speed on its
way down, where does the gravitational potential energy go? Why is there a
risk of brake failure in this situation?
Reflection
1. I learned that ________________________________________________
____________________________________________________________
_______________________________________________________
2. I enjoyed most on _____________________________________________
____________________________________________________________
__________________________________________________
3. I want to learn more on _________________________________________
____________________________________________________________
__________________________________________________
127
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References
https://www.s-cool.co.uk/gcse/physics/energy-calculations/revise-it/gravitationalpotential-energy
https://sharemylesson.com/teaching-resource/gravitational-potential-energy152290
https://sharemylesson.com/teaching-resource/gravitational-potential-energycalculations-187633
128
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Answer Key
Activity 1- Explore and Discover
Ball A
GPE = mgh
= 3kg x 10ms-2 x 4m
= 120J
Ball B
GPE = mgh
= 3kg x 10ms-2 x 2m
= 60J
Ball C
GPE = mgh
= 2kg x 10ms-2 x 4m
= 80J
• Ball A and B have the same weight but have different height above
the ground level so Ball A has greater GPE than ball B.
• Ball A and C have the same height but have different weight. Ball
A has more weight than C so it has grater GPE than ball C.
• Ball C weighs less than ball B but it has greater GPE than ball B
because it is higher than B.
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Therefore, we see that gravitational potential energy depends on the
weight and height of the object.
Mass (kg)
Weight (N)
Height (m)
5
2
8
20
5000
0.2
67
50
20
80
600
500 000
2
670
2
6
5
0.6
2
10
44
Gravitational Potential
energy (J)
100
120
400
360
100 000
20
29 80
Answer Key Activity 2 – Keep Moving
2.Acceleration due to gravity = 10 N/kg
3.Acceleration due to gravity = 4 N/kg
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Answer Key Activity 3 – Apply Your Skills
1. GPE = mgh = 60 x 9.8 x 2 = 1176 = 1200 J
2. Weight = mg = 24 x 9.8 = 235.2 = 240 N
Change in GPE = mgh = 24 x 9.8 x 3.0 = 705.6 = 710 J
3. Change in GPE = mgh = 3000 x 9.8 x 1000 = 2.94 x 107 = 2.9 x 107 J (29
MJ)
4. Since the bus gains no kinetic energy (its speed stays the same) it must be
using its brakes, and all the GPE lost by the bus in converted to heat in the
brakes. There is a risk of brake failure if the brakes overheat.
Mass (kg)
5
2
8
20
5000
0.2
67
Weight (N)
Height (m)
20
8
32
80
20 000
0.8
268
Prepared by:
2
6
5
0.6
2
10
44
Gravitational Potential
energy (J)
40
48
160
48
40 000
8
11 792
LEONOR C. NATIVIDAD
Baggao National High School
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GENERAL PHYSICS 1
Name: ____________________________
Grade Level: _________
Date: _____________________________
Score: ______________
LEARNING ACTIVITY SHEET
Planetary and Satellite Motion
Background Information for the Learners (BIL)
A satellite is any object that is orbiting the earth, sun or other massive body. It
maybe natural like the moon or man-made like those launched in space for the
specific purposes like communication, researches, weather forecasts, etc. This
module will discuss the underlying principles and mathematical equations in the
motion of planets and satellites.
Satellite Motion
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Satellites follow a projectile motion where it is acted upon by the force of
gravity. It was Newton who first theorized that if an object is launched with sufficient
speed, it would orbit the Earth.
Let’s take a look on Figure 1. Consider
a
cannonball fired from the top of a mountain. It
will
follow a trajectory similar to a projectile motion.
As
the projectile moves horizontally in a direction
tangent to the earth, the force of gravity would
pull
it downward. Paths A and B illustrate the path
of a
projectile with insufficient launch speed for
orbital motion. But if launched with sufficient
speed, the projectile would fall towards the
earth at the same rate that the earth curves.
This would cause the projectile to stay the
same height above the earth and to orbit in a
circular path (such as path C). And at even
greater launch speed, a cannonball would
Figure 1. Satellite Motion
once more orbit the earth, but now in an
https://cdn1.byjus.com/wpelliptical path (as in path D). At every point
content/uploads/2018/11/physics/2015/12/2
along its trajectory, a satellite is falling toward
the
earth. Yet because the earth curves, it never
0072839/32.png
reaches the earth.
Therefore, what should be the launch speed so that a projectile will orbit the
Earth? The answer lies on the curvature of the Earth. For every 8000 meters
measured along the horizon of the earth,
the earth's surface curves downward by
approximately 5 meters. For a projectile to
orbit the earth, it must travel horizontally a
distance of 8000 meters for every 5 meters
of vertical fall. For this reason, a projectile
launched horizontally with a speed of
http://www.physicsclassroom.com/Class/circle
about 8000 m/s will be capable of orbiting
the earth in a circular path. If shot with a
s/u6l4b2.gif
speed greater than 8000 m/s, it would orbit
the earth in an elliptical path.
The motion of objects is governed by Newton's laws. The same simple laws
that govern the motion of objects on earth also extend to the heavens--to govern the
motion of planets, moons, and other satellites.
Orbital Speed Equation
Consider a satellite with mass Msat orbiting a central body with a mass of mass
MCentral. The central body could be a planet, the sun or some other large mass
capable of causing sufficient acceleration on a less massive nearby object. If the
satellite moves in circular motion, then the net centripetal force (Fc), acting upon this
orbiting satellite is given by the relationship
𝑀
× 𝑣2
𝐹𝑐 = 𝑠𝑎𝑡𝑟
equation (1)
This net centripetal force is the result of the gravitational force (Fg) that attracts the
satellite towards the central body and can be represented as
𝐺×𝑀𝑠𝑎𝑡 ×𝑀𝑐𝑒𝑛𝑡𝑟𝑎𝑙
𝐹𝑔 =
equation (2)
𝑟2
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Since Fc = Fg, then we have
𝑀𝑠𝑎𝑡 × 𝑣 2
𝐺 × 𝑀𝑠𝑎𝑡 × 𝑀𝑐𝑒𝑛𝑡𝑟𝑎𝑙
=
𝑟
𝑟2
We now have the equation for orbital speed,
𝑣=√
where:
𝐺×𝑀𝑐𝑒𝑛𝑡𝑟𝑎𝑙
equation (3)
𝑟
G = 6.673 x 10-11 N•m2/kg2
Mcentral = the mass of the central body where the satellite orbits
r = the radius of orbit for the satellite
The Acceleration Equation
The equation for the acceleration of gravity is given as
𝐺 × 𝑀𝑐𝑒𝑛𝑡𝑟𝑎𝑙
𝑔=
𝑟2
Therefore, the acceleration of a satellite in a circular motion is given by the equation,
𝑎=
where:
𝐺×𝑀𝑐𝑒𝑛𝑡𝑟𝑎𝑙
𝑟2
equation (4)
G = 6.673 x 10 N•m /kg
r = the average radius of orbit of the satellite
-11
2
2
Orbital Period Equation
Note that the speed (v) is the ratio of the distance (2𝜋𝑟) travelled in one
revolution and the period (T),
𝑣=
2𝜋𝑟
𝑇
or we also have
𝑇2
𝑟3
=
4𝜋2
𝐺×𝑀𝑐𝑒𝑛𝑡𝑟𝑎𝑙
equation(5)
Then period of a satellite (T) and the mean distance from the central body (r)
are related by the following equation:
𝑇=
where:
3
2𝜋𝑟 ⁄2
√𝐺𝑀𝑐𝑒𝑛𝑡𝑟𝑎𝑙
equation (6)
T = period of the satellite
r = average radius of orbit for the satellite
G = 6.673 x 10-11 N•m2/kg2
A geosynchronous satellite is a satellite that orbits the earth with an orbital
period of 24 hours, thus matching the period of the earth's rotational motion. A special
class of geosynchronous satellites is a geostationary satellite. A geostationary
satellite orbits the earth in 24 hours along an orbital path that is parallel to an
imaginary plane drawn through the Earth's equator. Such a satellite appears
permanently fixed above the same location on the Earth.
Sample Problem #1
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A satellite wishes to orbit the earth at a height of 100 km (approximately 60
miles) above the surface of the earth. Determine the speed, acceleration and orbital
period of the satellite. (Given: Mearth = 5.98 x 1024 kg, rearth = 6.37 x 106 m)
Given:
Mearth = 5.98 x 1024 kg
G
= 6.673 x 10-11 N•m2/kg2
height = 100 km= 100,000 m
r
= rearth + height = 6.37 x 106 m + 100,000 m = 6.47 x 106 m
Unknown:
v, a, T
Solution:
For the speed, we use equation (3)
𝐺 × 𝑀𝑐𝑒𝑛𝑡𝑟𝑎𝑙
6.673 x 10−11 N • m2 /kg 2 × (5.98 x 1024 kg)
𝑣=√
=√
= 7.85 𝑥 103 𝑚⁄𝑠
𝑟
6.47 x 106 m
For the acceleration, we use equation (4)
𝐺 × 𝑀𝑐𝑒𝑛𝑡𝑟𝑎𝑙 6.673 x 10−11 N • m2 /kg2 × (5.98 x 1024 kg)
𝑎=
=
= 9.53 𝑚⁄ 2
2
𝑠
6
𝑟2
(6.47 x 10 m)
For the period, we use equation (6)
𝑇=
2𝜋𝑟
3⁄
2
3⁄
2
√𝐺𝑀𝑐𝑒𝑛𝑡𝑟𝑎𝑙
2𝜋 × (6.47 x 106 m)
=
√6.673 x 10−11 N •
m2
kg2
= 5176 𝑠 = 1.44 ℎ𝑟𝑠
× (5.98 x 1024 kg)
Sample Problem #2
The period of the moon is approximately 27.2 days (2.35 x 10 6 s). Determine
the radius of the moon's orbit and the orbital speed of the moon. (Given: Mearth = 5.98
x 1024 kg, rearth = 6.37 x 106 m)
=27.2 days = 2.35 x 106 s
= 6.37 x 106 m
= 5.98 x 1024 kg
= 6.673 x 10-11 N•m2/kg2
Given:
T
rearth
Mearth
G
Unknown:
r and v
Solution: For the radius of moon’s orbital, rearranging equation (5)
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3
𝑟=√
𝑇2
3
× 𝐺 × 𝑀𝑐𝑒𝑛𝑡𝑟𝑎𝑙 √
=
4𝜋 2
(2.35𝑥106 𝑠)2 × (6.673 x 10−11 N •
m2
) × (5.98 x 1024 kg)
kg 2
4𝜋 2
𝑟 = 3.82 𝑥 108 𝑚
For the orbital speed, we use equation (3)
𝐺 × 𝑀𝑐𝑒𝑛𝑡𝑟𝑎𝑙
6.673 x 10−11 N • m2 /kg 2 × (5.98 x 1024 kg)
𝑣=√
=√
= 1.02 𝑥 103 𝑚⁄𝑠
𝑟
3.82 x 108 m
Kepler’s Laws of Planetary Motion
In the early 1600s, Johannes Kepler proposed three laws of planetary motion.
Kepler was able to summarize the carefully collected data of his mentor, Tycho
Brahe, with three statements that described the motion of planets in a sun-centered
solar system.
Kepler's three laws of planetary motion are as follows:
• The Law of Ellipses. The path of the planets about the sun is elliptical in
shape, with the center of the sun being located at one focus.
• The Law of Equal Areas. An imaginary line drawn from the center of the sun
to the center of the planet will sweep out equal areas in equal intervals of time.
• The Law of Harmonies. The ratio of the squares of the periods of any two
planets is equal to the ratio of the cubes of their average distances from the
sun.
The Law of Ellipses
In his first law of planetary motion, the law of ellipses, Kepler
described that the path of the planets orbiting the sun
follows an elliptical shape or an ellipse. An ellipse is a
special curve in which the sum of the distances from
every point on the curve to two other points is a
constant. The two other points (represented here by the
Sun and point B) are known as the foci of the ellipse.
Kepler's first law is rather simple - all planets orbit the
sun in a path that resembles an ellipse, with the sun being
located at one of the foci of that ellipse.
Sun
B
Planet
The Law of Equal Areas
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Kepler’s second law describes the speed at
which any given planet will move while orbiting the
sun. The speed at which any planet moves
through space is constantly changing. This is due
to the planets’ elliptical orbit and the fact that the
sun is not in the center of the orbital path. If a line
were drawn from the center of the planet to the
center of the sun, that line would sweep out the
same area in equal periods of time.
At equal periods of time, a planet will sweep
out the same area as it orbits around the sun. A
planet moves fastest when it is closest to the sun
and slowest when it is farthest from the sun.
http://www.physicsclassroom.com/Class/circles/u
6l4a2.gif
The Law of Harmonies
Kepler’s third law compares the orbital period and radius of orbit of a planet to
those of other planets. The comparison being made is that the ratio of the squares
of the periods to the cubes of their average distances from the sun is the same for
every planet.
Let us consider the orbital period and average distance from sun (orbital
radius) for Earth and Mars as given in the table.
Planet
Period
(s)
Average
Distance (m)
T2/R3
(s2/m3)
Earth
3.156 x 107 s
1.4957 x 1011
2.977 x 10-19
Mars
5.93 x 107 s
2.278 x 1011
2.975 x 10-19
Observe that the T2/r3 ration of Earth and Mars are the same.
𝑇2
𝑟3
2
= 𝑘 = 2.977 𝑥 10−19 𝑚 ⁄ 3
𝑠
where:
equation (7)
T = orbital period
r = orbital radius
Let’s take a look on the T2/r3 ratio of the planets on the solar system. Take note that
they have almost the same T2/r3 ratio.
Mercury
Period
(years)
0.241
Average distance
(au)
0.39
T2/r3 ratio
(yr2/au3)
0.98
Venus
.615
0.72
1.01
Planet
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Earth
1.00
1.00
1.00
Mars
1.88
1.52
1.01
Jupiter
11.8
5.20
0.99
Saturn
29.5
9.54
1.00
Uranus
84.0
19.18
1.00
Neptune
165
30.06
1.00
Pluto
248
39.44
1.00
* 1 astronomical unit (au) = 1.4957 x 1011 meters = distance of Earth from Sun
** 1 year is the time of Earth to orbit the Sun = 3.156 x 107 seconds
Therefore,
𝑇1 2
𝑟1 3
(𝑇 ) = (𝑟 )
2
equation (8)
2
This is the only one of Kepler's three laws that deals with more than one planet
at a time. It has been calculated that this ratio holds for all the planets in our solar
system, in addition to moons and other satellites. It was this law that inspired Newton,
who came up with three laws of his own to explain why the planets move as they do.
Sample problem #1
The average orbital distance of Mars is 1.52 times the average orbital distance of the
Earth. Knowing that the Earth orbits the sun in approximately 365 days, use Kepler's
law of harmonies to predict the time for Mars to orbit the sun.
Given: 𝑟𝑀𝑎𝑟𝑠 = 1.52 ∙ 𝑟𝐸𝑎𝑟𝑡ℎ
;
𝑇𝐸𝑎𝑟𝑡ℎ = 365 𝑑𝑎𝑦𝑠
Solution: Using Kepler’s 3rd Law equation (8),
2
𝑇
3
𝑟
( 𝑇𝐸𝑎𝑟𝑡ℎ ) = ( 𝑟𝐸𝑎𝑟𝑡ℎ )
𝑀𝑎𝑟𝑠
𝑇𝑀𝑎𝑟𝑠
𝑀𝑎𝑟𝑠
(𝑇𝐸𝑎𝑟𝑡ℎ) 2 𝑥 (𝑟𝑀𝑎𝑟𝑠) 3
(365 𝑑𝑎𝑦𝑠) 2 𝑥 (1.52𝑟𝐸𝑎𝑟𝑡ℎ )3
√
=√
=
= 684 𝑑𝑎𝑦𝑠
(𝑟𝐸𝑎𝑟𝑡ℎ )3
𝑟𝐸𝑎𝑟𝑡ℎ 3
Learning Competency:
138
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Calculate quantities related to planetary or satellite motion. (STEM_GP12Red-IIb20)
ACTIVITY 1: PROBLEM SOLVING
Directions: Solve for the following problems. Write your answers with complete
solutions on a separate sheet of paper.
1. One of Saturn's moons is named Mimas. The mean orbital distance of Mimas is
1.87 x 108 m. The mean orbital period of Mimas is approximately 23 hours
(8.28x104 s). Use this information to estimate the mass for the planet Saturn. What
is the acceleration of Mimas?
2. Consider a satellite which is in a low orbit about the Earth at an altitude of 220 km
above Earth's surface. Determine the orbital speed of this satellite. Use the
information given below.
G = 6.673 x 10-11 Nm2/kg2
Mearth = 5.98 x 1024 kg
rearth = 6.37 x 106 m
3. Use the information below and the relationship above to calculate the T 2/r3 ratio
for the planets about the Sun, the moon about the Earth, and the moons of Saturn
about the planet Saturn. The value of G is 6.673 x10-11 N•m2/kg2.
Sun
M = 2.0 x 1030 kg
Earth
M = 6.0 x 1024 kg
Saturn
M = 5.7 x 1026 kg
a. T2/r3 for planets about sun
b. T2/r3 for the moon about Earth
c. T2/r3 for moons about Saturn
4. A geosynchronous orbit is an orbit in which the satellite remains over the same
spot on the planet as the planet turns. This is accomplished by matching the velocity
of the satellite to the velocity of the turning planet. The orbital radius of a
geosynchronous satellite is 4.23 × 107 m (measured from the center of Earth). What
is its period?
