# L S S S Phys G11 -1Pd-converted ```Liberia Secondary School System (LSSS)
Grade 11, Physics 1st Marking Period Lecture Note
Instructor: Mr. N. W. Keeta (0770772276/0886419872)
Topic: Motion in two Dimensions
Quantities (Lecture 1)
Scalars quantity is a physical quantity that has magnitude only. For example: speed,
distance, temperature, energy, density, mass, work, etc.
Vector quantity is a physical quantity that has both magnitude and direction. For example:
displacement, velocity, acceleration, force, momentum, electric field, magnetic field lines etc.
Basic Equations of motion (Motion in one dimension)
Equations of motion in one dimension: They are often referred to as the “SUVAT”
equations, eponymous from to the variables: s = displacement (s0 = initial displacement), u =
initial velocity, v = final velocity, a = acceleration and t = time (Kinematic Equation)
(1) 𝑣 = 𝑢 + 𝑎𝑡
1
(2) 𝑠 = 𝑢𝑡 + 𝑎𝑡 2
2
1
(3) 𝑠 = (𝑢 + 𝑣)𝑡
2
(4) 𝑣 2 = 𝑢2 + 2𝑎𝑠
1
(5) 𝑠 = 𝑣𝑡 − 𝑎𝑡 2
2
Types of motions (Lecture 2)
We recognize motion as a change in the position of an object with time. There are several
types of motions, but we will be discussing the following: projectile, rotary (rotational),
circular, simple harmonic, and oscillatory.
Projectile motion
Projectile motion is a form of motion where the particle (called projectile) is thrown
indirectly near the earth’s surface and it moves along a curved path under the action of
gravity. The path followed by a projectile motion is called its trajectory. Projectile motion
only occurs when there is one force applied at the beginning of the trajectory after which
there is no interference apart from gravity. VY
V = 0(at maximum height)
H
Angle of projection
θ
Projection plane
Projection path
(Trajectory)
VX
Horizontal plane
Fig 1.1
The vertical component of velocity uy = u sinθ and the horizontal velocity ux = u cosθ
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Example 1: A basket ball player attempts to make a basket from a distance of 8.0 m. The
basket is 3.0 m high and the ball leaves his hands from a height of 1.0 m. If he throws the ball
at an angle of 45°, what does the initial velocity of the basket need to be? [g = 9.8 m/s2]
Example 2: A boy stands on the roof of a 50.0 m tall building and throws a baseball
horizontally at 20.0 m/s always from the building. How far away from the foot of the building
will the ball land? [g = 9.8 m/s2] Answer 63.9 m
Vertical projection
If a piece of stone or ball is projected vertically upwards, its projection velocity tends to
decrease more slowly until it comes momentarily (shorty) to a stop. It then falls back faster
and faster till it hits the plane of projection. The decrease in velocity as the ball rises higher
up, is due to the effect of gravity. However, the maximum height an object under this
condition can achieve depends on the magnitude of the initial upward velocity used in
projecting it. Therefore, if hmax = maximum height attained, u = initial upward velocity, g =
acceleration of free fall then, time, t taken to attain the maximum height would be obtained
from this equation (v = u – gt). Note that at maximum height, the object usually comes to a
momentary stop. Hence, v = 0; 0 = u – gt that is, gt = u given t = u/g
Final velocity, v = 0
(at max height)
hmax
g
u
Plane of projection
To determine the time of flight, T
Fig 1.2 Vertical projection of ball
The time of flight is required for the object to be in air or in flight. This time, T is twice the
time required for the object to reach the maximum height. Hence, T = 2t that is,
2𝑢
𝑔
To determine the maximum height, ‘hmax’ attained
Apply the following equation (h = ut ± 1/2gt2 or v2 = u2 ± 2gh). For easier approach apply
equation (v2 = u2 ± 2gh) that is, v2 = u2 ± 2ghmax at maximum height v = 0 therefore
v2 = u2 - 2ghmax ↔2ghmax = u2
𝒖𝟐
∴ 𝒉𝒎𝒂𝒙 =
𝟐𝒈
Example 3: A tennis ball is thrown vertically upwards from the ground with a velocity of
50m/s. Calculates; (i) the maximum height reached, (ii) the time to reach the maximum
height and (iii) the time of flight.
