Physics
Lecture 3
1
Using Newton’s Law – Uniform Circular
Motion
Uniform Circular Motion

・ motion in a circle of constant radius at constant speed
・ acceleration is defined as
magnitude
centripetal acceleration
An object moving in a circle of radius r at constant speed v has an
acceleration whose direction is toward the center of the circle and
whose magnitude is
2
Using Newton’s Law – Uniform Circular
Motion
Circular motion is often described in terms of frequency f,

f 
1
T
For an object revolving in a circle at constant speed v,

2r
v
T
Example :
A 150g ball at the end of a string is revolving uniformly in a horizontal circle of
radius 0.600 m. The ball makes 2.00 revolutions in a second. What is its centripetal
acceleration?

v = 7.54 m/s, so a = 94.7 m/s2
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Using Newton’s Law – Uniform Circular
Motion

an object which is in uniform circular motion, there must be a net force acting on
it

the force is written as
when an object moving in a circle

Example :

Estimate the force a person must exert on a string attached to a 0.150kg ball to
make the ball revolve in a horizontal circle of radius 0.600 m. The ball makes 2.00
revolutions per second.
FT = 14N
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Using Newton’s Law – Uniform Circular
Motion
Example :

A small ball of mass m, suspended by a cord of length l, revolves in a circle of
radius r = l sin θ, where θ is the angle the string makes with the vertical.
Calculate the speed and period (time required for one revolution) of the ball in
terms of l, θ, g, and m.
v
5
gl sin 2 
l cos 
, T  2
cos 
g
Work and Energy
a) Work done by a constant force
the distance moved multiplied by the component of the force in the direction of
displacement
W = F||d [J]

In the SI system, the units of work are joules : 1[J]=1[Nm]
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Work and Energy
Example 1:

The force shown has magnitude FP = 20 N and makes an angle of 30° to the
ground. Calculate the work done by this force, when the wagon is dragged 100 m
along the ground.
W = 1732 [J]
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Work and Energy
Example 2 : Work on a backpack
(a) Determine the work a hiker must do on a 15.0kg backpack to carry it up a hill
of height h = 10.0 m, as shown. Determine also (b) the work done by gravity on
the backpack, and (c) the net work done on the backpack. For simplicity, assume
the motion is smooth and at constant velocity (i.e., acceleration is zero).

(a) WH = 1470 [J]
(b) WG = -1470 [J]
(c) Wnet = 0
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Work and Energy

b) Work done by a varying force

• a particle acted on by a variable force, F, moves along the path
• work can be approximated by dividing the distance up into small pieces, finding
the work done during each, and adding them up
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Work and Energy

•
b) Work done by a varying force
the work done by a variable force in moving an object between 2 points is equal
to the area under the curve between those 2 points
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Work and Energy

Example 3 : Work done by a spring force
•
spring either stretched or compressed an amount x
FP
from its normal, requires a force
FP  kx
•
the spring itself exerts a force in opposite direction
FS
FS  kx
•
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k : spring constant
Work to stretch or compress a spring
Work and Energy

Example 4 : Work done by a spring force
(a) A person pulls on a spring, stretching it 3.0 cm, which requires a maximum force of 75
N. How much work does the person do? (b) If, instead, the person compresses the spring
3.0 cm, how much work does the person do?
(a) WP = 1.1 [J]
(b) WP = 1.1 [J]

Example 5 : Force as a function of x
A robot arm that controls the position of a video camera in an automated surveillance
system is manipulated by a motor that exerts a force on the arm.The force is given by
where F0 = 2.0 N, x0 = 0.0070 m, and x is the position of the end of the arm. If the arm
moves from x1 = 0.010 m to x2 = 0.050 m, how much work did the motor do?
WM = 0.36 [J]
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Work and Energy

c) Kinetic Energy and Work-Energy Principle
•
•
Kinetic Energy; the units of kinetic energy are joules
Work-Energy Principle;
the net work done on an object is equal to the change in the object’s kinetic energy
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Work and Energy

Example 6 : Kinetic energy and work done on a baseball
A 145g baseball is thrown so that it acquires a speed of 25 m/s. (a) What is its kinetic
energy? (b) What was the net work done on the ball to make it reach this speed, if it
started from rest?
(a) K = 45 [J]
(b) Wnet = 45 [J]

Example 7 : A compressed spring
A horizontal spring has spring constant k = 360 N/m. (a) How much work is required to
compress it from its uncompressed length (x = 0) to x = 11.0 cm? (b) If a 1.85kg block is
placed against the spring and the spring is released, what will be the speed of the block
when it separates from the spring at x = 0? Ignore friction. (c) Repeat part (b) but assume
that the block is moving on a table and that some kind of constant drag force FD = 7.0 N is
acting to slow it down, such as friction.
(a) W = 2.18 [J]
(b) v = 1.54 m/s
(c) v = 1.23 m/s
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Conservation of Energy

a) Potential Energy
•
energy associated with forces that depend on the position or configuration of objects relative
to the surroundings
•

the units of potential energy are joules
Gravitational Potential Energy
Therefore, the gravitational potential energy at a
height y is:
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Conservation of Energy

Elastic Potential Energy
A spring has potential energy, called elastic potential
energy, when it is compressed. The force required to
compress or stretch a spring is :
Then, potential energy is :
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Conservation of Energy

b) total Mechanical Energy
• the sum of the kinetic energy plus the potential energy of the system at any moment:
•
as long as no nonconservative forces do work; E at initial is equal to E at any later time :
•
the principle of conservation of mechanical energy:
If only conservative forces are doing work, the total mechanical energy of a system neither
increases nor decreases in any process. It stays constant - it is conserved
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Conservation of Energy

Example 8 : Roller-coaster
Assuming the height of the hill is 40 m, and the roller-coaster car starts from rest at the
top, calculate (a) the speed of the roller-coaster car at the bottom of the hill, and (b) at
what height it will have half this speed.Take y = 0 at the bottom of the hill.
(a) v = 28 [m/s]
(b) y = 30 [m]

Example 9 : Toy dart gun
A dart of mass 0.100 kg is pressed against the spring of a toy dart
gun. The spring (with spring stiffness constant k = 250 N/m and
ignorable mass) is compressed 6.0 cm and released. If the dart
detaches from the spring when the spring reaches its natural
length (x = 0), what speed does the dart acquire?
v = 3.0 [m/s]
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The Law of Conservation of Energy

The law of conservation of energy is :
The total energy is neither increased nor decreased in any process. Energy can be transformed from one
form to another, and transferred from one object to another, but the total amount remains constant.
K  U  [changein all otherformsof energy] 0

Example 10 : Friction on the roller-coaster car
The roller-coaster car shown reaches a vertical height of only 25 m on the second hill
before coming to a momentary stop. It traveled a total distance of 400 m. Determine (a)the
thermal energy produced and (b)estimate the average friction force on the car, whose mass
is 1000 kg.
K  U  Ffrl  0
(a) thermal energy = 1.47 x 105 [J]
(b) 367.5 [N]
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Power

Power is the rate at which work is done.

Average power :

Instantaneous power :

Power can also be described as the rate at which energy is transformed :

Power is also needed for acceleration and for moving against the force of friction

The unit of power is Watt
20
Power

Example 11 : Power needs of a car
Calculate the power required of a 1400kg car under the following circumstances: (a) the
car climbs a 10°hill at a steady 80 km/h; and (b) the car accelerates along a level road
from 90 to 110 km/h in 6.0 s to pass another car. Assume that the average retarding force
on the car is FR = 700 N throughout.
(a) P = 6.78 x 104 [W]
(b) P = 6.13 x 104 [W]
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