5. Galileo is often credited with the early discovery of four of Jupiter's many moons.
The moons orbiting Jupiter follow the same laws of motion as the planets orbiting the
sun. One of the moons is called Io - its distance from Jupiter's center is 4.2 units and
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it orbits Jupiter in 1.8 Earth-days. Another moon is called Ganymede; it is 10.7 units
from Jupiter's center. Make a prediction of the period of Ganymede using Kepler's
law of harmonies.
6. It was once postulated that there exists a planet called Vulcan between Mercury
and Sun, whose presence would explain the anomalous precession of Mercury (this
was later debunked and Mercury’s precession was later explained by Einstein’s
theory of general relativity). Assuming that Vulcan has a circular orbit around the sun
with a radius equal to 2/3 of the average orbital of Mercury, what would be the orbital
period of Vulcan? The orbital radius of Mercury is 5.79 x 1010 m.
7. Orbital radius and orbital period data for the four biggest moons of Jupiter are listed
in the table below. The mass of the planet Jupiter is 1.9 x 1027 kg.
𝑇 2⁄
Jupiter's Moon
Period (s)
Radius (m)
𝑟3
Io
1.53 x 105
4.2 x 108
a. _______
Europa
3.07 x 105
6.7 x 108
b. _______
Ganymede
6.18 x 105
1.1 x 109
c. _______
Callisto
1.44 x 106
1.9 x 109
d. _______
What pattern do you observe in the last column of data? Which law of Kepler's does
this seem to support?
ACTIVITY 2. WORD SEARCH PUZZLE
Directions: Search and underline the word/s being described in the following
statements below. The answers maybe found horizontally, vertically, or diagonally.
1. Any object that is orbiting the earth, sun or other massive body.
2. A satellite that orbits the earth in 24 hours along an orbital path that is parallel to
an imaginary plane drawn through the Earth's equator.
3. A force exerted by any object moving in a circle (or along a circular path).
4. It is a force that attracts any two objects with mass.
5. It is a form of motion experienced by an object or particle that is projected near the
Earth's surface and moves along a curved path under the action of gravity only.
6. The path of the planets about the sun is elliptical in shape, with the center of the
sun being located at one focus.
7. The ratio of the squares of the periods of any two planets is equal to the ratio of
the cubes of their average distances from the sun.
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NOTE: Practice personal hygiene protocols at all times
8. It is the speed at which satellites/planets orbits around the center of a system.
9. The time it takes for a body to orbit a system.
10. An imaginary line drawn from the center of the sun to the center of the planet will
sweep out equal areas in equal intervals of time.
ACTIVITY 3. TEST YOURSELF
Directions: Encircle the letter that you think best answers the question.
1. Kepler’s first law of planetary motion says that the paths of the planets are
a. Parabolas
b. Hyperbolas
c. Ellipses
d. Circles
2. A planet orbiting the sun
a. Maintains the same distance from the sun
b. Moves at a constant speed
c. Moves faster when it is farther from the sun
d. Moves faster when it is closer to the sun
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3. Newton’s law of universal gravitation states that the gravitational force
between two bodies is
a. Directly proportional to the distance between them
b. Directly proportional to the square of the distance between them
c. Inversely proportional to the distance between them
d. Inversely proportional to the square of the distance between them
4. Two moons orbit a planet. The average orbital radius of the outer moon is 1.8
times that of the inner moon. The orbital period of the outer moon is
a. 0.56 times that of the inner moon
b. 1.8 times that of the inner moon
c. 2.4 times that of the inner moon
d. 5.8 times that of the inner moon
5. All of the following were proposed by Kepler on planetary motion except
a. Law of Harmonies
b. Law of Inertia
c. Law of Equal Areas
d. Law of Ellipses
6. The mass of Earth is approximately 6.0 x 1024 kg. What would the acceleration
due to gravity be on the surface of planet Q which has a mass of 5.0 x 1024 kg
and a radius equal to the radius of Earth?
a. 6.8 m/s2
b. 8.2 m/s2
c. 9.8 m/s2
d. 12 m/s2
7. Two satellites orbit the Earth at the same altitude in circular orbits. One satellite
has a mass of 250 kg, and the other has a mass of 150 kg. The orbital speed
of the larger satellite is
a. The same as the speed of the smaller satellite
b. 1.7 times the speed of the smaller satellite
c. 3.0 times the speed of the smaller satellite
d. 9.0 times the speed of the smaller satellite
8. Suppose that in the distant future, astronauts are exploring a planet in another
solar system. They find that the radius of the planet is 6.8 x 103 km and the
acceleration due to gravity on its surface is 15.3 m/s2. What is the mass of the
planet?
a. 1.1 x 1019 kg
b. 6.0 x 1024 kg
c. 1.1 x 1025 kg
d. 6.0 x 1026 kg
9. A satellite orbits Earth 1,250,000 m above the Earth’s surface. What is the
satellite’s orbital speed?
a. 6440 m/s
142
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b. 7240 m/s
c. 7920 m/s
d. 1110 m/s
10. A satellite orbits a planet in a circular orbit. If the orbital radius is 7.8 x 108 m
and the time required for a complete revolution is 3.5 x 106 s, what is the orbital
speed?
a. 140 m/s
b. 1400 m/s
c. 4500 m/s
d. 7000 m/s
For items 11-15, refer to this equation to answer the following questions:
𝑣=√
𝐺 × 𝑀𝐸𝑎𝑟𝑡ℎ
𝑟
11. If the mass of the satellite is increased, then the orbital speed would
_________________ (increase, decrease, be the same).
12. If mass of the earth is increased, then the orbital speed would
_______________ (increase, decrease, be the same).
13. If the radius of orbit of a satellite is increased, then the orbital speed would
________________ (increase, decrease, be the same).
14. -15. If the radius of orbit of a satellite is increased by a factor of 2 (i.e., doubled),
then the orbital speed would _________________ (increase, decrease) by a
factor of _______________.
Reflection:
1. I learned that ______________________________________________________
__________________________________________________________________
__________________________________________________________
2. I enjoyed most on __________________________________________________
__________________________________________________________________
__________________________________________________________
3. I want to learn more on ______________________________________________
__________________________________________________________________
__________________________________________________________
References
143
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Diwa Senior High School Series: General Physics 1. Diwa Publishing
Circular Motion and Satellite Motion. The Physics Classroom. Retrieved May 20,
2020 from https://www.physicsclassroom.com/class/circles/Lesson-4
Keplers Laws of Planetary Motion. Ck-12 Organization. Retrieved May 20, 2020
from https://www.ck12.org/book/cbse_physics_book_class_xi/section/7.9/
ANSWER KEY
144
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Activity 1: Test yourself
1.
Given: T=8.28x104 s, r=1.87 x 108 m, G = 6.673 x 10-11 Nm2/kg2
Unknown: Mcentral or MSaturn, and a
Solution: For the mass of Saturn
𝑇2
𝑟3
4𝜋2
=
, rearranging
𝐺×𝑀𝑐𝑒𝑛𝑡𝑟𝑎𝑙
4𝜋 2 × 𝑟 3
𝑀𝑆𝑎𝑡𝑢𝑟𝑛 =
=
𝐺 × 𝑇2
4𝜋 2 × (1.87 𝑥 108 𝑚)3
= 5.64 𝑥 1026 𝑘𝑔
𝑚2
−11
4
2
6.673 x 10
N ∙ 2 × (8.28 𝑥 10 𝑠)
𝑘𝑔
For the acceleration of Mimas,
𝑎=
2.
𝐺 × 𝑀𝑐𝑒𝑛𝑡𝑟𝑎𝑙
=
𝑟2
(6.673 x 10−11 N ∙
𝑚2
26
2 ) × (5.64 𝑥 10 𝑘𝑔)
𝑘𝑔
2
(1.87 𝑥 108 𝑚)
= 1.08
𝑚
𝑠2
G = 6.673 x 10-11 Nm2/kg2
Mearth = 5.98 x 1024 kg
rearth = 6.37 x 106 m
r= rearth + height= 6.37 x 106 m + 220,000 m = 6.59 x 106 m
v and T
Given:
Unknown:
Solution:
𝑚2
For v, 𝑣 = √
𝐺×𝑀𝑐𝑒𝑛𝑡𝑟𝑎𝑙
𝑟
=√
6.673 x 10−11 N∙ 2 ×(5.98 𝑥 1024 𝑘𝑔)
𝑘𝑔
6.59 𝑥 106 𝑚
= 7,782 𝑚⁄𝑠
3. a. T2/r3 for planets about sun
𝑇2
4𝜋 2
=
=
𝑟 3 𝐺 × 𝑀𝑐𝑒𝑛𝑡𝑟𝑎𝑙
4𝜋 2
6.673 x 10−11 N ∙
𝑚2
𝑘𝑔2
= 2.96 𝑥 10−19
𝑠2
𝑚3
= 9.86 𝑥 10−14
𝑠2
𝑚3
= 1.04 𝑥 10−15
𝑠2
𝑚3
× (2.0 𝑥 1030 𝑘𝑔)
b. T2/r3 for the moon about Earth
𝑇2
4𝜋 2
=
=
𝑟 3 𝐺 × 𝑀𝑐𝑒𝑛𝑡𝑟𝑎𝑙
4𝜋 2
6.673 x 10−11 N ∙
𝑚2
× (6.0 𝑥 1024 𝑘𝑔)
𝑘𝑔2
c. T2/r3 for moons about Saturn
𝑇2
4𝜋 2
=
=
𝑟 3 𝐺 × 𝑀𝑐𝑒𝑛𝑡𝑟𝑎𝑙
4.
Given:
Unknown:
Solution:
4𝜋 2
6.673 x 10−11 N ∙
𝑚2
× (5.7 𝑥 1026 𝑘𝑔)
𝑘𝑔2
r=4.23 × 107 m
G = 6.673 x 10-11 Nm2/kg2,
Mearth = 5.98 x 1024 kg
T
4𝜋 2 𝑟 3
𝑇=√
=
𝐺 × 𝑀𝑐𝑒𝑛𝑡𝑟𝑎𝑙 √
4𝜋 2 (4.23 𝑥107 )3
6.673 x 10−11 N ∙
𝑚2
× (5.98 𝑥 1024 𝑘𝑔)
𝑘𝑔2
= 86,833 𝑠 = 24.04 ℎ𝑟𝑠
145
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5.
Given:
Unknown:
rIo= 4.2 units
TIo= 1.8 Earth-days
rGanymede= 10.7 units
TGanymede=?
Solution: From law of harmonies, equation (8)
2
3
𝑇𝐼𝑜
𝑟𝐼𝑜
(
) =(
)
𝑇𝐺𝑎𝑛𝑦𝑚𝑒𝑑𝑒
𝑟𝐺𝑎𝑛𝑦𝑚𝑒𝑑𝑒
𝑇𝐺𝑎𝑛𝑦𝑚𝑒𝑑𝑒
6.
(𝑇𝐼𝑜) 2 𝑥 (𝑟𝐺𝑎𝑛𝑦𝑚𝑒𝑑𝑒) 3
(1.8 𝑑𝑎𝑦𝑠) 2 𝑥 (10.7)3
√
√
=
=
= 7.32 𝐸𝑎𝑟𝑡ℎ − 𝑑𝑎𝑦𝑠
(𝑟𝐼𝑜 )3
4.23
rMercury=5.79 x 1010 m
rVulcan=2/3 x (5.79 x 1010 m)= 3.86 x 1010 m
Unknown: TVulcan=?
Solution: Using equation (7)
𝑇2
2
= 𝑘 = 2.977 𝑥 10−19 𝑚 ⁄𝑠 3
3
𝑟
√2.977
𝑇𝑉𝑢𝑙𝑐𝑎𝑛 =
𝑥 10−19 × (𝑟𝑉𝑢𝑙𝑐𝑎𝑛 )3 = √2.977 𝑥 10−19 × (3.86𝑥 1010 )3
= 4.14 𝑥 106 𝑠 𝑜𝑟 47.89 𝑑𝑎𝑦𝑠 ≅ 48 𝑑𝑎𝑦𝑠
Alternately, using equation (8)
TMercury = 87.965 days
𝑇𝑀𝑒𝑟𝑐𝑢𝑟𝑦 2
𝑟𝑀𝑒𝑟𝑐𝑢𝑟𝑦 3
(
) =(
)
𝑇𝑉𝑢𝑙𝑐𝑎𝑛
𝑟𝑉𝑢𝑙𝑐𝑎𝑛
Given:
𝑇𝑉𝑢𝑙𝑐𝑎𝑛
(𝑇𝑀𝑒𝑟𝑐𝑢𝑟𝑦) 2 𝑥 (𝑟𝑉𝑢𝑙𝑐𝑎𝑛) 3
(87.965𝑑𝑎𝑦𝑠) 2 𝑥 (3.86 𝑥 1010 )3
√
=√
=
(𝑟𝑀𝑒𝑟𝑐𝑢𝑟𝑦 )3
(5.79 𝑥 1010 )3
= 47.88 𝑑𝑎𝑦𝑠 ≅ 48 𝑑𝑎𝑦𝑠
7. a. a. 3.16 x 10-16 , b.3.13 x 10-16 , c.2.89 x 10-16 , d.3.02 x 10-16
The value of T2/r3 is almost the same for all the moons of Jupiter. This is in
accordance with Kepler’s law of harmonies.
Activity 2. Word Search Puzzle
1. satellite
6. law of ellipses
2. geostationary
7. law of harmonies
3. centripetal
8. orbital speed
4. gravitational
9.period
5. projectile
10. law of equal areas
146
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Activity 3: Test Yourself.
1. C
4. C
7. A
10. B
13. decrease
2. D
5. B
8. C
11. the same
14-15.
3. D
6. B
9. B
12. increase
Decrease; √2
Prepared by:
Rosemarie C. Fernandez
Itawes National High School
147
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GENERAL PHYSICS 1
Name: _______________________________ Grade Level: _____________
Date: ________________________________ Score: __________________
LEARNING ACTIVITY SHEET
KEPLER’S LAWS OF PLANETARY MOTION
AND NEWTON’S LAW OF UNIVERSAL GRAVITATION
Background Information for the Learners (BIL)
When you look at the sky at night, the stars would appear like they are fixed
in their patterns. Their rotation through the sky over the seasons seem to be
unchanging that most cultures have used the presence of one or another
constellation to tell time. However, the planets seem to have distinct motion
compared to the stars. They move slowly and seemingly unpredictably across the
sky. Efforts to look for possible explanation on why planets move in such a way
resulted to modern science’s understanding of gravity and motion.
Our recent understanding of planetary motion has a rich history.
Contradicting the thousand-year old idea of Aristotle of a stationary Earth at
the center of a revolving universe, Copernicus proposed the idea that the Earth was
a planet (like Venus or Saturn) and that all planets rotate and revolve around the
Sun. Despite criticisms proofs of a heliocentric solar system gradually intensified.
A Danish Astronomer Tycho Brahe made astronomical observations with his
naked eyes. Brahe was able to record accurate measurements of the motion of the
planets around the Sun. His astronomical observations were later handed down to
his assistant, Johannes Kepler.
Kepler analyzed and studied Brahe’s observations and measurements which
laid the foundation of his three laws of planetary motion. Meanwhile, Isaac Newton
discovered a physical law that governs the attraction between bodies in the universe.
Using this idea, Newton formulated his law of universal gravitation.
Both Kepler’s laws of planetary motion and Newton’s law of universal
gravitation will help us understand how heavenly bodies go about in motion.
148
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KEPLER’S LAWS OF PLANETARY MOTION
Kepler’s first law of planetary motion is called law of ellipses. It states the orbit
of a planet around the sun is an ellipse, having the sun as one of the foci. The sun
therefore is not the center of the ellipse but is instead one focus. Planets follow the
ellipse making the distance between the Earth and the Sun constantly changing.
Image retrieved from https://www.google.com/search?q=law+of+ellipses&tbm=isch&hl=en-
US&chips=q:law+of+ellipses,g_1:first+law:a5VUM9e6Lzw%3D&authuser=1&sa=X&ved=2
ahUKEwj3k5Hn1zsAhV0zIsBHUC7BakQ4lYoAXoECAEQFw&biw=532&bih=600#imgrc=YSIrO8Wu0rpcDM
The second law is called the law of equal areas. It states that a planet moves
around the sun in such a way that a line drawn from the sun to the planets sweeps
equal areas in equal periods of time. The planet moves faster when it is nearer the
sun. Thus, the planet moves fastest at the perihelion (shortest distance) and slowest
at the aphelion (farthest distance). This law is a consequence of the conservation of
angular momentum.
149
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Image retrieved from
https://www.google.com/search?q=law+of+equal+areas&tbm=isch&ved=2ahUKEwj4xH21-zsAhWLuJQKHaKFBakQ2cCegQIABAA&oq=law+of+equal+areas&gs_lcp=CgNpbWcQAzICCAAyAggAMgQIABBDM
gYIABAFEB4yBAgAEBgyBAgAEBgyBAgAEBg6BQgAELEDUKgjWPQ5YMtCaABwAHgAg
AGfAYgBpAuSAQQwLjExmAEAoAEBqgELZ3dzLXdpei1pbWfAAQE&sclient=img&ei=GJyk
X_jbLIvx0gSii5bICg&authuser=1&bih=600&biw=532&hl=en-US#imgrc=tQ0d0r3GKrQ-rM
The third law is called the harmonic law or the law of periods. It states that the
ratio of the squares of the periods P (or T in other references) of any two planets
revolving around the sun is equal to the ratio of the cubes of their mean distance R
(or d in other references) from the sun. Period is the time for a planet to travel one
revolution around the sun.