Given u = 50m/s, acceleration due to gravity = - 10m/s2 (Negative value is due to the
retarding effect of gravity as ball rises). Answer i. 125 m, ii. 5.0 sec, and iii 10.0 sec.
𝑇=
To determine the height, h, from which an object is projected
Note that the projection velocity is horizontal, as such; velocity in the vertical is zero. By
applying the equation (h = ut + ½ gt2) and noting that, u along h, is 0; h = ½ gt2, the time,
taken to reach the ground would be
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𝑡=√
2ℎ
𝑔
To determine the horizontal distance (range) covered with the horizontal velocity u. Note: g
in the horizontal direction is zero so ½ gt will be equal to zero. By taken h = Range (R), the
equation (h = ut + ½ gt2) becomes, R = ut + 0 i.e.
𝑅 = 𝑢 × √
𝟐𝒉
𝒈
Example 4: A bullet is fired horizontal with velocity of 40m/s from the top of a building 50m
high. How far from the foot of the building will the bullet be assumed to touch the ground? (g
= 10m/s2). Calculate the time of flight. Answer R = 126.5 m, T = 3.162 sec,
Resultant velocity of a projectile (Lecture 3)
When a projectile is in flight, its resultant velocity, v which is tangential (indirect) to the
direction of flight is made up of
(i)
Horizontal component velocity, Vx = horizontal velocity of projection, (u)
(ii)
Vertical component velocity, Vy = vertically downward velocity, (gt). (Since v = 0
in the vertical direction) u = horizontal velocity of projection.
O
*The resultant velocity, v = √ (V2x + V2y)
Projection
height, h
θ
VY
Range, R
Fig 1.3 Horizontal projection of ball
indicating velocity after time, t
𝟐
𝟐
VX after time, ‘t’ 𝑉 = √𝒖 + (𝒈𝒕) if V is
inclined at an angle of θ to Vx then, tanθ =
V
𝒈𝒕
Vy/Vx = gt/u therefore, 𝜽 = 𝒕𝒂𝒏−𝟏 ( )
𝒖
Example 5: A stone is thrown
horizontally with a velocity of 15 m/s
from a height of 100 m. Calculate the
speed at which the stone hits the ground.
(g = 10 m/s2). Horizontal velocity of
projection, u = 15 m/s, height of
projection, h = 100 m. Answer v =
47.17m/s
Motion down an inclined plane
If the friction between an object and the inclined plane in which it is in contact with is
neglected, the object would tends to slip down the inclined surface faster because of the
component effect of acceleration due to gravity.
By Resolution, component of g along the plane, gx =
sinθ and that perpendicular to the plane, gy = g cosθ.
Inclined plane
Object
If s = length of the inclined plane
h = vertical height of fall
S
gx
θ gy
t = time taken to slip down the plane
h
v = final velocity attained on reaching the base of
g
plane
θ
θ = angle of inclination of plane
Considering u = 0 at rest then, final velocity, v = gxt
Angle of inclination
Fig 1.4 Motion down an inclined plane when u = 0. That is, v = gt sinθ
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To determine the vertical height of fall
Object
Object
S
S
gx
θ gy
g
h = vertical height of fall
h
60°
θ
Fig 1.5
Use this to solve example 6 Fig 1.6
Angle of inclination
Fig 1.6
From Fig 1.5
h/s = sinθ that is, h = s sinθ
Substituting for s using equation (s = v2/(2gsinθ)) the vertical height of fall, h becomes.