Kepler's Third Law implies that the period for a planet to orbit the Sun
increases rapidly with the radius of its orbit. Thus, we find that Mercury, the innermost
planet, takes only 88 days to orbit the Sun but the outermost planet (Pluto) requires
248 years to do the same. (Note that the subscripts “1” and 2” distinguish quantities
for planet 1 and 2 respectively.
Image retrieved from
https://www.google.com/search?q=the+ratio+of+the+squares+of+the+periods+P+(or+T+in+other+re
ferences)+of+any+two+planets+revolving+around+the+sun+is+equal+to+the+ratio+of+the+cubes+o
f+their+mean+distance+R+(or+d+in+other+references)+from+the+sun&tbm=isch&ved=2ahUKEwjm
gKjO3OzsAhUHe5QKHYClBBAQ2-
150
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cCegQIABAA&oq=the+ratio+of+the+squares+of+the+periods+P+(or+T+in+other+references)+of+an
y+two+planets+revolving+around+the+sun+is+equal+to+the+ratio+of+the+cubes+of+their+mean+di
stance+R+(or+d+in+other+references)+from+the+sun&gs_lcp=CgNpbWcQA1DPrgNYz64DYLG3A2
gAcAB4AIABAIgBAJIBAJgBAaABAaoBC2d3cy13aXotaW1nwAEB&sclient=img&ei=AqGkXb3BIf20QSAy5KAAQ&bih=657&biw=1366&hl=en-US#imgrc=ketjHMHpOi4IMM
Concept Check:
What are the 3 Kepler’s laws of reflection?
_____________________________________
_____________________________________
_____________________________________
Sample Problem 1:
The mean solar distance of Mercury is 0.387 AU.
What is its period?
Solution:
a. Let subscripts 1 and 2 refer to Mercury
and Earth, respectively.
R1 = 0.387 AU
R2 = 1 AU
P2 = 1 y
P2 = ?
b. We will use the equation
𝑷𝟐𝟏
𝑷𝟐𝟐
=
𝑹𝟑𝟏
𝑹𝟑𝟐
to solve
for the period of Mercury (P2).
Did You Know….
The period of planets is compared to
that of the period of the Earth. A unit
of measurement for this period is
called Earth year or simply year. A
unit distance from the sun is referred
to as astronomical unit (AU). Hence,
the average distance between the
Sun and the Earth is one AU.
c. Substitute the values of the given quantities and solve:
𝑷𝟐𝟏
𝑷𝟐𝟐
=
𝑹𝟑𝟏
𝑷𝟐𝟏
𝑹𝟐
𝟏𝒚
𝟑 →
=
(𝟎.𝟑𝟖𝟕 𝑨𝑼)𝟑
(𝟏 𝑨𝑼)𝟑
→ 0.27 y ≈ 88 days
d. Therefore, it takes approximately 88 days for Mercury to be able to orbit
around the sun once.
151
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𝒎𝟏 𝒎𝟐
In equation, 𝑭𝑮 = 𝑮 𝟐
𝒓
where G is the universal gravitational constant
𝒎𝟐
equal to 𝟔. 𝟔𝟕𝟒 𝒙 𝟏𝟎−𝟏𝟏 𝑵. 𝒌𝒈𝟐 .
Concept Check:
What are the 3 Kepler’s laws of reflection?
_____________________________________
_____________________________________
_____________________________________
Sample Problem 2:
Suppose two planets A and B revolve around the same star in circular orbits.
The distance of A from the star is twice that of B. The mass of B is three times the
mass of A. Find the ratio of the gravitational force exerted by the star on the two
planets.
Solution
a. Let M be the mass of the star. Let rA and rB be the distance of planets A and B
from the star, respectively. Since we are given that the distance of A from the
star is twice that of B, then rA = 2rB. We also know that the mass of B is three
times the mass of A, then mB = 3mA.
b. The force exerted by the star on the two planets. A and B are FA and FB.
𝒎 𝒎
c. We will use equation 𝑭𝑮 = 𝑮 𝒓𝟏𝟐 𝟐 to solve the problem.
d. Manipulating the equation,
𝑭𝑨 = 𝑮
𝑭𝑩 =
𝒎𝑨 𝑴
𝒓𝑨 𝟐
𝑮𝒎𝒃 𝑴
𝒓𝑩 𝟐
→ (a)
=
𝑮(𝟑𝒎𝑨 )𝑴
𝒓
( 𝑨 )𝟐
→ (b)
𝟐
152
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𝑭
e. We divide (a) by (b) to get the ratio 𝑭𝑨 .
𝑩
𝑭𝑨
𝑭𝑩
→
𝑮𝒎𝑨 𝑴
𝒓𝑨 𝟐
𝑮(𝟑𝒎𝑨 )𝑴
𝒓
( 𝑨 )𝟐
𝟐
=
𝟏
𝟏𝟐
f. Therefore, FB = 12 FA. This means that the gravitational force exerted by the
star on the more massive planet is greater than on the less massive one.
Learning Competency
For circular orbits, relate Kepler's third law of planetary motion to Newton's law of
gravitation and centripetal acceleration (STEM_GP12G-IIc-22)
ACTIVITY 1: PUZZLE UP
Directions: Complete the crossword by filling in a word that fits each clue.
153
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Activity 2: Proving Kepler’s Constant
Kepler’s third law relates the radius of an orbit to its period of orbit. The square
of the period of orbit, divided by the cube of the radius of the orbit, is equal to a
constant (Kepler’s Constant) for that one object being orbited. The equation for this
is 𝑲 =
𝑻𝟐
𝒓𝟑
; where T is the period of the planet and r is its radius.
Directions: Using this equation, compute for Kepler’s constant from the information
of the planets given on the table below.
Planet
Period, T (days)
Radius, r (m)
Kepler’s constant
5
Mercury
88
2.44 x 10
5.33 x 10-13
5
Venus
225
6.05 x 10
2.28 x 10-13
Earth
365
6.38 x 105
5.13 x 10-13
5
Mars
684
3.40 x 10
1.66 x 10-11
6
Jupiter
4331
7.14 x 10
5.15 x 10-14
Questions:
1. What do you notice on the period of the planets if it is farther away from the
sun?
2. What happens to the radius of the orbit of the planets if it is farther away from
the sun?
3. What is the meaning of the Kepler’s constant in terms of planet’s revolution
around the sun?
Activity 3: Let’s Solve
Directions: Solve the following problems
1. Compute for the value of the acceleration due to gravity g of an object at an
altitude equal to twice the radius of the Earth? (radius of Earth = 6.4 x 10 6 m)
2. Scientists once hypothesized the existence of a planet called Vulcan to
explain Mercury’s precession. Vulcan is supposed to be between Mercury and
the Sun with a solar distance equal to 2/3 of that of Mercury. What would be
its supposed period?
3. What is the period T of a planet which radius is as twice as of Earth when it
completes one revolution in 875 days? ( Earth radius = 6.38 x 10 5)
154
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Activity 4: Think critically
Directions: Read the statement and write your analysis
1. Planet A is lighter than planet B and they orbit the same star. How do you
compare the gravitational force exerted by the star on the two planets?
2. Suppose two planets of the same mass orbit the same star but the distance
of Planet A from the star is thrice that of Planet B, which gravitational force is
greater? Explain.
REFLECTION:
1. I learned that ___________________________________________________
________________________________________________________________
________________________________________________________________
________________________________________________________________
___________________
2. I enjoyed most on ________________________________________________
________________________________________________________________
________________________________________________________________
________________________________________________________________
____________
3. I want to learn more on ___________________________________________
________________________________________________________________
________________________________________________________________
________________________________________________________________
________________
155
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References
Silverio, Angelina A. Exploring Life Through Science: Physics: Phoenix Publishing House,
Inc., 2017.
Hewitt, Paul G. Conceptual Physics. San Francisco: Addison Wesley, 2002
The
Science:
Orbital
Mechanics.
Retrieved
https://earthobservatory.nasa.gov/features/OrbitsHistory/page2.php
from
Kepler’s
Laws
of
Planetary
Motion.
Retrieved
https://www.britannica.com/science/Keplers-laws-of-planetary-motion
from
The
Universal
Law
of
Gravitation.
Retrieved
from
http://physics.weber.edu/amiri/physics1010online/WSUonline12w/OnLineCourseMo
vies/CircularMotion&Gravity/reviewofgravity/ReviewofGravity.html
https://www.google.com/search?q=law+of+ellipses&tbm=isch&hl=enUS&chips=q:law+of+ellipses,g_1:first+law:a5VUM9e6Lzw%3D&authuser=1&sa=X&ved=2
ahUKEwj3k5Hn1zsAhV0zIsBHUC7BakQ4lYoAXoECAEQFw&biw=532&bih=600#imgrc=YSIrO8Wu0rpcDM
https://www.google.com/search?q=law+of+equal+areas&tbm=isch&ved=2ahUKEwj4xH21-zsAhWLuJQKHaKFBakQ2cCegQIABAA&oq=law+of+equal+areas&gs_lcp=CgNpbWcQAzICCAAyAggAMgQIABBDM
gYIABAFEB4yBAgAEBgyBAgAEBgyBAgAEBg6BQgAELEDUKgjWPQ5YMtCaABwAHgAg
AGfAYgBpAuSAQQwLjExmAEAoAEBqgELZ3dzLXdpei1pbWfAAQE&sclient=img&ei=GJyk
X_jbLIvx0gSii5bICg&authuser=1&bih=600&biw=532&hl=en-US#imgrc=tQ0d0r3GKrQ-rM
https://www.google.com/search?q=the+ratio+of+the+squares+of+the+periods+P+(or+T+in+other+re
ferences)+of+any+two+planets+revolving+around+the+sun+is+equal+to+the+ratio+of+the+cubes+o
f+their+mean+distance+R+(or+d+in+other+references)+from+the+sun&tbm=isch&ved=2ahUKEwjm
gKjO3OzsAhUHe5QKHYClBBAQ2cCegQIABAA&oq=the+ratio+of+the+squares+of+the+periods+P+(or+T+in+other+references)+of+an
y+two+planets+revolving+around+the+sun+is+equal+to+the+ratio+of+the+cubes+of+their+mean+di
stance+R+(or+d+in+other+references)+from+the+sun&gs_lcp=CgNpbWcQA1DPrgNYz64DYLG3A2
gAcAB4AIABAIgBAJIBAJgBAaABAaoBC2d3cy13aXotaW1nwAEB&sclient=img&ei=AqGkXb3BIf20QSAy5KAAQ&bih=657&biw=1366&hl=en-US#imgrc=ketjHMHpOi4IMM
156
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ANSWER KEY / POSSIBLE ANSWERS
ACTIVITY 1: PUZZLE UP
Across
2. Law of periods
4. harmonic law
6. astronomical unit
7. year
9. Tycho Brahe
Down
1. Law of equal areas
3. ellipse
5. perihelion
8. gravitation
10. law of ellipses
Activity 2: Proving Kepler’s Constant
Planet
Period, T (days)
Mercury
88
Venus
225
Earth
365
Mars
684
Jupiter
4331
Radius, r (m)
2.44 x 105
6.05 x 105
6.38 x 105
3.40 x 105
7.14 x 106
Kepler’s constant
5.33 x 10-13
2.28 x 10-13
5.13 x 10-13
1.66 x 10-11
5.15 x 10-14
Activity 3: Let’s Solve
1. g ≈ 1.1 m/s2
2. T ≈ 19 hours
Activity 4: Think critically
1. Gravitational force between Planet A and the star is lesser compared to the
gravitational force of Planet A and the star. This is because the universal law
of gravitation states that gravitational force that is directly proportional to the
mass of each object. Planet A is lighter, hence the force between it and the
star is lesser.
2. The universal law of gravitation states that gravitational force is inversely
proportional to the square of the distance between them. Planet A is farther
from the star than Planet B, hence, the gravitational force between the star
and Planet A is weaker than that of the gravitational force between the star
and Planet B.
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Prepared by:
KARLA CHRISTIANA R. MARAMAG
Camalaniugan National High School
GENERAL PHYSICS 1
Name: ____________________________
Date: _____________________________
Grade Level: _________
Score: ______________
LEARNING ACTIVITY SHEET
Periodic Motion
Background Information for the Learners (BIL)
Periodic motion refers to motion that is repeated at regular intervals of time.
Examples of periodic motion are the movement of hands of a clock, the pendulum in
a grandfather’s clock, a rocking chair, heartbeat, the rotation of the blades of an
electric fan, and the movement of earth about its axis and about the sun.
A body undergoing periodic motion always has a stable equilibrium position.
The equilibrium position, otherwise known as resting position, is the position
assumed by the body when it is not vibrating. This equilibrium position is represented
by position O of the girl in the swing in figure 8-1.
Fig. 8-1. The motion of the swing is an example of periodic motion
Source: Silverio,Angelina.”Exploring Life Through Science Series: General Physics 1.” In Teachers
Wraparound Edition. Quezon City, Phoenix Pulishing House, Inc., 2017
When the girl is displaced from its equilibrium position to position A, a restoring
force (gravity) acts on it to pull it back toward position O. A restoring force is a force
that tends to restore a body from its displacement to its equilibrium position. By the
time the girl reaches position O, the body has gained kinetic energy, overshoots this
position, moves, stops somewhere on the other side (position B). The body is again
pulled back toward equilibrium. Vibrations about this equilibrium position results only
158
NOTE: Practice personal hygiene protocols at all times
from the action of the restoring force. The amplitude (A) of vibration is the maximum
displacement of a body from its equilibrium position. This is represented by the
displacement from position O to position A or from position O to position B.
The period (T) of a body in periodic motion is the time required to make a
complete to-and-fro motion is called a cycle. Referring to figure 8-1, the motion of
the swing from position A to position B and back to position A is one cycle. Period is
usually expressed in seconds.
Frequency (f) is the number of cycles per unit of time. Its SI unit is the hertz,
abbreviated as Hz. One hertz is equal to one cycle per second. Frequency is the
reciprocal of period.
Sometimes, angular frequency (ω) is used instead of frequency. Angular frequency
is commonly expressed in radians per second. The relationship between angular
frequency is given by:
or
Learning Competency:
Relate the amplitude, frequency, angular frequency, period, displacement, velocity,
and acceleration of oscillating systems (STEM_GP12PM-IIc-24)
Activity 1: Finding My Reciprocal
Directions: Find for the period and frequency (in Hz) of each problem below. Write
your answer on the space provided.
1
A very tall skycraper
What is the period?________________
sways back and
________________________________
forth every 4.0 seconds.
What is the frequency?______________
________________________________
2
What is the period?________________
A tuning fork has
________________________________
a frequency of 252 Hz
What is the frequency?______________
________________________________
3
In 1940, Tacoma
What is the period?________________
________________________________
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Narrow bridge oscillated
What is the frequency?______________
up and down 5 times
________________________________
per second.
4
5
At an amusement park,
What is the period?________________
the pirate ship swings
________________________________
back and forth every 20
What is the frequency?______________
seconds.
________________________________
A smoke alarm battery
What is the period?________________
is beeping 2 times
________________________________
per minute
What is the frequency?______________
________________________________
6
A speaker vibrates
What is the period?________________
at 200 cycles per
________________________________
second.
What is the frequency?______________
________________________________
7
A pendulum takes
What is the period?________________
0.5 second to
________________________________
complete one cycle.
What is the frequency?______________
________________________________
8
An oscillator makes
What is the period?________________
4 vibrations in
________________________________
1 second.
What is the frequency?______________
________________________________
9
A swing takes 2
What is the period?________________
seconds to complete
________________________________
one cycle
What is the frequency?______________
________________________________
10
A string virates
What is the period?________________
at a frequency
________________________________
of 25 Hz.
What is the frequency?______________
________________________________
160
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Activity 2: Bingo Choice Card
Directions: The Bingo Choice card shows some terms/concepts related to periodic
motion. Choose words in either horizontal, vertical or diagonal pattern and relate your
choosen concepts/terms to one another. Write your answer on the space provided
below the Bingo card.
BINGO CHOICE CARD
Restoring
force
Period
(T)
Pendulum
Displacement
Hertz
Amplitude
Periodic
motion
Velocity
Frequency
ω
A
Resting
Position
f
Angular
Frequency
Radians per
second
Answer:
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
______
__________________________________________________________________
__________________________________________________________________
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__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
_________
Activity 3: The Swinging Pendulum
Directions: Analyze the problem below and answer logically the questions that
follow. Write your answer on the space provided.
1. A pendulum takes 10 seconds to swing
through 2 complete cycles.
a. How long does it take to complete
one cycle?
_____________________________________
_______________________________
b. What is its period? ________________
__________________________________
c. What is its frequency?_______________
__________________________________
d. What is the angular frequency? ________
_______________________________
e. What does the position N represents?
Source: Silverio,Angelina.”Exploring Life
Through Science Series: General Physics
1.” In Teachers Wraparound Edition.
Quezon City, Phoenix Pulishing House, Inc., 2017
_________________________________
f. When the pendulum is displaced from position N to position Y, what factor
tends to restore the pendulum from itsdisplacement to its equilibrium
position?
_____________________________________________________________
_____________________________________________________________
__
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g. What does the displacement from position N To position Y or position N to
position X represents?
_____________________________________________________________
_____________________________________________________________
h. What does the motion of the pendulum represents when it swings from
position Y to position X and back to position Y?
_____________________________________________________________
_____________________________________________________________
i.
How many cycle/s will it make when it swings only once from position Y to
position X and back to position Y?
_____________________________________________________________
_____________________________________________________________
Activity 4: Fact or Bluff
Directions: Write Fact if the statement is true. If the statement is False, write Bluff.
Write your answer on the space provided.