𝒗𝟐
𝒗𝟐
𝒉=
× 𝒔𝒊𝒏𝜽 𝒊. 𝒆. 𝒉 =
𝟐𝒈𝒔𝒊𝒏𝜽
𝟐𝒈
Example 6: A smooth body slides down a plane inclined at an angle of 60 degree to the
horizontal. Calculate (a) the component of the acceleration of the body
(i)
Along the inclined plane and
(ii)
Perpendicular to the plane
(b) The velocity of the body at the base of the plane after 10 second
(c) The vertical height of fall (g = 10 m/s2)
Angle of inclination of plane, θ = 60°, Time taken to slip down, t = 10 sec, if gx = component
of g along the plane and gy = component of g, perpendicular to plane
Solution
(a) (i) gx = g sinθ
= 10 sin 60 = 8.66m/s2 ans.
(ii) gy = g cosθ
= 10 cos 60 = 5m/s2 ans.
(b) Velocity of base of
plane v = gxt = g sinθ × t
= 8.66 × 10 = 86.6 m/s ans.
(c) Vertical height of fall
h = v2/2g = (86.6)2/ (2 × 10)
= 374.98 m ans.
Inclined projection
When an object is projected into space at an angle of θ° to the horizontal:
(i) It describes a path that is fully parabolic as shown in fig 1.7. Another practical example
of this parabolic path can be described, when an arrow is shot into space or when a
rocket is launched into space.
(ii) Its velocity of projection decreases slowly and slowly as it gains maximum height.
(iii)At the maximum height reached, the final velocity becomes zero.
(iv) Because of gravity, the object returns to the same projection plane as describe R from
the point of projection.
Note: An object that moves through space on projection is called a projectile, the path taken
VY
is called trajectory.
V = 0(at maximum height)
H
Angle of projection
Projection plane
θ
Fig 1.7
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Projection path
(Trajectory)
VX
Horizontal plane
To determine the time taken to attain the maximum height
Vertical component velocity, = uy
Final velocity at maximum height = v = 0
Acceleration of free fall due to gravity = - g from v = uy – gt; 0 = uy – gt, therefore t = uy/g
that is,
𝒖 𝒔𝒊𝒏 𝜽
𝒕=
𝒈
Time of flight T
The time required for the projectile to be in flight. This time is twice the time attain (achieve)
to reach the maximum height H. that is, T = 2t, therefore,
𝟐𝒖 𝐬𝐢𝐧 𝜽
𝑻=
𝒈
To determine the maximum height, Hmax
Apply either equations (h = ut – 1/2gt2 or v2 = u2 – 2gh). If h = Hmax
Initial velocity, u = vertical component, velocity required to attain the maximum height = ux
Hmax = uxt – 1/2gt2 and noting that
𝒕=
𝒖 𝒔𝒊𝒏 𝜽
𝒈
It implies that
2
𝐻𝑚𝑎𝑥 = 𝑢 𝑠𝑖𝑛 𝜃 (
𝐻𝑚𝑎𝑥 =
(𝒖𝒔𝒊𝒏𝜽)𝟐
𝒈
2
𝑢𝑠𝑖𝑛𝜃
1 𝑢𝑠𝑖𝑛𝜃 2 (𝑢𝑠𝑖𝑛𝜃)
1 (𝑢𝑠𝑖𝑛𝜃)
)− 𝑔(
) =
−
𝑔
2
𝑔
𝑔
2 𝑔
𝟐
(𝒖𝒔𝒊𝒏𝜽)
𝟏
𝟏
𝒖𝟐 𝒔𝒊𝒏𝟐 𝜽
× (𝟏 − ) =
× ∴ 𝑯𝒎𝒂𝒙 =
𝟐
𝒈
𝟐
𝟐𝒈
Or if equation (v2 = u2 – 2gh) is applied and noting that v = 0, at maximum height then
0 = u2y – 2gHmax therefore, 2gHmax = u2y that is, 2gHmax = (usinθ)2
∴ 𝑯𝒎𝒂𝒙 =
𝒖𝟐 𝒔𝒊𝒏𝟐 𝜽
𝟐𝒈
To determine the range, R
The range, R is the horizontal distance from the point of projection to the point where the
projectile hits the horizontal plane. The velocity that enables the projectile to cover this range
is the horizontal component velocity ux = u cosθ.
Since, g = 0 along this distance and time required to cover the range = time of flight, T
Then, R = uxT = (u cosθ × 2usinθ)/g. that is,
𝟐𝒖𝟐 𝒔𝒊𝒏𝜽𝒄𝒐𝒔𝜽
𝑹=
𝒈
But sin 2θ = 2 sinθcosθ
𝒖𝟐 𝒔𝒊𝒏 𝟐𝜽
∴𝑹=
𝒈
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The maximum Range of a projectile can be attained when the angle of projection is 45
degree. Therefore,
(𝟐 × 𝟒𝟓)
𝒖𝟐
𝑹𝒎𝒂𝒙 = 𝒖𝟐 𝒔𝒊𝒏
(𝐬𝐢𝐧𝐜𝐞 𝐬𝐢𝐧 𝟗𝟎 = 𝟏) 𝒊. 𝒆. 𝑹𝒎𝒂𝒙 =
𝒈
𝐠
Example 7: A ball is projected at an angle of elevation of 60 degree with an initial velocity
of 100 m/s. calculate
(i)
The time of flight
(ii)
Maximum height reached
(iii) The range covered.
Solution
Given: initial velocity, u = 100 m/s, angle of projection = 60°
(i)
Time of flight,
𝑇 =
2𝑢 𝑠𝑖𝑛 𝜃
2 × 100 × 𝑠𝑖𝑛 60
=
= 𝟏𝟕. 𝟑𝟐 𝒔𝒆𝒄 𝒂𝒏𝒔.
𝑔
10
(ii) Maximum height, 𝐻𝑚𝑎𝑥
(iii)
The range cove, 𝑅
=
=
𝑢2 𝑠𝑖𝑛2 𝜃
2𝑔
𝑢2 𝑠𝑖𝑛 2𝜃
𝑔
𝐻𝑚𝑎𝑥 =
∴𝑅=
1002 (𝑠𝑖𝑛60)2
1002 𝑠𝑖𝑛 2(60)
10
2𝑔
∴ 𝑯𝒎𝒂𝒙 = 𝟑𝟕𝟓 𝐦
= 𝟖𝟔𝟔. 𝟎𝟐𝟓 𝒎 ans.
Rotary motion (Lecture 4)
Rotary motion is the motion of a body that moves about an internal axis. For rotary motion to
be constant, the object must spin about a fixed point at a steady rate.
Angular velocity is defined as the time rate of angular displacement.
∆𝜃
𝜔=
∆𝑡
Conversion factor
1 revolution = 360° = 2ᴨ radians
360° 180°
1 𝑟𝑎𝑑𝑖𝑎𝑛 =
=
= 57.3°
2𝜋
𝜋
Angular acceleration is the constant rate of change in angular velocity, and designated as α
(alpha).
∆𝜔
𝛼=
∆𝑡
Example 8: Find the angular displacement after 15.0 sec (in Radians) for a wheel that
accelerates at a constant rate from rest 725 rev/min in 10.0 min. [∆θ = 1/2α∆t2] Answer 14.2
Example 9: A flywheel has a constant angular acceleration of 8.4 rad/s2. How long will it
take the wheel to attain an angular velocity of 6.0 rev/sec starting from rest? [α = ∆ω/∆t]
Centrifugal force
The Centrifugal force is the reaction force that tends to move a body away from the center. In
other words, it acts in opposite direction to the centripetal force
𝑚𝑣 2
(−
)
𝑟
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Centrifuge
A centrifuge is a device used to separate particles in suspension from the liquid in which they
are contained. It consists of two tubes which are whirled (rotated) by electrical means in a
circle at uniform speed.
Circular motion (Lecture 5)
Circular motion is a movement of an object along the circumference of a circle or rotation
along a circular path. It can be uniform, with constant angular rate of rotation (and constant
speed), or non-uniform with a changing rate of rotation.