________________1. Period is directly proportional to frequency.
________________2. Heartbeat is an example of periodic motion
________________3. A body undergoing periodic motion always has an unstable
equilibrium position.
________________4. A pendulum with a frequency of 2 hertz has a period of 0.5 s.
________________5. The amplitude of a vibration is not related to the equilibrium
position.
________________6. One complete to-and-fro motion is called a cycle.
________________7.The resting position is otherwise known as the equilibrium
position.
________________8. Angular frequency is represented by greek letter α.
________________9. An object is undergoing periodic motion when it moves
repeatedly at regular intervals of time.
________________10. The motion of the swing is an example of rotational motion.
Activity 5: Solve It!
Directions: Use the relationships of the concepts of periodic motion to solve for
problems below. Write your answer on the space povided.
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1. What will be the period of a string if it makes 6 vibrations in just one second ?
What will be the angular frequency?
_______________________________________________________________
_______________________________________________________________
_______________________________________________________________
_______________________________________________________________
____
_______________________________________________________________
_
2. A swing takes 0.5 minute to sway back and forth. What is the period in
seconds? What is the frequency? What is the angular frequency?
_______________________________________________________________
_______________________________________________________________
_______________________________________________________________
_______________________________________________________________
____
_______________________________________________________________
_
164
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References
Source: Silverio,Angelina.”Exploring Life Through Science Series: General Physics
1.” In Teachers Wraparound Edition. Quezon City, Phoenix Pulishing House, Inc.,
2017
https://www.physicsclassroom.com/class/waves/Lesson-0/Properties-of-PeriodicMotion
165
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ANSWER KEY
ACTIVITY 1:
1. Period:
Frequency:
4.0s
0.25 Hz
2. Period:
Frequency:
0.00397s
252 Hz
3. Period:
Frequency:
0.2s
5 Hz
4. Period:
Frequency:
0.05s
20 Hz
5. Period:
Frequency:
0.0083s
120 Hz
6. Period:
Frequency:
0.005s
200 Hz
7. Period:
Frequency:
0.5s
2.0 Hz
8. Period:
Frequency:
0.25s
4 Hz
9. Period:
Frequency:
2s
0.5 Hz
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10. Period:
Frequency:
0.04s
25 Hz
ACTIVITY 2:
Checking the varied answers of the learners may be based on the following
relationships of the concepts below:
• Restoring force is a force that tends to restore a body from its displacement to
its equilibrium position.
• Period is the time required to make a complete to-and-fro motion.
• T represents the period of a body in periodic motion.
• Pendulum is an object that exhibits period motion.
• Hertz is the SI unit for frequency, equivalent to 1 cycle per second.
• Amplitude is the maximum displacement of a body from its equilibrium position.
• Periodic motion refers to motion that is repeated at regular intervals of time.
• Frequency is the number of cycles per unit of time.
• (ω) is the symbol for angular frequency
• A is the symbol for amplitude.
• Resting position is the position assumed by the body when it is not vibrating.
• f is the symbol for frequency.
• Angular frequency is commonly expressed in radians per second.
• Radians per second is the unit for angular frequency.
• Displacement – a measure of how far an object has moved in particular
direction from its original position.
• Velocity – the rate of change in displacement of an object at a given time
interval.
ACTIVITY 3:
a. The time to make one complete cycle is 5 seconds
b. The period is 5 seconds
c. The frequency is 0.2 Hz
d. The angular frequency is 1.26 Hz.
e. The equilibrium position
f. A restoring force (gravity)
g. The amplitude
h. The period
i. 1 cycle
ACTIVITY 4:
1. BLUFF
2. FACT
3. BLUFF
6. FACT
7. FACT
8. BLUFF
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4. FACT
5. BLUFF
9. FACT
10. BLUFF
ACTIVITY 5:
1. The period is 0.17 s
The angular frequency is 3.75 Hz
2. The period in seconds is 30.
The frequency is 0.03 Hz.
The angular frequency is 0.188
Prepared by:
Kimberly Anne C. Pagdanganan
Licerio Antiporda Sr. National High School Dalaya Extension
GENERAL PHYSICS 1
Name: ____________________________
Date: _____________________________
Grade Level: _________
Score: ______________
LEARNING ACTIVITY SHEET
Simple Harmonic Motion (or SHM)
Background Information for the Learners (BIL)
Simple Harmonic Motion (or SHM)
SHM –Terminologies and Description
Amplitude (A) - is defined as the maximum magnitude of the displacement
from the equilibrium position. Its unit is meter (m).
Period (T) - is defined as the time taken for one cycle.
T=1/f
Frequency (f) - is defined as the number of cycles in one second. Its unit
is hertz (Hz) :
1 Hz = 1 cycle s-1 = 1 s-1
f = 1 / T=ω = 2π × f = ω / 2π
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Equilibrium Position -- a point where the acceleration of the body undergoing
oscillation is zero. At this point, the force exerted on
the body is also zero.
Restoring Force -- the force which causes simple harmonic motion to occur. This
force is proportional to the displacement from equilibrium
& always directed towards equilibrium. Fs =−k x
Simple Harmonic Motion - Oscillatory motion where the net force on the
system is a restoring force
An object is said to be in simple harmonic motion if the following occurs:
• It moves in a uniform path.
• A variable force acts on it.
• The magnitude of force is proportional to the displacement of the
mass.
• The force is always opposite in direction to the displacement
direction.
•
The motion is repetitive and a round trip, back and forth, is always made in
equal time periods.
SHM Visually
Examples:
• Spring
• Pendulum
https://upload.wikimedia.org/wikipedia/commons/e/ea/Simple_Harmonic_Mo
tion_Orbit.gif
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SHM – Hooke’s Law
•
•
SHM describes any periodic motion that
results from a restoring force (F) that is
proportional to the displacement(x) of an
object from its equilibrium position.
Frest= - kx, where k = spring constant
Note:
Elastic limit – if exceeded, the spring does not return
to its original shape
•
Law applies equally to horizontal and vertical
models
Hook’s Law- Horizontal Spring
•
At max displacement (2 & 4), spring force and
acceleration reach a maximum and velocity
(thus KE) is zero
•
At zero displacement (1 & 3) PE is zero, thus
KE and velocity are maximum
The larger the k value the stiffer the spring
Negative sign indicates the restoring force is
opposite the displacement
•
•
Hooke’s Law – Vertical Springs
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•
•
Hooke’s Law applies equally to a vertical model
of spring motion, in which the weight of the
mass provides a force.
@ Equilibrium position with no motion:
• Spring force↑ = weight↓
Practice
A load of 50 N stretches a vertical spring by 0.15 m. What is the spring
constant?
Solve F = -kx for k
50 = -k*0.15
k = - 50/0.15 = 333.3 N/m (drop the – sign)
Mass-Spring System - Period
The period of a mass-spring can be calculated as follows:
T= 2
mass
Spring constant
T = 2
m
k
Practice
1. What is the spring constant of a mass spring system that has a mass of
0.40 kg and oscillates with a period of 0.2 secs?
Solve
T=2π
m
k
0.2 = 2π*√(0.4/k)
k = 394.8 N/m
Practice
2. If a mass of 0.55 kg stretches a vertical spring 2 cm from its rest
position, what is the spring constant (k)?
Solve F = -kx for k (or ΔF = -k*Δx)
k = F/x, where F = weight (mg) of the mass
k = mg/x = 0.55 x 9.8/0.02
k = 269.5 N/m
SHM – Simple Pendulum
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If a pendulum of length l is distributed through an angle θ (1 or 3), the restoring
force component drives the bob back and through then rest at position 2.
2π length = 2π l
Grav
g
Period (T) =
Practice
1. What period would you expect from a pendulum of length 0.5 m on the
moon where g = 1.6 m/s²?
Solve T = 2π
l
g
T = 2π √(0.5/1.6)
T = 3.51 seconds
Learning Competency
Recognize the necessary conditions for an object to undergo
simple harmonic motion (STEM_GP12PMIIc-25)
Learning Activity 1 -Simple Harmonic Match
Directions: Draw a line to connect to its corresponding answer.
Column A
Column B
1. What is the time taken for one
oscillation?
A. Simple harmonic motion
(SHM)
2. What provides the restoring force
for a pendulum?
B. Amplitude
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3. When is the potential energy a
maximum?
C. Restoring Force
4. When is the kinetic energy a
maximum?
D. At the equilibrium position
5. I It is a type of periodic motion
where the restoring force is
proportional to the displacement of
the body from its equilibrium
position.
E. force of gravity
6. What is the total energy in a
system undergoing simple
harmonic motion?
F. Towards zero displacement
7. What happens to the direction of
acceleration when the mass
undergoing simple harmonic motion
passes through the equilibrium
point?
G. At maximum displacement
8. It is a force acting
opposite to displacement to bring
the system back to equilibrium,
which is its rest position
H. Potential and Kinetic
Energy
9. In simple harmonic motion what
is the maximum value of x?
I. period
10. In what direction does the
restoring force acts?
J. P.E is zero
Activity 2 – SHM Crossword
Directions: Fill in the crossword puzzle with the correct vocabulary word by
reading the clues below.
1
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2
3
4
5
6
7
8
9
10
https://wordmint.com/puzzles/2550608
Across
2. Number of oscillations in unit time
4. The highest point
6. A system that undergoes simple
harmonic motion
7. This is a property of a spring
8. Simple harmonic motion graphs are
similar to this function
9. Maximum displacement
10. The force that brings the object to its
equilibrium position
Down
1. A back and forth vibration
3. The length of a complete wave
5. Time taken for one complete oscillation
Activity 3 – SHM Scramble Word
Directions: Study the scrambled letters and try to unscramble or rearrange the
letters to form a word.
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1. DOSTWRA EORZ CASPEMEILDNT ___________________
2. UTALEPIMD ______________________________________
3. OGIRNRSTE FRCEO _______________________________
4. IOPDRE __________________________________________
5. AT ETH ULEIBUIIMQR ONIOPITS _____________________
6. FREOC FO VARYITG _______________________________
7. EP. SI OREZ ______________________________________
8. PSEMIL HOCRMNIA OOMITN S)MH(___________________
9. AT MAMIMXU MTPLDEEISNAC _______________________
10. LEAITOPTN DNA KINICTE YGEREN ______________________
Activity 4- SHM-Springs and Pendulums
TS = 2π
m
k
Tᵖ= 2π
L
g
Directions: Show your work clearly on a separate page. Make a sketch of
the problem. Start each solution with a fundamental concept equation written
in symbolic variables. Solve for the unknown variable in a step-by step
sequence.
1. What is the period of a simple pendulum 50 cm long?
a. On Earth
b. On a freely falling elevator
c. On the moon (gMoon = 1/6thgEarth)
2. The length of a simple pendulum is 0.66 m, the pendulum bob has
a mass of 310 g, and it is released at an angle of 120 to the vertical
a. With what frequency does it oscillate? Assume SHM.
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NOTE: Practice personal hygiene protocols at all times
b. What is the pendulum bob’s speed when it passes through the
lowest point of the swing? (Energy is conserved)
c. What is the total energy stored in the oscillation assuming no
losses?
3. Suppose you notice that a 5-kg weight tied to a string swings back
and forth 5 times in 20 seconds. How long is the string?
4. A mass of 400 g is suspended from a spring hanging vertically,
and the spring is found to stretch 8.00 cm.
a. Find the spring constant.
b. How much will the spring stretch if the suspended mass is
575 g?
5. A 3.00-kg mass is attached to a spring and pulled out horizontally
to a maximum displacement from equilibrium of 0.500 m.
a. What spring constant must the spring have if the mass is
to achieve an acceleration equal to that of gravity?
b. What is its period of vibration?
Reflection
1I learned that ________________________________________________
____________________________________________________________
_______________________________________________________
2.I enjoyed most on _____________________________________________
____________________________________________________________
__________________________________________________
3.I want to learn more on _________________________________________
____________________________________________________________
__________________________________________________
176
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References:
•
•
•
•
•
•
Young, H., Freedman, R., Ford, A., & Young, H. (2012). Sears and
Zemansky's University physics. Boston, MA: Pearson Learning Solutions.
Baltazar and Tolentino. Exploring Life Through Science General Physics 1.
Teachers Wraparound Edition. Phoenix Publishing House, Inc., 2017
https://www.augusta.k12.va.us/cms/lib01/VA01000173/Centricity/Domain/39
6/Simple_Harmonic_Motion_(SHM).pdf
https://www.livingston.org/cms/lib4/NJ01000562/Centricity/Domain/1357/HW
8.2%20-SHM.pdf
https://wordmint.com/puzzles (Note: My Puzzles in My Account)
https://sharemylesson.com/teaching-resource/shm-198980
177
NOTE: Practice personal hygiene protocols at all times
ANSWER KEY:
Activity 1 -Simple Harmonic Match
1. Period
2. Force of gravity
3. At maximum displacement
4. P.E is zero
5. Restoring force
6. Potential and Kinetic Energy
7. At the equilibrium position
8. Simple Harmonic Motion (SHM)
9. Amplitude
10. Towards zero displacement
Activity 2 – SHM Crossword
1.
2.
3.
4.
5.
simple harmonic motion
frequency
wavelength
Crest
period
6. simple pendulum
7. spring constant
8. cosine
9. amplitude
10. restoring force
Activity 3 -SHM-Scramble Word
1. DOSTWRA EORZ CASPEMEILDNTTowards zero displacement
2. UTALEPIMD __Amplitude______________________________
3. OGIRNRSTE FRCEO __Restoring Force__________________
4. IOPDRE __period____________________________________
5. AT ETH ULEIBUIIMQR ONIOPITS __At the equilibrium position
6. FREOC FO VARYITG __force of gravity________________
7. EP. SI OREZ __P.E is zero__________________________
8. PSEMIL HOCRMNIA OOMITN S)MH(
(SHM)
Simple harmonic motion
9. AT MAMIMXU MTPLDEEISNAC __At maximum displacement__
10. LEAITOPTN DNA KINICTE YGEREN Potential and Kinetic Energy
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Activity 4 -SHM-Springs and Pendulums
1. a) T=1.42s
b) T is infinite
c) T = 3.51s
2. a) 0.613 Hz
b) 0.532 m/s
c) 0.0439 J
3. a) 4.0 m
4. a) k = 49N/m
b) x = 11.5cm
5. a) k = 58.8N/m b) T=1.42s
Prepared by:
LEONOR C. NATIVIDAD
Baggao National High School
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NOTE: Practice personal hygiene protocols at all times
GENERAL PHYSICS 1
Name: ____________________________
Date: _____________________________
Grade Level: _________
Score: ______________
LEARNING ACTIVITY SHEET
Simple harmonic motion: spring mass system, simple pendulum,
physical pendulum
Background Information for the Learners (BIL)
Our world is filled with oscillations in which objects move back and forth
repeatedly. Many oscillations are merely amusing or annoying, but many others are
dangerous. Here are a few examples: When a bat hits a baseball, the bat may
oscillate enough to sting the batter’s hands or even to break apart. When wind blows
past a power line, the line may oscillate so severely that it rips apart, shutting off the
power supply to a community. When an airplane is in flight, the turbulence of the air
flowing past the wings makes them oscillate, eventually leading to metal fatigue and
even failure. When a train travels around a curve, its wheels oscillate horizontally.
When an earthquake occurs near a city, buildings may be set oscillating so severely
that they are shaken apart. When an arrow is shot from a bow, the feathers at the
end of the arrow manage to snake around the bow staff without hitting it because the
arrow oscillates. When a coin drops into a metal collection plate, the coin oscillates
with such a familiar ring that the coin’s denomination can be determined from the
sound.
One important property of oscillatory motion is its frequency, or number of
oscillations that are completed each second. The symbol for frequency is f, and its
SI unit is the hertz (abbreviated Hz), where:
1 hertz= 1 Hz= 1 oscillation per second= 1s-1
Related to the frequency is the period T of the motion, which is the time for
one complete oscillation (or cycle); that is,
𝑻=
𝟏
𝒇
Any motion that repeats itself at regular intervals is called periodic motion
or harmonic motion.
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The Period and Frequency of a Mass on a spring
One interesting characteristic of the Simple Harmonic Motion of an object attached
to a spring is that the angular frequency, and therefore the period and frequency of
the motion, depends on only the mass and the force constant, and not on other
factors such as the amplitude of the motion. We can use the equations of motion and
Newton’s second law (𝐹⃗ net = m𝑎⃗) to find equations for the angular frequency,
frequency, and period.
Fx= -kx; where Fx is the applied force, k is the spring constant and x is the
length
ma= -kx;
m
𝑑2 𝑥
𝑑𝑡 2
𝑑2 𝑥
𝑑𝑡 2
= -kx
𝑘
= -𝑚x
Since:
x (t)= A cos(ꞷt + 𝜑)
Substituting the equations of motion for x and a gives us
𝑘
-Aꞷ2 cos (ꞷt +𝜑)= - 𝑚 A cos(ꞷt + 𝜑)
Cancelling out like terms and solving for the angular frequency yields
ꞷ= √
𝒌
𝒎
The angular frequency depends only on the force constant and the mass, and not
the amplitude. The angular frequency is defined as ω = 2π/T, which yields an
equation for the period of the motion:
𝒎
T= 𝟐𝝅√
𝒌
The period also depends only on the mass and the force constant. The greater the
mass, the longer the period. The stiffer the spring, the shorter the period. The
frequency is
𝟏
𝟏
𝑻
𝟐𝝅
f= =
√
𝒌
𝒎
1
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The diagram on the right shows
the mass-spring system. When
mass M attached to a linear
spring is pulled and released, its
up-and-down motion above and
below the equilibrium level is
called
“simple
harmonic
motion.” Figure (a) shows a
spring that is not loaded. Figure
(b) shows the same spring but
loaded and stretched a distance
(-h), and Figure (c) shows the
loaded spring stretched further
a distance (-A) and released. It
The Mass-Spring System. Credit: www.pstcc.edu
shows that the attached mass M oscillates up and down to (+A) and (-A) above and
below the equilibrium level.