A. Uniform Velocity (Speed)
If the displacement along the circular path = S; then, from Geometry S = r θ; where r = the
radius of the circle described by the object. Assuming the time to bring about the angular
displacement, θ = t then, the velocity of the object
𝐷𝑖𝑠𝑝𝑙𝑎𝑛𝑐𝑚𝑒𝑛𝑡 𝑎𝑙𝑜𝑛𝑔 𝑡ℎ𝑒 𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟 𝑝𝑎𝑡ℎ
𝑠
𝑉=
𝑜𝑟 𝑉 =
𝑇𝑖𝑚𝑒
𝑡
Since
𝑟𝜃
𝑆 = 𝑟𝜃, 𝑡ℎ𝑒𝑛 𝑉 =
𝑡
But angular velocity,
𝜃
𝐻𝑒𝑛𝑐𝑒 𝑡ℎ𝑒 𝑒𝑞𝑢𝑡𝑖𝑜𝑛 𝑏𝑒𝑐𝑜𝑚𝑒𝑠 𝑉 = 𝑟𝜔
𝑡
Note: Angular velocity, ω is defined as the rate of change of angular displacement. It is
Generally, the velocity of an object in circular motion,
(i) Changes continually in direction, as the object moves in the circle.
(ii) Remains constant in magnitude as it moves round
(iii)Is observed to be tangential to the direction of motion
𝜔=
Acceleration of the object (a)
The acceleration of the object moving in a circle
(i) Is always directed towards the center of the circle described
(ii) Has a constant magnitude
The mathematical relationship can be derived from the definition and vector diagram
VB
VB
θ
A
Figure 1.8
●
∆V = VB - VA
●B
VA
θ
- VA
Figure 1.9
Vector Diagram showing relative change in velocity
The velocity change, ∆V is directed towards the center and as such perpendicular to V. Since
the magnitude of the velocity at A and B are the same, despite their direction. Therefore the
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velocity change ∆𝑉 = 𝑣𝑠𝑖𝑛𝜃 as the angular velocity change and the time to make the
change become smaller, therefore,∆𝑉 = 𝑣𝜃
Since,
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
∆𝑣
𝑎=
=
𝑡𝑖𝑚𝑒
𝑡
Then, the centripetal acceleration
𝑣𝜃
𝑎=
𝑡
Centripetal acceleration, 𝑎 = 𝑉𝜔; Noting that 𝑉 = 𝑟𝜔; then the centripetal acceleration can
be expressed as
𝑣2
𝑎 = 𝑟𝜔2 𝑜𝑟 ( )
𝑟
Centripetal force (f)
The Centripetal force is defined as the force that tends to keep an object of mass, m round a
That is
𝑚𝑣 2
( 𝑜𝑟 𝑚𝑟𝜔2 )
𝐶𝑒𝑛𝑡𝑟𝑖𝑝𝑒𝑡𝑎𝑙 𝑓𝑜𝑟𝑐𝑒, 𝐹 =
𝑟
Example 10: An object of mass 2.0 kg moves in a circle of radius 4.0 m at uniform speed of
16.0 ms-1. Calculate
(i) The angular velocity
(ii) The acceleration and
(iii)The centripetal force
Given
unknown
Basic Equation
𝑣
m = 2.0 kg
ω
𝜔=
r = 4.0 m
a
𝑟
v = 16.0 ms-1
F
𝑎 = 𝑉𝜔
𝐹 = 𝑚𝑟𝜔2
Solution:
(ii) Angular acceleration
(i) Angular Velocity
𝑎 = 𝑉𝜔
(iii) Centripetal Force
𝜔 = 𝑣/𝑟
𝐹 = 𝑚𝑟𝜔2
𝑎 = (16)(4) = 𝟔𝟒 𝒎𝒔−𝟐
16
𝐹 = (2)(4)(4)2 = 𝟏𝟐𝟖 𝑵
𝜔=
= 4 𝑟𝑎𝑑/𝑠
4
Simple harmonic motion (Lecture 6)
Simple harmonic motion is a type of periodic motion where the restoring force is directly
proportional to the displacement. In order words, it means the to – and – fro of an object (or
particle) whose acceleration is directly proportional to its displacement from a fixed point and
is directed towards the center. If y = displacement (distance) from the fixed point and 𝜔 =
𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦. Then, acceleration
𝑎 = −𝜔2 𝑦
The minus (-) sign indicates that the object’s motion always decreases towards the center.
Graph of a simple Harmonic motion
When a particle is undergoing simple harmonic motion, the displacement time graph of the
motion is represented by a Sinusoidal curve (or sine curve). The equation of this motion is
express as
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𝑦 = 𝑎 sin 𝜔𝑡
Where a = amplitude of the motion (i.e. the maximum distance or displacement on either side
of the fixed point).
Amplitude
a
Fig.1.10
A
Circle
Simple Harmonic motion of a spiral spring with an attached weight
When a weight is attached to one end of a suspended
coiled spring it expands. According to Hook’s law, the
restoring force “F” in the spring would be directly
proportional to the extension “e” produced, provided
the spring is not deformed. The restoring force
𝐹 = −𝑘𝑒,
Where k = spring constant, with mass, m attached. (a)
𝑊𝑒𝑖𝑔ℎ𝑡 = 𝑘𝑒 𝑖. 𝑒 𝑚𝑔 = 𝑘𝑒
When attached weight is displaced vertically through
distance x (c) and allowed to oscillate,
𝑅𝑒𝑠𝑡𝑜𝑟𝑖𝑛𝑔 𝑓𝑜𝑟𝑐𝑒 = −𝑘(𝑒 + 𝑥)
Required net force towards center of oscillation
= 𝑘(𝑒 + 𝑥) − 𝑚𝑔
𝑖. 𝑒 𝑚𝑎 = −[𝑘(𝑒 + 𝑥) − 𝑚𝑔]
Figure 1.11
𝑆𝑖𝑛𝑐𝑒 𝑚𝑔 = 𝑘𝑒 𝑡ℎ𝑒𝑛, 𝑚𝑎 = −𝑚𝑔 − 𝑘𝑥 + 𝑚𝑔
𝑚𝑎 = −𝑘𝑥, ℎ𝑒𝑛𝑐𝑒, 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛, 𝑎 =
𝑘𝑥
𝑚
Comparing this equation with equation of simple harmonic motion we obtained
𝑘
𝜔2 =
𝑚
2𝜋
𝑚
𝑒
∴ 𝑝𝑒𝑟𝑖𝑜𝑑 𝑜𝑓 𝑚𝑜𝑡𝑖𝑜𝑛, 𝑇 =
= 2𝜋√ = 2𝜋√
𝜔
𝑘
𝑔
Oscillation (Lecture 7)
Oscillation is the repetitive variation, typically in time, of some measure about an essential
value (often a point of equilibrium) or between two or more different states. Familiar
examples include a swinging pendulum and alternating current power.