Sample Problem:
A block whose mass m is 680 g is fastened to a spring whose spring constant k is
65 N/m. The block is pulled a distance x =11 cm from its equilibrium position at x =0
on a frictionless surface and released from rest at t =0. What are the angular
frequency, the frequency, and the period of the resulting motion?
Given:
m= 680 g= 0.68 kg
k= 65 N/m
Calculations:
a. To solve for the angular frequency, we use the formula:
𝑘
65 𝑁/𝑚
ꞷ= √𝑚 = √ 0.68 𝑘𝑔 = 9.78 rad/s
b. To find the period (T), we use the formula:
ω=
2π
𝑇
or T=
2π
ω
2π
= 9.78 rad/s= 0.64 s
c. Finally, to get the frequency (f), we use the formula:
The Simple
1
1 Pendulum
f
=
=
= 1.56isHzdefined to have a point mass, also known as the pendulum
A simple
pendulum
𝑇
0.64
𝑠
bob, which is suspended from a string of length L with negligible mass. Here, the only
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forces acting on the bob are the force of gravity (i.e., the weight of the bob) and
tension from the string. The mass of the string is assumed to be negligible as
compared to the mass of the bob.
A simple pendulum has a smalldiameter bob and a string that has a
very small mass but is strong enough
not to stretch appreciably. The linear
displacement from equilibrium is s,
the length of the arc. Also shown are
the forces on the bob, which result in
a net force of –mg sin θ toward the
equilibrium position- that is, the
restoring force.
We represent the forces on the mass
in terms of tangential and radial
components. The restoring force Fθ is the tangential component of the net force:
Fθ = –mg sin θ
The restoring force is provided by gravity; the tension T merely acts to make the point
mass move in an arc. The restoring force is proportional not to θ but to sin θ so the
motion is not simple harmonic. However, if the angle θ is small, sin θ is very nearly
equal to θ in radians. With this approximation, it becomes:
𝑥
Fθ = –mg sin θ = -mg 𝐿
or
Fθ= -
𝑚𝑔
𝐿
x
The restoring force is then proportional to the coordinate for small displacements,
and the force constant is k= mg/L. The angular frequency ω of a simple pendulum
with small amplitude is:
ꞷ= √
𝒌
𝒎
=√
𝒎𝒈/𝒍
𝒎
=√
𝒈
𝑳
The corresponding frequency and period relationships are:
f=
ꞷ
=
𝟐𝝅 𝟏
T=
𝟏
𝟐𝝅 𝟐𝝅
𝒈
√𝑳
𝑳
= = 𝟐𝝅√𝒈
ꞷ 𝒇
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Note that these expressions do not involve the mass of the particle. This is because
the restoring force, a component of the particle’s weight, is proportional to m. Thus
the mass appears on both sides of 𝛴𝐹⃗ = m𝑎⃗and cancels out.
Sample Problem:
Find the period and frequency of a simple pendulum 1.0 m long at a location where g=
9.8 m/s2.
Given:
L= 1.0 m
g= 9.8 m/s2
Calculations:
a. To solve for the period (T), we use the formula:
𝑳
𝒈
𝟏.𝟎 𝒎
=
𝟗.𝟖 𝒎/𝒔𝟐
T= = 𝟐𝝅√ =𝟐𝝅√
2.007 s
b. To get the frequency (f), we use the formula:
f=
1
𝑇
=
1
2.007 𝑠
= 0.498 or 0.50 Hz
Physical Pendulum
Any object can oscillate like a pendulum. Consider a coffee mug hanging on a hook
in the pantry. If the mug gets knocked, it oscillates back and forth like a pendulum
until the oscillations die out. We have described a simple pendulum as a point mass
and a string. A physical pendulum is any object whose oscillations are similar to
those of the simple pendulum, but cannot be modeled as a point mass on a string,
and the mass distribution must be included into the equation of motion.
As for the simple pendulum, the restoring force of the physical pendulum is the force
of gravity. With the simple pendulum, the force of gravity acts on the center of the
pendulum bob. In the case of the physical
pendulum, the force of gravity acts on the
center of mass (CM) of an object. The object
oscillates about a point O. Consider an object
of a generic shape as shown in the figure.
An example showing gravity acts through the
center of mass (CM) of the rigid body. Hence,
the length of the pendulum used in equations
is equal to the linear distance between the
pivot and the center of mass (h).
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The equation of torque gives:
𝜏=𝐼𝛼
where 𝛼 is the angular acceleration (𝛼 =
𝑑2 𝜃
𝑑𝑡 2
), 𝜏 is the torque, and I is the moment
of inertia.
The torque is generated by gravity so:
𝜏= -(mg) hsinθ,
where h is the distance from the center of mass to the pivot point and θ is the angle
from the vertical. The negative sign shows that the restoring torque is clockwise when
the displacement is counterclockwise, and vice versa.
𝜏=𝐼𝛼
-h (mg) sinθ =𝐼𝛼
We assume the angle θ is small, for then we can approximate sin θ with θ (expressed
in radian measure). With that approximation and some rearranging, we then have:
𝛼=−
𝑚𝑔ℎ
𝐼
θ
We see that the angular frequency of the pendulum is:
ꞷ= √
𝒎𝒈𝒉
𝑰
Since ω = 2π/T, the period is:
𝑰
T= 𝟐𝝅√𝒎𝒈𝒉
and the frequency is:
𝟏
f= =
𝟏
𝑻 𝟐𝝅
√
𝒎𝒈𝒉
𝑰
In case we know the moment of inertia of the rigid
body, we can evaluate the above expression of the
period for the physical pendulum. For illustration, let
us consider a uniform rigid rod, pivoted from a frame
as shown in the figure. Clearly, the center of mass is
at distance L/2 from the point of suspension:
𝐿
h= 2
The moment of inertia of the rigid rod about its center
is:
Ic =
𝑚𝐿2
12
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However, we need to evaluate the moment of inertia about the pivot point, not the
center of mass, so we apply the parallel axis theorem:
Io= Ic + mh2=
𝑚𝐿2
12
𝐿
+ m( 2)2 =
𝑚𝐿2
3
Plugging this result into the equation for period, we have:
𝑰
2𝑚𝐿2
2𝐿
T= 𝟐𝝅√𝒎𝒈𝒉 = 𝟐𝝅√3𝑚𝑔𝐿 = 𝟐𝝅√3𝑔
The important thing to note about this relation is that the period is still independent
of the mass of the rigid body. However, it is not independent of the mass distribution
of the rigid body. A change in shape, size or mass distribution will change the moment
of inertia. This, in turn, will change the period.
As with simple pendulum, a physical pendulum can be used to measure g.
Sample Problem:
A
uniform rod with length L (1.0 m), pivoted at one end, what is the period of its
motion as a pendulum?
Given:
L= 1.0 m
g= 9.8 m/s2
Calculations:
a. To solve for the period (T), we use the formula:
𝟐𝑳
𝟐 (𝟏.𝟎 𝒎)
=𝟐𝝅√
=
𝟑𝒈
𝟑 (𝟗.𝟖 𝒎/𝒔𝟐 )
T= = 𝟐𝝅√
1.64 s
Learning Competency:
Calculate the period and the frequency of spring mass, simple pendulum, and
physical pendulum. (STEM-GP12PMIIc-27)
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Activity No. 1: Conceptual Questions
Directions: Read and analyze each item and answer what is asked.
1. Which of the following mass-spring systems have the highest frequency of
vibration?
Case A: A spring with a k= 300 N/m and a mass of 200 g suspended from it.
Case B: A spring with k=400 N/m and a mass of 200g suspended from it.
2. Which of the following mass-spring systems will have the highest frequency of
vibration?
Case A: A spring with a k=300 N/m and a mass of 100 g suspended from it.
Case B: A spring with a k=300 N/m and a mass of 200 g suspended from it.
3. Which would have the highest frequency of vibration?
Pendulum A: A 200-g mass attached to a 1.0-m length string
Pendulum B: A 400-g mass attached to 0.5-m length string
Activity No. 2: Problem Set
Directions: Solve the following problems systematically. Show your solution.
1. What is the period of a mass-spring oscillation system with a spring constant of
120 N/m and mass of 0.5 kg?
2. A 0.70 kg object vibrates at the end of a horizontal spring (k = 75 N/m) along a
frictionless surface. What is the frequency of the vibration?
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3. A 2.0 kg mass attached to an ideal spring oscillates horizontally with a amplitude
of 0.30 m. The spring constant is 85 N/m. What is the frequency of the mass’ motion?
4. A body of unknown mass is attached to an ideal spring that is mounted horizontally
with its left and held stationary. The spring constant of the spring is 120 N/m and it
vibrates with a frequency of 6.0 Hz. Assuming that there is no friction, find the period
of the motion; (b) the angular frequency; (c) the mass of the body.
5. When a 0.750-kg mass oscillates on an ideal spring, the frequency is 1.33 Hz.
What will the frequency be if 0.220 kg is added to the original mass?
6. The pendulum on a cuckoo clock is 5.00-cm long. What is its frequency?
7. A simple pendulum with a length of 2.6 m oscillates on the Earth’s surface. What
is the frequency of oscillations?
8. A simple pendulum with a length of 1 m oscillates on the Moon’s surface where
acceleration due to gravity is 1.7m/s2. What is the period of oscillations?
9. How long does it take a child on a swing to complete one swing if her center of
gravity is 4.00 m below the pivot?
10. All walking animals, including humans, have a natural walking pace—a number
of steps per minute that is more comfortable than a faster or slower pace. Suppose
that this pace corresponds to the oscillation of the leg as a physical pendulum. Treat
the leg as a uniform rod pivoted at the hip joint. Fossil evidence shows that T. rex, a
two-legged dinosaur that lived about 65 million years ago, had a leg length,L= 3.1 m
and a stride length S= 4.0 m (the distance from one footprint to the next print of the
same foot.) Find the period of oscillation of the leg of the T. rex.
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Activity No. 3: Mass on Spring Interactive
Goal: To determine what factors affect the frequency and the period of a vibrating
mass on a spring and to state the relationship between those variables and the
frequency or period.
Getting Ready: Navigatetothe Vibrating Mass on a Spring
InteractiveatThePhysicsClassroomwebsite:http://www.physicsclassroom.com/Physi
cs--‐Interactives/Waves--‐and--‐Sound/Mass--‐on--‐a--‐Spring
Navigational Path:
www.physicsclassroom.com===>Physics Interactives ==> Waves and Sound==>
Vibrating Mass on a Spring
Getting Acquainted:
Once you've launched the Interactive and resized it, experiment with the interface.
Place a mass on the end of the spring and observe the vibration. Click/tap the Start
button to view the plot of its vertical position as a function of time. Reset the system
and place a mass on each spring and observe that their graphs are color coded --‐
consistent with the color of the spring. Notice that the time, height, and velocity of the
mass are reported below the graphs. And most importantly for this lab, observe how
the vertical line on the graph can be moved along the axis in order to obtain values
of height and velocity at various times on the graph.
The Challenge:
Your challenge is to determine what factors affect the frequency and the period of a
vibrating mass on a spring. Make your study of this question very systematic --‐
varying one factor at a time while you hold others constant. You can easily test the
mass and the stiffness of the spring as possible factors. If you are "quick," you might
also be able to test damping as a possible factor. Conduct several trials for each
variable under study. For each trial, measure the period by recording the difference
in time from the start and the end of one cycle or of several cycles. Use the provided
tables. Not all columns or rows will necessarily be used.
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Factor #1:
Factor:
# of
Start Time
Stop Time
______________
Cycles
(s)
(s)
Factor:
# of
Start Time
Stop Time
______________
Cycles
(s)
(s)
Period (s)
Frequency
(Hz)
Factor #2
Period (s)
Frequency
(Hz)
Conclusion:
Identify the factors that affect the frequency and the period of a vibrating mass on a
spring. For each factor having an effect, describe the effect (e.g., state something
like ... "As the
frequency
increases, the period
and the
.")
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Activity No. 4: Pendulum Lab
Goal: Investigate how the period of a simple pendulum depends on the length of
the string and the mass of the pendulum bob.
Procedure
How to open the simulation:
1. Go to the simulation page: http://phet.colorado.edu/en/simulation/pendulum-lab
2. Click
to start.
3. It will take time to load and then this screen appears:
Predictions:
1. Does the mass of the bob affect the number of swings? Explain.
2. Does the length of the pendulum affect the number of swings? Explain.
Explore!
For the next 5 minutes become familiar with the simulation. Change various features
such as mass, length etc.
Next: Click
and conduct the following investigation.
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Part 1: Mass and Number of Swings
Hypothesis:
As the mass of the pendulum _________________ the number of swings
__________________.
For this activity keep the length of both pendulums the same but different mass.
Click
. Start both pendulums at 90 degrees. Check the other tools
button and use the timer to keep track of the time.
Click play on the timer and then again
so that the pendulums are
released. Count the number of full swings for 30 seconds.
Record the data on the table below.
Mass (kg)
Length (m)
Number of
swings for 30
seconds
Pendulum 1
Pendulum 2
Did mass effect the number of full swings? Write a conclusion based on the data you
collected.
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Part 2: Length and number of swings
Write a hypothesis:
As the length of the pendulum _______________________ the number of
swings_____________________.
For this activity the mass should stay the same but the length will change each time.
Click
and conduct the following investigation.
Use
the photogate timer and record the period it takes for each
length. Remember the period of a pendulum is the time it takes the pendulum
to make one full back-and-forth swing. Click reset again.
Next, use the timer (by clicking other tools) and observe the number of swings
the pendulum makes each time you change the length.
Each time you adjust the length, count the number of full swings in a 30
second interval.
Make sure that the pendulum is released at the same position each time.
Record the data on the table below.
Length of the
pendulum (m)
Period (s)
Number of swings in
30 sec
0.5
1
Write a conclusion on how the length of the pendulum affects the number of full
swings?
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NOTE: Practice personal hygiene protocols at all times
Reflection:
1. I learned that
2. I enjoyed most on
3. I want to learn more on
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References:
Ling, S.J., Loyola, J.S., Moebs, W. (2016). University Physics (Volume 1). Houston,
Texas: OpenStax
Resnick, D., Halliday, R. (2011). Fundamentals of Physics (9th Edition). Hoboken,
NJ: John Wiley & Sons Inc.
s
Young, H.D., Freedman, R.A., Ford, A.L. (2012). University Physics with Modern
Physics (13th Edition). San Francisco, CA: Pearson Education, Inc.
http://phet.colorado.edu/en/simulation/pendulum-lab
http://www.physicsclassroom.com/Physics--Interactives/Waves--and--Sound/Mass-on--a--Spring
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ANSWER KEY
Activity No. 1: Conceptual Questions
1. Case B has the highest frequency. Both springs have the same mass; only the
spring constant (k) is different. A spring with higher spring constant will have a shorter
period. Frequency and period are inversely related. The highest frequency will have
the shortest period.
2. Case A has the highest frequency. Both springs have the same spring constant
(k) but different mass. A spring with the greater the mass, has a longer
period.Frequency and period are inversely related. The highest frequency will have
the shortest period.
3. Pendulum B. The mass of the bob is not an important variable; only the length of
the string will affect the period (and thus the frequency). Frequency and period are
inversely related. Thus, the pendulum with the shorter string will have a higher
frequency of vibration.
Activity No. 2: Problem Set
1. Given:
2. Given:
k= 120 N/m
k= 75 N/m
m= 0.5 kg
m=0.70 kg
T=?
f=?
Calculation:
𝑚
T= 2𝜋√ 𝑘
Calculation
1
f=
0.5 𝑘𝑔
2𝜋
√
1
𝑘
𝑚
75 𝑁/𝑚
√ 0.70 𝑘𝑔
2𝜋
T= 2𝜋√120 𝑁/𝑚
f=
T= 𝟎. 𝟒𝟏 𝒔
f= 1.65 Hz
3. Given:
4. Given:
m= 2.0 kg
k= 120 N/m
k= 85 N/m
f= 6.0 Hz
f=?
T=? , ꞷ=?, m=?
Calculation:
f=
1
2𝜋
√
𝑘
𝑚
Calculations:
1
1
𝑓
6.0 𝐻𝑧
T= =
= 0.17 s
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f=
1
85 𝑁/𝑚
ω = 2𝜋f= 2𝜋 (6.0 𝐻𝑧) = 37.7 rad/s
√ 2.0 𝑘𝑔
2𝜋
ꞷ= √
f= 1.04 Hz
𝑘
𝑚
𝑘
m=
5. Given:
= 0.08 kg
f=?
Calculation:
2𝜋
(37.7 𝑟𝑎𝑑/𝑠)2
g=9.8 m/s2
f2=?
√
120𝑁/𝑚
L= 5.00 cm→ 0.05 m
f= 1.33 Hz
1
=
6. Given:
m= 0.750 kg, 0.220kg
f=
𝜔2
Calculation:
𝑘
1
𝑔
√
2𝜋 𝐿
f=
𝑚
k= 4𝜋 2 f2m
f=
1
2𝜋
k= 4𝜋 2 (1.33 Hz)2(0.750 kg)
√
9.8 𝑚/𝑠 2
0.05 𝑚
f= 2.23 Hz
k= 52.4 N/m
m2= 0.750 kg + 0.220 kg= 0.97 kg
f2=
1
𝑘
1
52.4 𝑁/𝑚
√𝑚 = 2𝜋 √ 0.97 𝑘𝑔 = 1.17 Hz
2𝜋
2
7. Given:
L= 2.6 m
L= 1 m
g=9.8 m/s2
g= 1.7 m/s2
f=?