Conical Pendulum
Conical pendulum is a type of pendulum that swings in a horizontal circle. Such that the
suspended string appears to describe a right circular cone of angle θ to the vertical axis in the
Page 9 of 11
diagram below (fig. 1.12), if a small body of mass, m, is suspended from one end of the string
of length, L, the forces on the body will be
(i)
Its weight, mg, and
(ii)
The tension, T (as shown in the fig. 1.12)
The resultant force, F, acting on the body is that which keeps the body moving in the
circle. By resolving T into its components, the vertical component force 𝑇𝑦 = 𝑇𝑐𝑜𝑠𝜃
why the horizontal component, 𝑇𝑥 = 𝑇𝑠𝑖𝑛𝜃. Therefore, the resultant force, 𝑇𝑠𝑖𝑛𝜃 =
centripetal force, that is
𝑚𝑣 2
𝑇 sin 𝜃 =
𝑜𝑟 (𝑇 sin 𝜃 = 𝑚𝑟𝜔2 )
𝑟
Also, since there is no vertical acceleration,
𝑇𝑐𝑜𝑠𝜃 = 𝑚𝑔
To determine the speed of body
Simplifying the equation:
B
𝑚𝑣 2
𝑇𝑠𝑖𝑛𝜃
= 𝑟 ⁄𝑚𝑔
θ
(b) θ
𝑇𝑐𝑜𝑠𝜃
(a)
𝑠𝑖𝑛𝜃
but tan 𝜃 =
T (Tension)
L
cos 𝜃
2
𝑣
Tcosθ
tan 𝜃 =
Tcosθ
𝑟𝑔
Note:
T Sinθ
∴ 𝑣 2 = 𝑟𝑔 tan 𝜃
O
O
OA
r
F
𝑖. 𝑒 𝑣 = √𝑟𝑔 tan 𝜃
mg
mg
Fig. 1.12
To determine the period T, of
motion
Recall, circular speed,
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑐𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑜𝑓 𝑐𝑖𝑟𝑐𝑙𝑒
𝑉=
=
𝑡𝑖𝑚𝑒
𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛
2𝜋𝑟
2𝜋𝑟
2𝜋𝑟
𝑟
𝑖. 𝑒 𝑉 =
∴𝑇=
=
= 2𝜋√
𝑇
𝑣
𝑔 tan 𝜃
√𝑟𝑔 tan 𝜃
Assuming the period, T, is required when radius, r, is not given, then from (fig. 1.12)
𝑟 = 𝑂𝐴 = 𝐿 sin 𝜃
By substitution, period
𝐿 sin 𝜃
𝐿𝑐𝑜𝑠𝜃
𝑇 = 2𝜋√ ×
= 2𝜋√
𝑔 tan 𝜃
𝑔
The angular velocity, ω, can be obtained by recalling that
2𝜋
𝑇=
𝜔
Substituting in
2𝜋√
𝐿𝑐𝑜𝑠𝜃 2𝜋
𝑔
=
𝑡ℎ𝑒𝑛, 𝜔 = √
𝑔
𝜔
𝐿𝑐𝑜𝑠𝜃
Page 10 of 11
Example 11: A pendulum bob of mass, 2.0 kg attached to a string 1.0 m in length, is made to
swinging in horizontal circle of radius 50 cm. Find
(i) The period of the motion and
(ii) Tension in the cord
Given
Unknown
Basic equation
𝒐𝒑𝒑
L = 1m
𝑺𝒊𝒏𝜽
𝒔𝒊𝒏𝜽 = 𝒉𝒑𝒚
M = 2kg
𝑪𝒐𝒔𝜽
𝐬𝐢𝐧𝟐 𝜽 + 𝐜𝐨𝐬𝟐 𝜽 = 𝟏
r = 50cm = 0.5m
Period, T
𝑻 = 𝟐𝝅√
Tension, T
𝑻𝒄𝒐𝒔𝜽 = 𝒎𝒈
g=
10ms-2
𝑳𝒄𝒐𝒔𝜽
𝒈
𝑜𝑝𝑝 𝑟 0.5
= =
= 0.5 = sin−1(0.5) = 30°
ℎ𝑦𝑝 𝐿
1
𝐵𝑢𝑡 𝑐𝑜𝑠𝜃 = √1 − 𝑠𝑖𝑛2 𝜃 = 0.866
(𝑖) sin 𝜃 =
Solution
Visualized
(a)
∴ 𝑃𝑒𝑟𝑖𝑜𝑑, 𝑇 = 2𝜋√
θ
(𝑖𝑖)
L
Figure 1.13
(1)(0.866)
𝐿𝑐𝑜𝑠𝜃
= 2𝜋√
= 2𝜋(0.294) = 𝟏. 𝟖𝟓 𝒔𝒆𝒄
𝑔
10
𝑇𝑒𝑛𝑠𝑖𝑜𝑛, 𝑇 =
T (Tension)
r
20 N
Page 11 of 11
(2)(10)
𝑚𝑔
=
= 𝟐𝟑. 𝟎𝟗𝑵
𝑐𝑜𝑠𝜃
0.866
```