T=?
Calculation:
1
𝑔
f=
1
2𝜋
√
Calculation:
𝐿
√
2𝜋 𝐿
f=
8. Given:
T=2𝜋√𝑔
9.8 𝑚/𝑠 2
2.6 𝑚
1𝑚
T=2𝜋√1.7 𝑚/𝑠2
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f= 0.31 Hz
T= 4.82 s
9. Given:
10. Given:
L= 4.0 m
L= 3.1 m
g= 9.8 m/s2
g= 9.8 m/s2
Calculation:
Calculation:
𝟐𝑳
𝟑𝒈
T= = 𝟐𝝅√
𝟐𝑳
𝟑𝒈
T= = 𝟐𝝅√
𝟐 (𝟒.𝟎 𝒎)
𝟑 (𝟗.𝟖 𝒎/𝒔𝟐 )
T= = 𝟐𝝅√
T= 3.28 s
T= 2.86 s
T= = 𝟐𝝅√
𝟐 (𝟑.𝟏 𝒎)
𝟑 (𝟗.𝟖 𝒎/𝒔𝟐 )
Activity No. 3: Mass on Spring Interactive
Sample Data:
Factor # 1
Factor:
Mass
# of
Cycles
Start Time
(s)
Stop Time
(s)
Period (s)
Frequency
(Hz)
1 kg
1
0.2
0.77
0.57
1.75
2 kg
1
0.1
0.92
0.72
1.39
3 kg
1
0.1
1.09
0.99
1.01
4 kg
1
0.02
1.32
1.3
0.77
(Note: Spring stiffness and dumping should be constant)
Factor # 2
Factor:
Spring stiffness
# of
Cycles
Start Time
(s)
Stop Time
(s)
Period (s)
Frequency
(Hz)
1
1
0.43
1.32
0.89
1.12
2
1
0.02
0.63
0.61
1.64
3
1
0.4
0.93
0.53
1.89
4
1
0.29
0.68
0.39
2.56
(Note: Mass and dumping should be constant. Spring stiffness is the spring
constant.)
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Conclusion:
"As the mass increases, the period increases and the frequency decreases.”
"As the spring stiffness increases, the period decreases and the frequency
increases.”
Activity No. 4: Pendulum Lab
Predictions:
1. Students’ responses may vary.
2. Students’ responses may vary.
Explore!
Part 1: Mass and Number of Swings
Sample Hypothesis:
As the mass of the pendulum increases the number of swings increases.
As the mass of the pendulum increases the number of swings decreases.
As the mass of the pendulum increases the number of swings remains the same.
Sample Data:
Pendulum 1
Pendulum 2
Mass (kg)
Length (m)
1
1.5
0.70
0.70
Number of
swings for 30
seconds
15
15
Conclusion:
Based on the recorded data, it is concluded that mass does not affect the number of
full swings of a pendulum.
Part 2: Length and number of swings
Sample Hypothesis:
As the length of the pendulum increases the number of swings increases.
As the length of the pendulum increases the number of swings decreases.
As the length of the pendulum increases the number of swings remains the same.
Sample Data:
Length of the
pendulum (m)
Period (s)
Number of swings in
30 sec
0.5
0.80
18
1
2.25
13
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Conclusion:
Based on the recorded data, it is concluded that length affects the number of full
swings of a pendulum. The shorter the length, the shorter the period and the greater
the number of swings. The longer the length, the longer the period and the lesser
number of swings.
Prepared by:
IVY MISTICA A. VILLANUEVA
Casambalangna national High School
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GENERAL PHYSICS 1
Name: _______________________________
Grade Level: ____________
Date: _______________________________
Score: _________________
LEARNING ACTIVITY SHEET
DAMPED OSCILLATION
Background Information for the Learners (BIL)
In the real world, oscillations seldom follow true SHM. Friction of some sort
usually acts to dampen the motion, so it dies away, or needs more force to continue.
There are no non-conservative forces, the total mechanical energy is constant, and
a system set into motion continues oscillating forever with no decrease in amplitude.
A guitar string stops oscillating a few seconds after being plucked. To keep
swinging on a playground swing, you must keep pushing. Although we can often
make friction and other non-conservative forces small or negligible, completely
undamped motion is rare. In fact, we may even want to damp oscillations, such as
with car shock absorbers.
Real-world systems always have some
dissipative forces, however, and oscillations die out
with time unless we replace the dissipated
mechanical energy (Fig. 1). A mechanical pendulum
clock continues to run because potential energy
stored in the spring or a hanging weight system
replaces the mechanical energy lost due to friction in
the pivot and the gears. But eventually the spring
runs down, or the weights reach the bottom of their
travel. Then no more energy is available, and the
pendulum swings decrease in amplitude and stop.
The decrease in amplitude caused by dissipative
forces is called damping, and the corresponding
motion is called damped oscillation.
Figure 1: A swinging bell left to itself
Figure 2 shows a mass m attached to a spring will eventually stop oscillating due to
with a force constant k. The mass is raised to a
damping forces (air resistance and
position A0, the initial amplitude, and then released.
friction at the point of suspension).
The mass oscillates around the equilibrium position
in a fluid with viscosity, but the amplitude decreases Source: http://wccsystems.com/
201
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for each oscillation. For a system that has a small amount of damping, the period and
frequency are constant and are nearly the same as for SHM, but the amplitude
gradually decreases as shown. This occurs because the non-conservative damping
force removes energy from the system, usually in the form of thermal energy.
Consider the forces acting on the mass. Note that the only contribution of the
weight is to change the equilibrium position, as discussed earlier in the chapter.
Figure 2.For a mass on a spring oscillating in a viscous fluid, the period remains
constant, but the amplitudes of the oscillations decrease due to the damping caused by
the fluid.
Source:https://cnx.org/contents/ffd9e0b7-b2ac-495e-b20b-e2a4a3444150@7/DampedOscillations
Therefore, the net force is equal to the force of the spring and the damping force
(FD). If the magnitude of the velocity is small, meaning the mass oscillates slowly,
the damping force is proportional to the velocity and acts against the direction of
motion (FD=−b). The net force on the mass is therefore
ma=−bv−kx.
(Eq. 1)
Writing this as a differential equation in x, we obtain
𝑑2𝑥
m
𝑑𝑡 2
+𝑏
𝑑𝑥
𝑑𝑡
+ 𝑘𝑥 = 0
(Eq. 2)
To determine the solution to this equation, consider the plot of position versus time
shown in Figure 3The curve resembles a cosine curve oscillating in the envelope of
𝑏
an exponential function A0e-αt where α=2𝑚. The solution is
𝑏
𝑥(𝑡) = 𝐴0 𝑒 −2𝑚𝑡 cos(𝜔𝑡 + 𝜙)
(Eq. 3)
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𝑘
𝑏
2
As b increases, 𝑚 − (2𝑚) becomes smaller and eventually reaches zero when
b = √4𝑚𝑘. If b becomes any larger,
𝑘
𝑏
2
− (2𝑚) becomes a negative number and
𝑚
2
√ 𝑘 − ( 𝑏 ) is a complex number.
𝑚
2𝑚
Figure 4. The position versus time for three systems consisting of a mass and a spring
in a viscous fluid. (a) If the damping is small (b < √4mk), the mass oscillates, slowly
losing amplitude as the energy is dissipated by the non-conservative force(s). The
limiting case is (b) where the damping is (b = √4mk). (c) If the damping is very large
(b > √4mk), the mass does not oscillate when displaced, but attempts to return to
the equilibrium position.
Source: https://s3-us-west-2.amazonaws.com/courses-images/wpcontent/uploads/sites/2952/2018/01/31200736/CNX_UPhysics_15_06_DampedOscB.jpg
If you gradually increase the amount of damping in a system, the period and
frequency begin to be affected, because damping opposes and hence slows the back
and forth motion. (The net force is smaller in both directions.) If there is very large
damping, the system does not even oscillate—it slowly moves toward equilibrium.
Figure 4 shows the displacement of a harmonic oscillator for different amounts of
damping. When we want to damp out oscillations, such as in the suspension of a car,
we may want the system to return to equilibrium as quickly as possible Critical
damping is defined as the condition in which the damping of an oscillator results in it
returning as quickly as possible to its equilibrium position The critically damped
system may overshoot the equilibrium position, but if it does, it will do so only once.
Critical damping is represented by Curve B in Figure 4. With less-than critical
damping, the system will return to equilibrium faster but will overshoot and cross over
one or more times. Such a system is underdamped; its displacement is represented
by the Curve A in Figure 4. Curve C in Figure 4 represents an overdamped system.
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As with critical damping, it too may overshoot the equilibrium position, but will reach
equilibrium over a longer period of time.
Figure 4 shows the displacement of a harmonic oscillator for different amounts
of damping.
1. When the damping constant is small, b < √4𝑚𝑘, the system oscillates while
the amplitude of the motion decays exponentially. This system is said to
be underdamped, as in curve (a). Many systems are underdamped, and
oscillate while the amplitude decreases exponentially, such as the mass
oscillating on a spring. The damping may be quite small, but eventually the
mass comes to rest.
2. If the damping constant is b = √4𝑚𝑘, the system is said to be critically
damped, as in curve (b). An example of a critically damped system is the
shock absorbers in a car. It is advantageous to have the oscillations decay as
fast as possible. Here, the system does not oscillate, but asymptotically
approaches the equilibrium condition as quickly as possible.
3. Curve (c) in Figure 4 represents an overdamped system where b >√4𝑚𝑘. An
overdamped system will approach equilibrium over a longer period of time.
Critical damping is often desired, because such a system returns to
equilibrium rapidly and remains at equilibrium as well. In addition, a constant force
applied to a critically damped system moves the system to a new equilibrium position
in the shortest time possible without overshooting or oscillating about the new
position. For example, when you stand on bathroom scales that have a needle
gauge, the needle moves to its equilibrium position without oscillating. It would be
quite inconvenient if the needle oscillated about the new equilibrium position for a
long time before settling. Damping forces can vary greatly in character. Friction, for
example, is sometimes independent of velocity (as assumed in most places in this
text). But many damping forces depend on velocity—sometimes in complex ways,
sometimes simply being proportional to velocity.
Sample Problem:
A 2-kilogram mass attached to a spring of spring constant k= 10 N/m
oscillates through a fluid that exerts a damping force F d = - (4 N ꞏ s/m) v on the mass,
where v is the velocity of the mass. What is the oscillatory behavior of the system?
Solution:
Given:
m= 2kg.
k = 10N/m
b or Fd = 4 N.s/m
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Find: Oscillatory behavior (underdamped, critically damped or overdamped)
Equation:b = √4𝑚𝑘
Since all units are in the standard SI we omit them here.
(b)2= (√4𝑚𝑘)2
b2 = 4mk
Substitute the given values to the equation
(4)2 = (4)(2)(10)
16 < 80
Since b2< 4km, the system is underdamped; there is not enough damping to stop the
oscillation due to the large spring force and mass.
Figure 5. An automobile shock
absorber. The viscous fluid causes a
damping force that depends on the
relative velocity of the two ends of the
unit.
Source:https://en.wikipedia.org/wiki/Sho
ck_absorber
In a vibrating tuning fork or guitar
string, it is usually desirable to have as little
damping as possible. By contrast, damping
plays a beneficial role in the oscillations of an
automobile’s suspension system. The shock
absorbers provide a velocity dependent
damping force so that when the car goes over
a bump, it does not continue bouncing forever
(Fig. 5). For optimal passenger comfort, the
system should be critically damped or slightly
underdamped. Too much damping would be
counterproductive; if the suspension is
overdamped and the car hits a second bump
just after the first one, the springs in the
suspension will still be compressed
somewhat from the first bump and will not be
able to fully absorb the impact.
Summary
•
Damped harmonic oscillators have non-conservative forces that dissipate their
energy.
•
Critical damping returns the system to equilibrium as fast as possible without
overshooting.
•
An underdamped system will oscillate through the equilibrium position.
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•
An overdamped system moves more slowly toward equilibrium than one that is
critically damped.
Since different damped oscillations were already discussed in the first part of
this module, there are different learning activities which were prepared in order to
test your understanding with regards to the topic. Let’s get started!
Learning Competency:
Differentiate underdamped,
(STEM_GP12PM-IId-28)
overdamped,
and
critically
damped
motion
Activity 1: Concept Check
Direction: Answer the following questions completely
1. Why are completely undamped harmonic oscillators so rare?
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
2. Describe the difference between overdamping, underdamping, and critical
damping.
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
3. Most harmonic oscillators are damped and, if undriven, eventually come to a
stop. How is this observation related to the second law of thermodynamics?
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
4. Consider a door that uses a spring to close the door once open. How can this
lead to any of the above types of damping depending on the strength of the
damping?
a. Overdamping
_______________________________________________________
_______________________________________________________
_______________________________________________________
b. Underdamping
_______________________________________________________
_______________________________________________________
_______________________________________________________
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c. Critical damping
_______________________________________________________
_______________________________________________________
_______________________________________________________
Activity 2: Testing your Understanding!
Directions: Encircle the letter that corresponds to the correct answer.
1. What is the name of the quantity represented by the symbol ω?
a. Angular momentum
b. Angular frequency
c. Phase Constant
d. Uniform Circular
2. What term is used to describe an oscillator that “runs down” and eventually
stops?
a. Tired oscillator
b. Out of shape oscillator
c. Damped Oscillator
d. Driven oscillator
3. The damping force on an oscillator is directly proportional to the velocity. What
is the units of the constant of proportionality?
a. Kg m s-1
b. Kg m s-2
c. Kg s-1
d. Kg s
4. When a damped harmonic oscillator completes 100 oscillations, its amplitude
is reduced to 1/3 of its initial value. What will be its amplitude when it
completes 200 oscillations?
a.1/5
c. 1/6
b. 2/3
d. 1/9
5. In case of a forced vibration, why is it that the resonance wave becomes very
sharp?
a. Applied periodic force is small
b. Quality factor is small
c. Damping force is small
d. Restoring force is small
6. Energy is supplied to the damped oscillatory system at the same rate at which
it is dissipating energy, then the amplitude of such oscillations would become
constant. What do we call such oscillations?
a. Damped oscillations
c. Coupled oscillations
b. Undamped oscillations
d. Maintained oscillations
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7. What happens to the amplitude of Simple Harmonic Motion at resonance
when an ideal case of zero damping?
a. Maximum
c. zero
b. Minimum
d. Infinite
8. Under what condition angular frequency, ω of the damped oscillator would be
equivalent to the angular frequency, ω0 of the undamped oscillator?
a. Velocity of oscillator is small
b. Damping constant, b is small
c. Damping constant, b is large
d. Force applied is small
9. What determines the natural frequency of a body?
a. Position of the body with respect to force applied
b. Elastic properties and dimensions of the body
c. Mass and speed of the body
d. Oscillations of the body
10. In what way should the angular frequency of a damped system must relate
to the angular frequency of the corresponding simple harmonic system?
a. The frequency of the simple harmonic system must be larger
b. The frequency of the damped system must be larger
c. They must be equal
d. Choices A, B, and C
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Activity 3: Venn Diagram
Directions: Using the Venn Diagram, identify the similarities and differences of
underdamped, critically damped, and overdamped. You can search additional
information about interference and diffraction via textbook, video, books on tape,
classroom library, school library, and or Internet. Remember to cite the references
you used. Refer to the Rubrics below on how your Graphic Organizer will be graded.
Venn Diagram
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Rubrics
CRITERIA
Full Credit
(20 points)
- Compares &
contrasts
items clearly
- Only includes
relevant and
accurate
information
Partial Credit
(15 points)
- Compares
and contrasts
clearly, but
supporting
information is
general
- Only
includes
relevant
information
- Whole-towhole
similarities
- Whole-towhole
differences
Organization
- Similarities& Structure
to-differences
- Consistent
order when
discussing the
comparison
- Beaks
information
into one of the
structures
- Does not
follow
consistent
order when
comparing
Purpose &
Supporting
Details
Grammar &
Spelling
Transitions
- No errors in
grammar
or spelling
- Moves
smoothly from
one idea to the
next
- Comparison
and contrast
transition
words to show
relationships
- Variety of
sentence
structures &
transitions
- 1-2 errors in
grammar or
spelling that
distract the
reader
- Moves from
one idea to
the next, but
with little
variety
- Uses
comparison
and contrast
transition
words to
show
relationships
between
ideas
Limited Credit
(10 points)
Minimal Credit
(5 points)
- Compares
and contrasts
clearly, but
supporting
information is
incomplete.
- May include
irrelevant
information
- Compares or
contrasts, but
does not do
both
- No supporting
information, or
incomplete
information
- Breaks
information
into structure,
but some
information
is in wrong
section
- Some details
are not in
logical or
expected order
- Many details
are not in
logical
order
- Little sense
that the writing
is organized
- 3-4 errors
that distract
the reader
- Some
transitions
work well, but
connections
between other
ideas are fuzzy
RATING
- Excessive
errors that
distract the
reader from the
content
- Transitions
are unclear or
nonexistent
TOTAL:
210
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Activity 4: Real-life Application of Damped Oscillations
Directions: Every phenomenon that is happening around us have a Physics
Concept behind it. There are phenomena that are application of Damped
Oscillations. Give at least three application for each. Give a brief explanation how
that damped oscillations applies the given phenomena. An example for each is
provided below as your reference. Refer to the Rubrics below on how output will be
graded.
Note: Cite the references used in this activity.
Underdamped
Critically damped
Oscillation
Oscillation
Automatic
door
Overdamped Oscillation
and The automobile shock
window closer is under absorber is an example of
damped it will close with a critically damped
considerable velocity or the device. As a car goes over
a bump, the spring in its
door will swing too and fro
shock-absorber assembly is
before closing at its normal
compressed, but the elastic
positions. Here the system potential energy of the
Any example of public
transportation braking
systems would be good
examples of overdamped
where the desire is to
oscillates with a gradual spring immediately forces it
provide the rider with
decrement to zero
back to a position of
comfort over the speed of
Source:
equilibrium, thus ensuring
https://www.quora.com/What
-are-over-damped-criticallyand-under-damped-
that the bump is not felt
throughout the entire
vehicle. However, springs
coming to a stop. Like a
train, elevator or
automobile.
alone would make for a
systems#:~:text=Underdampe
bouncy ride; hence, a
Source:
d-
modern vehicle also has
https://physics.stackexchange.
,The%20system%20oscillates%
shock absorbers. The shock
com/questions/353000/what-
20(at%20reduced%20frequenc
absorber, a cylinder in
are-practical-uses-of-over-
y%20compared%20to%20the
which a piston pushes down
damping
%20undamped,amplitude%20
on a quantity of oil, acts as
gradually%20decreasing%20to
a damper—that is, an
%20zero.&text=The%20syste
inhibitor of the springs'
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m%20oscillates%20at%20its,cl
oscillation.
ose%20the%20door%20once%
Read
20open.
more: http://www.sciencecla
rified.com/everyday/RealLife-Physics-Vol2/Oscillation-Real-lifeapplications.html#ixzz6Snck
auR5
1.
1.
1.
2.
2.
2.
3.
3.
3.
Rubrics
CRITERIA
LEVEL 1
(1 point)
LEVEL 2
(3 points)
The paper is
organized,
makes good use
of transition
statements and
in most
instances
follows a logical
progression.
ORGANIZATION
The paper is poorly
organized and
difficult to follow.
COMPLETION
One product was
only given and
explained.
Two products
were given with
explanation.
Grammar &
Spelling
More than 5 errors
in punctuation and
spelling.
3-5 errors in
punctuation and
spelling.
LEVEL OF
CONTENT
Shows some
thinking and
reasoning but most
ideas are
underdeveloped
and
unoriginal.
Content
indicates
original
thinking and
develops ideas
with sufficient
and firm
evidence.
LEVEL 3
(5 points)
RATING
The paper is well
organized, uses
transition
statements
appropriately
and follows a
logical
progression.
Three or more
products were
given with
explanation.
Minimal errors in
punctuation and
spelling.
Content
indicates
synthesis of
ideas, in depth
analysis and
evidences
original
thought and
support for the
topic.
TOTAL:
212
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Reflection:
1. I learned that
_______________________________________________________________
_______________________________________________________________
_______________________________________________________________
__
2. I enjoyed most on
_______________________________________________________________
_______________________________________________________________
_______________________________________________________________
__
3. I want to learn more on
_______________________________________________________________
_______________________________________________________________
_______________________________________________________________
___
213
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References
Damped Harmonic Motion. https://courses.lumenlearning.com/physics/chapter/167-damped-harmonicmotion/#:~:text=An%20overdamped%20system%20moves%20slowly,without%20o
scillating%20about%20the%20equilibrium.
Harmonic Oscillations and Damping. http://ipl.physics.harvard.edu/wpuploads/2013/03/Lab5_2011.pdf
Damped Oscillations.
https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_P
hysics_(OpenStax)/Map%3A_University_Physics_I__Mechanics%2C_Sound%2C_Oscillations%2C_and_Waves_(OpenStax)/15%3A_
Oscillations/15.06%3A_Damped_Oscillations
214
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Answer Key:
Activity 1
1. Friction often comes into play whenever an object is moving. Friction causes
damping in a harmonic oscillator.
2. An overdamped system moves slowly toward equilibrium. An underdamped
system moves quickly to equilibrium but will oscillate about the equilibrium
point as it does so. A critically damped system moves as quickly as possible
toward equilibrium without oscillating about the equilibrium.
3. The second law of thermodynamics states that perpetual motion machines
are impossible. Eventually the ordered motion of the system decreases and
returns to equilibrium.
4. a. Finally, if it is overdamped it will return to closed without oscillating but
more slowly depending on how overdamped it is.
b. If it is underdamped it will swing back and forth with decreasing size of the
swing until it comes to a stop.
c. If it is critically damped, then it will return to closed as quickly as possible
without oscillating.
Activity 2
1.
2.
3.
4.
5.
b
c
c
d
c
6. d
7. d
8. b
9. b
10. a
Activity 3
•
Students’ output may vary. See attached Rubrics below the activity for
scoring purposes.
Activity 4
•
Students’ output may vary. See attached Rubrics below the activity for
scoring purposes.
Prepared by:
JENNY VHIE S. VINAGRERA
Licerio Antiporda Sr. National High School- Main
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NOTE: Practice personal hygiene protocols at all times
GENERAL PHYSICS 1
Name: ____________________________
Grade Level: _________
Date: _____________________________
Score: ______________
LEARNING ACTIVITY SHEET
MECHANICAL WAVES
Background Information for the Learners (BIL)
A wave is a transfer of energy through a medium from one point to another.
Some examples of waves include; water waves, sound waves, and radio waves.
Waves come in two different forms; a Transverse Wave which moves the medium
perpendicular to the wave motion, and a Longitudinal Wave, which moves the
medium parallel to the wave motion.
Waves have several properties which are represented in the diagrams below.
In a Transverse wave the Crest and Troughs are the locations of maximum
displacement up or down. The Amplitude is the measurement of maximum
displacement. The Wavelength is the distance of one complete wave cycle. For
example; the distance from crest to crest or trough to trough would be 1 wavelength.
In a Longitudinal wave, areas of maximum displacement are known as
Compressions and Rarefactions. The stronger the wave, the more compressed and
spread out the wave medium becomes.
Transverse Wave
Longitudinal Wave
Compressions
Rarefactions
Mechanical Waves are waves which propagate through a material medium
(solid, liquid, or gas) at a wave speed which depends on the elastic and inertial
properties of that medium. There are two basic types of wave motion for mechanical
waves: longitudinal waves and transverse waves.
Longitudinal Waves
In a longitudinal wave the particle displacement is parallel to the direction of
wave propagation. The illustration below shows a one-dimensional longitudinal plane
wave propagating down a tube. The particles do not move down the tube with the
wave; they simply oscillate back and forth about their individual equilibrium positions.
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The wave is seen as the motion of the compressed region (ie, it is a pressure wave),
which moves from left to right.
Longitudinal Wave
Examples of longitudinal waves include:
A. Sound waves
https://www.tuttee.co/blog/phys-sound-waves
Sound waves in air (and any fluid medium) are longitudinal waves because
particles of the medium through which the sound is transported vibrate parallel to the
direction that the sound wave moves.
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B. Ultrasound Waves
Ultrasound is sound waves with frequencies higher than the upper audible limit of
human hearing. Ultrasound is not different from "normal" (audible) sound in its physical
properties, except that humans cannot hear it.
Did you know?
Dolphins and porpoises use echolocation for
hunting and orientation. By sending out highfrequency sound, known as ultrasound, dolphins
can use the echoes to determine what type of object
the sound beam has hit.
C. Seismic P-waves
A Seismic P wave, or compressional wave, is a seismic body wave that
shakes the ground back and forth in the same direction and the opposite direction as
the direction the wave is moving.
https://earthquake.usgs.gov
218
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Transverse Waves
A transverse wave is a wave in which particles of the medium move in a
direction perpendicular to the direction that the wave moves. Suppose that a slinky
is stretched out in a horizontal direction across the classroom and that a pulse is
introduced into the slinky on the left end by vibrating the first coil up and down. Energy
will begin to be transported through the slinky from left to right. As the energy is
transported from left to right, the individual coils of the medium will be displaced
upwards and downwards. In this case, the particles of the medium move
perpendicular to the direction that the pulse moves. This type of wave is a transverse
wave. Transverse waves are always characterized by particle motion being
perpendicular to wave motion.
Examples of transverse waves include:
A. Ripples on the surface of water
The circular ripples produced on the
surface
of
the
water
expand
and
propagate through water. As the ripples
move horizontally across the surface of
water, the water particles vibrate up and
down. Thus, the water waves (ripples)
propagate horizontally, the particles of the
medium (water) vibrate perpendicular to
Google image
the direction of wave propagation.
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B. Vibrations in a guitar string
Plucking the string gave it energy,
which is moving through the string in a
mechanical wave. A mechanical wave
is a wave that travels through matter.
The matter a mechanical wave travels
Google image
through is called the medium. The type
of mechanical wave passing through
the
vibrating
guitar
string
is
a
transverse wave.
C. Seismic S-waves
S wave or secondary wave is the second
wave you feel in an earthquake. An S
wave is slower than a P wave and can
only move through solid rock, not
through any liquid medium. It is this
property
of
S
seismologists to
waves
that
conclude that
led
the
Earth's outer core is a liquid. S waves
move rock particles up and down, or
side-to-side--perpendicular
to
the
direction that the wave is traveling in (the
FIGURE ON THE RIGHT - AN S WAVE TRAVELS THROUGH A
MEDIUM.
PARTICLES
ARE REPRESENTED
BY CUBES IN THIS
direction
of wave
propagation).
MODEL. IMAGE ©2000-2006 LAWRENCE BRAILE.
Periodic Waves
A periodic wave is a wave with a repeating continuous pattern which
determines its wavelength and frequency. It is characterized by the amplitude, a
period (T) and a frequency(f). Amplitude wave is directly related to the energy of a
wave, it also refers to the highest and lowest point of a wave. Period defines as time
required to complete cycle of a waveform and frequency is number of cycles per
second of time.
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Periodic Wave Relationships
The relationship “distance = velocity x time” is the basic wave relationship.
With the wavelength as distance, this relationship becomes
using
. Then
gives the standard wave relationship.
Examples :
1. A radio wave has a frequency of 93.9 MHz (93.9 x106 Hz). What is its period?
–answer
2. A wave is traveling at a velocity of 12 m/s and its wavelength is 3m. Calculate the
wave frequency.
– answer
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Learning Competency:
Define mechanical waves, longitudinal wave, transverse wave, periodic wave, and
sinusoidal wave (STEM_GP12PM-IId-31)
ACTIVITY 1. Make Some Waves!
Directions: Answer the following questions.
1) What is a wave?
_______________________________________________________________
_______________________________________________________________
_______________________________________________________________
2) Describe a difference between longitudinal and transverse waves.
_______________________________________________________________
_______________________________________________________________
_______________________________________________________________
3) Give one example of a longitudinal wave and one example of a transverse
wave.
_______________________________________________________________
_______________________________________________________________
_______________________________________________________________
4) Think about the gold coin Angie and Harmon found on the sea floor. What kind
of wave behavior would bring a gold coin close to shore?
_______________________________________________________________
_______________________________________________________________
_______________________________________________________________
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Activity 2. Transverse and Longitudinal Waves
Directions: Answer the questions about transverse and longitudinal waves.
1. What kind of wave is pictured above?
Answer: ________________________
2. Label the following on the wave above: crest, trough, wavelength, amplitude,
direction of travel.
3. In what direction would the particles in this wave move, relative to the direction of
wave travel?
Answer: ________________________
4. What kind of wave is pictures above?
Answer: ________________________
5. Label the following on the wave above: compression, rarefaction, wavelength,
direction of travel.
6. In what direction would the particles in this wave move, relative to the direction of
wave travel?
Answer: ________________________
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Directions: For each wave described below, identify the wave as more like
transverse wave or a longitudinal wave.
7. The wave created by moving the end of a spring toy up and down.
Answer: __________________________
8. The wave created by moving the end of a spring toy back and forth parallel to the
length of the spring.
Answer: __________________________
9. A sound wave.
Answer: __________________________
10. An electromagnetic wave.
Answer: __________________________
Activity 3. Through the Waves!
Directions: Solve the following problems. Show your complete solution and
encircle your final answer.
1. A swimmer at the beach notices that three wave crests pass a certain point every
10.0 seconds. She also notes that each wave crest is about 2.0 meters apart.
a. What is the period of the wave that the swimmer is observing?
b. What is the frequency of the wave that the swimmer is observing?
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c. What is the speed of the waves that the swimmer is observing?
2. A submarine trying to detect an enemy destroyer notes that a sonar signal sent
through the water returns 0.40 seconds after it was sent. The frequency of the
sonar used by the submarine is 20 kilo-hertz. The speed of sound in sea water
is 1.56 x 103 meters per second.
e. How far away is the destroyer?
f. The sonar computers receive a reflection from the destroyer at a frequency of 19
kilo-hertz. What useful information about the motion of the destroyer does this
mean the computer can report?
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Reflection:
1. I learned that ______________________________________________________
__________________________________________________________________
____________________________________________________________
2. I enjoyed most on __________________________________________________
__________________________________________________________________
__________________________________________________________
3. I want to learn more on ______________________________________________
__________________________________________________________________
__________________________________________________________
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References
(1)Daniel A. Russell, 2016, Acoustic and Vibrations Animations, Pennsylvania.
(2)https://www.paulding.k12.ga.us/cms/lib/GA01903603/Centricity/Domain/2519/W
aveIntroductionWaveTypesWaveFrequency.pdf
(3) https://www.physicsclassroom.com/class/waves/Lesson-1/Categories-of-Waves
(4) http://www.geo.mtu.edu/UPSeis/waves.html
(5) https://www.eeweb.com/periodic-wave/
(6)https://www.teachengineering.org/activities/view/cub_soundandlight_lesson1_ac
tivity1
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Answer Key
Activity 1 (Students’ answers may vary)
Activity 2
1. Transverse wave
2. Student’s answer may vary
3. The particles move perpendicular to the direction of wave propagation
4. Longitudinal wave
5. Student’s answer may vary
6. The particles move parallel to the direction of wave propagation
7. Transverse wave
8. Longitudinal wave
9. Longitudinal wave
10. Transverse wave
Activity 3
a. . 3.3 s
b. 0.3 Hz
c. 0.6 m/s
d. 66.2 m
Prepared by:
ALDRIN FIGUEROA GRAGEDA
Pattao National High School- Main
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GENERAL PHYSICS 1
Name:___________________________
Grade Level:__________
Date:____________________________
Score:_______________
LEARNING ACTIVITY SHEET
SINUSOIDAL WAVE FUNCTION: Speed, Wavelength, Frequency,
Period, Direction and Wave number.
Background Information for the Learners (BIL)
The sinusoidal wave is the simplest example of a periodic continuous waves, it
can also be defined as a smooth repetitive oscillation. (Oscillation: act of regularly
moving from one position to another and back to the original position. This is
manifested through the string instruments, try strumming a guitar and take a video of
the string from the inside of the guitar. Observed the string as it oscillates, you can
see it forms little waves---sinusoidal waves).
The Sinusoidal Wave Equation is expressed as:
But before we break down the sine wave equation, let us first differentiate
between the motion of the wave and the motion of the elements of the medium.
Take a look at the figures below.
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Figure (a) shows a snapshot of a wave moving through a medium. While figure
(b) shows a graph of the position of one element of the medium as a function of time.
The point at which the displacement of the element from its normal position is highest
is called the crest of the wave. The distance from one crest to the next is called the
wavelength. More generally, the wavelength is the minimum distance between any
two identical points (such as the crests) on adjacent waves, as shown in figure (a).
If you count the number of seconds between the arrivals of two adjacent crests
at a given point in space, you are measuring the period T of the waves. In general,
the period is the time interval required for two identical points (such as the crests) of
adjacent waves to pass by a point. But what really is the difference between the two
figures? Notice the visual similarity between figures (a) and (b). The shapes are the
same, but (a) is a graph of vertical position versus horizontal position while (b) is
vertical position versus time. figure (a) is a pictorial representation of the wave for a
series of particles of the medium— this is what you would see at an instant of time.
Figure (b) is a graphical representation of the position of one element of the medium
as a function of time. The fact that both figures have the identical shape represents
Sinusoidal Wave Equation,a wave is the same function of both x (distance) and
t(time).
What’s in the Sinusoidal Wave Equation?
y = the height of wave at position (x) and
time (t); it is in meters where x is the
distance along the piece of string or along
the x-axis (also in meters) and t is the time
in seconds.
A = the amplitude which is also measured in
meter.
k = the wave number, the unit is rad/m. The wave number can be
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calculated
from the wavelength using the equation;
ω = the angular frequency; unit used is rad/s. Can be calculated using this equations;
ω =2πf (where f is measured in hertz (hz) or second-1)
phi is the phase constant or phase shift and is defined as; how far the function is
shifted horizontally from the usual position. The phase shift of a sine curve is how
much the curve shifts from zero. If the phase shift is zero, the curve starts at the
origin, but it can move left or right depending on the phase shift. A negative phase
shift indicates a movement to the right, and a positive phase shift indicates movement
to the left.
Note that in the given equation of sinusoidal wave above, where;
is a wave moving to the right. For a wave moving to the left, x -ω t will then change
to x + ω t.
SPEED
Using the equation;
Recall
that;  = 2f
and
Formula derivation:
We know that,
From the formula  = 2f
and
we can derive it to get the formula for frequency and wavelength.
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Getting
frequency,
both sides by
using this formula, 2
divide
and you will get
f =  / 2
Getting
wavelength,
 = 2 / 
using this
formula, cross multiply and you will get
It’s time to plug in the derive formulas with this equation;
to get equation or formula for of wave’s speed using wave number and angular
frequency.
v=
v
(2 /  ) ( / 2 )
=
 /
(with a unit in m/s)
WAVELENGTH
We may use
to get the formula for wavelength which is,
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FREQUENCY
Again, we may use this equation
and arrive with
f = v/
or
f = c/
Note that the value of c depends on the medium. Speed of sound in air at a
temperature of 20°C: c = 343 m/s or speed of radio waves and light in a vacuum: c =
299,792,458 m/s. (Speed of sound c = 343 m/s also equates to 1235 km/h, 767 mph,
1125 ft/s.)
Or we may also use
f = 1/ T
where T is the period (cycle duration of wave).
PERIOD
The period of the sine curve is the length of one cycle of the curve or can be defined
as the distance between two consecutive maximum points, or two consecutive
minimum points (these distances must be equal). The natural period of the sine curve
is 2π.
To better understand period look at the waves below,
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The red wave has the shortest period.
The green and black waves have equal periods. (Even though the green wave has
greater amplitude than the black wave, they both have the same period.)
The blue wave has the longest period.
Period can be calculated using any of these two equations;
T = 1/ f
WAVE NUMBER
The wave number is related to the angular frequency by:
where λ (lambda) is the wavelength, f is the frequency, and v is the linear speed.
Thus wave number’s equation is,
Let us try an example!
The equation of a wave is given by: y=(x,t) = 2.0m sin(3.0x-4.0t+π/2)
Where x is in meter and time is in seconds.
Find the amplitude, frequency, wavelength, speed and initial height at x = 1.0m
Determine the direction of the wave.
ANSWER/SOLUTION:
y=(x,t) = 2.0m sin(3.0x-4.0t+π/2)
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Amplitude = 2.0m (note that it is already given in the equation)
Frequency,
f =  / 2
f = 4.0rads − 1 / 2rad
f =
Wavelength,
0.64s-1 or 0.64Hz
 = 2 / 
 = 2rad / 3.0radm − 1
 = 2.1m
Speed,
v = f
v = (0.64 s − 1)(2.1m)
v = 1.3m / s
Or;
v =  /
v = 4.0rad • s − 1 / 3.0rad • m − 1
v = 1.3m / s
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Note that, the formula to be used will depend on what value related to the unknown
is given.
Initial height at x=1.0m,
y(1.0m,0s ) = 2.0m sin (3.0rad / m •1.0m − 4.0rad / s • 0s +  / 2)
y(1.0m,0s )2.0m sin (4.57 )
Remember, that your calculator should be in
rad when calculating for this, because most of the units are in rad.
= −2m
The direction of the wave is to the right or in the positive direction, recall that in the
equation given kx and ωt have opposite sign.
Learning Competency:
From a given Sinusoidal Wave Function, infer the speed, wavelength, frequency,
period and then wave number. (STEM_GP12PMIId-32)
ACTIVITY 1: TRUE OR FALSE
Directions: Read and analyze the following statement, write true if the given
statement is correct and false if it is incorrect.
1. A wave source that oscillates with simple harmonic motion (SHM) generates a
sinusoidal wave.
Answer:_____________
2. The period of the wave is the frequency of the oscillating source.
Answer:_____________
3. The period T is related to the wave frequency f by;
f = 1/ T
Answer:_____________
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4. The amplitude A of the wave is the maximum value of the displacement. The crests
of the wave have displacement Dcrest = A and the troughs have displacement
Dtrough = −A.
Answer:_____________
5. The distance spanned by one cycle of the motion is called the amplitude of the
wave.
Answer:_____________
ACTIVITY 2: MATCH MY FORMULA CORRECTLY
Directions: Match the following terms with their corresponding formula.
____1. Speed
A.
____2. Wavelength
B.
____3. Frequency
C.
____4. Period
D.
v =  /
 = 2f
f = 1/ T
____5. Wave Number
E.
____6. Angular Frequency
F.
T = 1/ f
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ACTIVITY 3: WHAT IS MY UNIT?
Directions: Write the correct units for the following formulas.
(Show your dimensional analysis)
v =  /
 = 2f
f = 1/ T
T = 1/ f
ACTIVITY 4: CALCULATE MY UNKNOWN
Directions: Given the sinusoidal function, compute for what is being asked in the
problem. Solution must be complete with the inclusion of units.
1. A sinusoidal wave traveling in the positive x direction has an amplitude of 15.0 cm,
a wavelength of 40.0 cm, and a frequency of 8.00 Hz.The vertical position of an
element of the medium at t=0 and x=0 is also 15.0 cm.Find the wave number k,
period T, angular frequency ω, and speed v of the wave.
You’d use these familiar equations: k=2π/λ, T=1/f, ω=2πf, v=λf
2. A sinusoidal electromagnetic wave of frequency 40.0 MHz travels in free space in
the x direction, determine the wavelength and period of the wave.
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Reflection:
1.I learned that
__________________________________________________________________
__________________________________________________________________
______________________________________________________________
2.I enjoyed most on
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
3.I want to learn more on
__________________________________________________________________
__________________________________________________________________
___________________________________________________________
239
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References:
https://www.youtube.com/watch?v=UFt7vP7OBEE&t=78s
https://cass.ucsd.edu/~rskibba/work/Teaching_files/Phys1C_8April.pdf
Thomson_-_Physics_For_Scientists_And_Eng.pdf
Physics_serway.pdf
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Answer Key:
ACTIVITY 1: 1T, 2F, 3T,4T,5F
ACTIVITY 2: 1A, 2E, 3D, 4F,5C,6B
ACTIVITY 3: 1.m/s, 2. m, 3. Hz or s-1, 4. s 5. rad/m, 6.rad/s
ACTIVITY 4:
1.
2.
.
Prepared by:
ANGELIKA B. TORRES
Santa Ana Fishery National High School
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GENERAL PHYSICS 1
Name: ____________________________ Grade Level: _________________
Date: _________________________
Score: ______________________
LEARNING ACTIVITY SHEET
THE INVERSE SQUARE RELATION INTENSITY OF WAVES AND
DISTANCE
Background Information for Learners (BIL)
Inverse Square Law, General
Any point source which spreads its influence equally in all directions without a
limit to its range will obey the inverse square law. This comes from strictly geometrical
considerations. The intensity of the influence at any given radius r is the source
strength divided by the area of the sphere. Being strictly geometric in its origin, the
inverse square law applies to diverse phenomena. Point sources of gravitational
force, electric field, light, sound or radiation obey the inverse square law. It is a
subject of continuing debate with a source such as a skunk on top of a flag pole; will
its smell drop off according to the inverse square law?
http://hyperphysics.phy-astr.gsu.edu/hbase/Forces/isq.html
Applications of the inverse square law:
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Inverse Square Law, Radiation
As one of the fields which obey the general inverse square law, a point
radiation source can be characterized by the relationship below whether you are
talking about Roentgens , rads, or rems . All measures of exposure will drop off by
inverse square law.
The source is described by a general "source strength" S because there are
many ways to characterize a radiation source - by grams of a radioactive isotope,
source strength in Curies, etc. For any such description of the source, if you have
determined the amount of radiation per unit area reaching 1 meter, then it will be one
fourth as much at 2 meters.
Radiation Hazards
According to the Occupational Safety and Health Administration (OSHA),
radioactive sources are found in a wide range of industrial settings. Examples
include, non-destructive testing of metals through radiographic testing, hospital XRay imaging centers and nuclear power generation.
There are two types of radiation, ionizing and non-ionizing. Ionizing radiation has the
ability to change atoms exposed to it, which makes it a health concern to humans.
Radiation protection programs are focused on keeping each worker's occupational
radiation dose As Low As Reasonably Achievable (ALARA).
The three basic fundamentals in an ALARA program are time, distance and
shielding. Time being the length time a person is exposed to a source, shielding
being the placement of barriers between a person and the source and distance being
how far a person is from the source. This lesson will focus on distance and how to
calculate its effect on the intensity of a radioactive source.
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The inverse square law states the intensity of a source such as radiation,
changes in inverse proportion to the square of the distance from the source. To put
it in simpler terms, this means that as you move away from an energy source, the
strength decreases and the decrease is directly related to the distance from the
source. Since the intensity and distance are inversely related, you can calculate the
change in intensity as the distance changes. This is represented by the formula:
Specifically, the intensity is proportional to the inverse of the square of the
distance. For example, if the source is two times as far away then the intensity is one
divided by the square of two:
If the source is three times as far away the intensity is one divided by the square of
three:
Calculating Intensity
In an industrial setting the intensity of a radioactive source is typically known
for a specific distance. In order to calculate the effect on a population, you will need
to solve for the intensity based on the distance the population is from the source. The
equation to solve for the second distance, as taken from the inverse law is:
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Where:
I1 = Intensity 1 at Distance 1
I2 = Intensity 2 at Distance 2
D1 = Distance 1 from source
D2 = Distance 2 from source
Sample Calculation
Now that we have the equation, let's solve for the intensity of a radioactive
source at a second distance. For this example, we have a source with an intensity of
500,000 milliroentgen/hour at one foot. We need to calculate the intensity at 100 feet
away from the source where people might be working.
I1 = 500,000 mR/hr
I2 = ? mR/hr
D1 = 1 foot
D2 = 100 feet
Now, let’s rework the formula to solve for I2 (Intensity at Distance 2)
Inverse Square Law, Light
As one of the fields which obey the general inverse square law, the light from
a point source can be put in the form
where E is called illuminance and I is called pointance.
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The source is described by a general "source strength" S because there are
many ways to characterize a light source - by power in watts, power in the visible
range, power factored by the eye's sensitivity, etc. For any such description of the
source, if you have determined the amount of light per unit area reaching 1 meter,
then it will be one fourth as much at 2 meters.
The fact that light from a point source obeys the inverse square law is used to
advantage in measuring astronomical distances. If you have a source of known
intrinsic brightness, then it can be used to measure its distance from the Earth by the
"standard candle" approach.
The inverse square law describes the intensity of light at different distances
from a light source. Every light source is different, but the intensity changes in the
same way. The intensity of light is inversely proportional to the square of the distance.
This means that as the distance from a light source increases, the intensity of light is
equal to a value multiplied by 1/d2, The proportional symbol, , is used to show how
these relate. The relationship between the intensity of light at different distances from
the same light source can be found by dividing one from the other. The formula for
this is shown below. Visible light is part of the electromagnetic spectrum, and the
inverse square law is true for any other waves or rays on that spectrum, for example,
radio waves, microwaves, infrared and ultraviolet light, x rays, and gamma rays. The
intensity of visible light is measured in candela units, while the intensity of other
waves is measured in Watts per meter squared (W/m2).
Proportional:
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I = light intensity (candela, W/m2)
means "is proportional to"
d = distance from a light source (m)
Intensity at different distances:
I1 = light intensity at distance 1
I2 = light intensity at distance 2
d1 = distance 1 from light source (m)
d2 = distance 2 from light source (m)
Inverse Square Law Formula Questions:
1) If a bright flashlight has a light intensity of 15.0 candela at a distance 1.00 m from
the lens, what is the intensity of the flashlight 100.0 m from the lens?
Answer : The intensity at the farther distance can be found using the formula:
If d1 = 1.00 m from the lens, and d2 = 100.0 m from the lens, then I1 = 15.0 candela,
and we need to solve for I2. This requires rearranging the equation:
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Now, substitute the values that are known in to the equation:
I2 = (0.0001)(15.0 candela)
I2 = 0.0015 candela
The intensity of the flashlight at a distance of 100.0 m is 0.0015 candela.
2) The intensity of a radio signal is 0.120 W/m2 at a distance of 16.0 m from a small
transmitter. What is the intensity of the signal 4.00 m from the transmitter?
Answer: The intensity at the near distance can be found using the formula:
If d1 = 4.00 m from the transmitter, and d2 = 16.0 m from the transmitter, then
I2 = 0.120 W/m2, and we need to solve for I1. This requires rearranging the equation:
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Now, substitute the values that are known in to the equation:
I1 = (16.0)(0.120 W/m2)
I1 = 1.92/m2
The intensity of the radio signal 4.00 m from the transmitter is 1.92 W/m2.
https://www.youtube.com/watch?v=3n02pw1kOjU
Learning Competency:
Apply the inverse square
(STEM_GP12MWS-IIe-34)
relation
intensity
of
waves
and
distance.
ACTIVITY 1: INVERSE SQUARE LAW
Objectives: Using simple materials, students create model to show how they
discover inverse square law. Explain why the world gets dark so fast outside the
circle of the campfire?
Materials: • A Mini Maglite flashlight (or other small bright light source),Ruler, A 3x5
index card, Scissors, A medium sized binder clip, Graphing paper with ½- inch or ¼inch squares (if using metric use graph paper with 1-cm squares), Cardboard box or
piece of foam core , tape
Time needed/location: Investigate modelling Inverse square law while the sun is out
or in the room with without light. Depending on the level of investigation, can take
between 20 – 45 minutes.
Procedure:
1. Use your ruler and scissor to measure and cut out a ½ x ½ square inch in
the center of the index card.
2. Clip the binder clip to the bottom of the card to make a stand.
3. Mount the graph paper on the side of the cardboard box or piece of foam
core to make a screen.
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.
4. Next, unscrew the front reflector assembly of the Mini Maglite to expose
the bulb. The bulb will come on and stay on when reflector assembly is
removed. (see the image below)
5. Prop the light so it is at the same height as the square hole that you cut in
the card.
6. Position the card one inch in front of the light source.
7. Line up the Mini Maglite, Square hole, and graph paper so when the light
shines through the hole you see a square of light on the graph paper.
If you are using metric units:
Follow the directions above, but instead of a ½ x ½ inch hole, cut – out 1x1 –
cm square hole in the center of the index card.
1. Position the card 2 centimetres in front of the light source.
2. Use the graph paper with 1 cm square printed on it.
To Do and Notice
Keep the distance between the bulb and the card with the square hole
constant at one inch. (If you are using metric graph paper, we recommend a distance
of 2 cm.) Put the graph paper at different distance from the bulb, and count how many
squares on the graph paper are lit at each distance (see the diagram below). The
result will be easier to understand if you make a table of “number of squares lit”
versus “distance.” Be sure to measure the distance from the graph paper to the bulb
each time.
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What’s going on?
As you move the graph paper from the Mini Maglite, what happen to the light? Why?
Explain your answer.
ACTIVITY 2: FIND ME
Directions: Choose the correct answer. Write the letter of your answer on the blank
before the number.
______1. If I1 = 100 IU at d1 = 1 cm, what is I2 at d2 = 10 cm?
A. 1 IU
B. 2 IU
C. 3 IU
D. 4 IU
______ 2. A reading of 12 lumens/hr is obtained at a distance of 3.5 cm from point
source. At what distance would a reading of 1000 lumens/hr be obtained?
A. 1.000 cm
B. 0.285 cm
C. 0.380 cm
D. 0.450 cm
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_______ 3. A reading of 287 mR/hr is obtained at a distance of 1.5 cm from a point
source. At what distance would a reading of 28.7 mR/hr be obtained?
A. 4.89 cm
B. 4.90 cm
C. 4.74 cm
D. 4.57 cm
_______ 4. A reading of 287 mR/hr is obtained at a distance of 1.5 cm from a point
source. What would be the reading at a distance of 15 cm?
A. 2.90 mR/hr
B. 2.87 mR/hr
C. 3.87 mR/hr
D. 3.90 mR/hr
_______ 5. A reading of 100 mR/hr is obtained at a distance of 1cm from a point
source. What would be the reading at a distance of 1nm?
A.100,000 mR/hr
B. 20,000 mR/hr
C. 30,000 mR/hr
D. 10,000 mR/hr
Activity 3: SOLVE ME
Directions: Read and analyze the problem very carefully. Then solve and show your
complete solutions.
1. The intensity of monochromatic light is in the ratio 16:1. Calculate the second
distance if the first distance is 6m?
a. 24 m
b. 21 m
c. 26 m
d. 27 m
2. If a bright flashlight has a light intensity of 15.0 candela at a distance 1.00 m
from the lens, what is the intensity of the flashlight 100.0 m from the lens?
a. 0.01500 candela
b. 0.00015 candela
c. 0.00165 candela
d. 0.00015 candela
3. The intensity of light is 800 W/m2 at 4 m away from the source. What is the
intensity of the light of a distance of 3m away from the light source?
a. 1522.7 W/m2
b. 1622.5 W/m2
c. 1422.2 W/m2
d. 1822.2 W/m2
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Reflection:
1. I learned that ______________________________________________________
__________________________________________________________________
__________________________________________________________
2. I enjoyed most on __________________________________________________
__________________________________________________________________
__________________________________________________________
3. I want to learn more on ______________________________________________
__________________________________________________________________
__________________________________________________________
253
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References:
http://creativecommons.org/licenses/bysa/3.0energyeducation.ca/encyclopedia/Inv
erse_square_law
http://creativecommons.org/licenses/by-sa/3.0
https://study.com/academy/lesson/inverse-square-law-for-radiation-definitionformula.html
JG The Organic Chemistry
http://hyperphysics.phy-astr.gsu.edu/hbase/Forces/isq.html
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ANSWER KEY
Activity 1. Answers may vary
Activity 2.
1.
2.
3.
4.
5.
a
c
c
b
d
Activity 3.
1. a
2. b
3. c
Prepared by:
MARIO T. BOLANDO JR.
Matucay National High School
255
NOTE: Practice personal hygiene protocols at all times
256
NOTE: Practice personal hygiene protocols at all times
257
NOTE: Practice personal hygiene protocols at all times