Kinematics
Kinematics
Kinematics
Kinematics
Kinematics
Kinematics is the study of relationships between positions, speed, acceleration which do not
consider the forces involved (ie. without regard to the causes of the motion.)
Displacement
A body moves from one position to another position. Displacement is a vector quantity
includes both magnitude (distance) and direction.
For example, a point was displaced 30 m due east and then 40 m due north, its final
displacement, being the sum of these two, would be 50 m 53 °8’ north of east. This would
indicate the resultant displacement of the point relative to the starting position.
Velocity is defined as the distance travelled in unit time in a given direction. Hence, it is a
vector.
Speed and velocity are both denoted by v and are given by the same equation
s
v (where s is the distance travelled in time t.)
t
Average velocity is to define velocities which cannot be maintained constant for a given
period of time.
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Followings are the unit commonly used to present speed and velocity:
meter per second (m/s)
kilometer per hour (km/hr)
Conversion of units:
e.g. To convert 36 km/hr to be the unit of m/s
36 1000
Answer: = 10 m/s
60 60
Velocity-Time Graph
If the velocity of a body at different times is plotted against the time, the graph obtained is a
velocity-time graph.
Constant velocity,
i.e. acceleration zero
Uniform acceleration
Body starting from rest
m
Slope = n= constant
c) In the case of constant acceleration,
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Constant acceleration
Body having an initial velocity u
Constant deceleration
From an initial velocity u
q larger than 90°
Acceleration-Time Graph
If the acceleration of a body at different
The area under the curve
times is plotted against the time, the graph
= the velocity of a body at t1.
obtained is an acceleration-time graph.
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Example 1
A car staring from rest, attains a velocity of 10 m s -1 in 5 s. Calculate the acceleration and the
distance travelled in this time.
Solution
Given: u = 0, v = 10 m /s, t = 5 s
Using: v =u+at
10 = 0 + 5 a
a = 2 ms-2
Using: v2 = u2 + 2 a s
102 = 0 + 2 (2) s
s = 25 m
Thus, the acceleration of the car is 2 m /s2 and the distance travelled in this time is 25 m.
Example 2
When a racing car is speeding at 66 m /s , the brakes are applied and it slows down uniformly
to 22 m/s in 5 s. Determine (a) the deceleration produced by the application of the brakes, (b)
the distance covered during the 5th second.
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(b) The “5th second” means the time interval between t = 4 s and t = 5 s
Using : s = u t + 1/2 a t2
= [66(5) + ½(-8.8)(52)] - [66(4) + ½(-8.8)(42)]
= 26.4 m
Thus, the distance covered in the 5th second is 26.4 m.
Example 3
A particle moving with uniform acceleration passes three posts A, B, C on a straight line. The
distance of separation between A and B is 60 m, while that between B and C is 80 m. The
particle takes 6 s to go from A to B and 5 s to go from B to C. From these observations,
determine the acceleration of the particle and its distance from A when its velocity is 22 m /s.
Let ‘u’ be the initial velocity (velocity when passing A) of the particle and ‘a’ its acceleration.
Arrange the given data as follows:
A to B A to C
s = 60 m s = 60 + 80 = 140 m
t=6s t = 6 + 5 = 11 s
12 2
Thus, the acceleration of the particle is m /s .
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When the particle has attained the final velocity of 22 m/s, we have, with A as a reference
point as before,
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u = 6 m /s
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v = 22 m/s
a = 12/11 m/s2
Using v2= u2 + 2 a s
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(22)2 = ( 6 )2 + 2 (12/11) s
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s = 201.1 m
Thus, the particle is 201.1 m from A when its velocity is 22 m/s.
The equations for motion due to gravity follow the same pattern as those for linear motion,
with the slight modification of introducing ‘g’ as the acceleration , instead of ‘a’. Hence the
motion equation applied to falling bodies become:
Equations
v=u-gt
s = u t - ½ g t2
v2=u2-2gs
Example 4
A body is projected upwards with a velocity of 50 m/s from the top of a tower 100 m high.
(a) Find the time taken for a body to reach the ground?
(b) Find the velocity with which the body strikes the ground?
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(a) using s = u t - ½ g t2
-100 = (50)t -(½)(9.81)t2
t2 -10.2 t -20.4 = 0
t = 11.1 s
(b) Using v 2 = u2 - 2 g s
v 2 = (50)2 -(2)(9.81)(-100)
= 4462
v = 66.8 m /s
Thus, body strikes the ground after 11.1 seconds, with a velocity of 66.8 m/s.
Example 5
An object is dropped from a helicopter ad strikes the ground 12 seconds later. Determine
(a) Find the height of the helicopter and
(b) Find the velocity with which the body strikes the ground.
(c) if a second object had been projected upwards from the ground with a velocity of 200
m/s at the same instant as the first object was dropped from the helicopter, where and
when they would meet?
Solution
Given u = 0 , t = 12 s
(a) using s = u t -½ g t2
s = 0 -½ (9.81) (12)2 = -706 m
(b) using v=u-gt
= 0 -(9.81)(12) = -118 m s-1
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- (706-s2) = s1
-706 + 200 t -½ g t2 = - ½ g t2
200t = 706
t = 3.53 s
Thus, the helicopter is 706 m high, and the object strikes the ground with a velocity of 118
m/s.
A second object projected upwards from the ground with a velocity of 200 m/s would meet
the first at a height of 645 m, 3.53 seconds after the instant of projection.
Relative Motion
The difference between the motions of two objects is termed “relative motion”. Relative
velocity is a term used when the velocity of one object is related to the velocity of another
reference object. Both of them can also be moving.
For example, an aircraft travelling due west at 400km/h just passes over another aircraft
travelling due north at 600km/h. What is the velocity and direction of the first aircraft relative
to the second?
Velocity of aircraft 1 is 721.11 km/h at 56.31o south of west relative to the aircraft 2.
Notation: V1 = the velocity of object 1 relative to earth (i.e. the velocity of object 1)
V2 = the velocity of object 2 relative to earth (i.e. the velocity of object 2)
V12 = the velocity of object 1 relative to object 2 (i.e. as if object 2 was at rest)
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Angular velocity
Angular velocity = the rate of change in angular displacement
It is properly expressed in terms of radians per second, using the notation ω (omega). In
practice, the number of revolution per second, or revolutions per minute (rpm), is often
quoted.
v r
v
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Angular Acceleration
Angular acceleration is the rate of change of angular velocity using notation α (alpha).
Angular retardation is a negative angular acceleration.
Since �=𝒅�/𝒅�
�=(� − ��)/�
𝑟�=(𝑟� −𝑟��)/� = (𝑣 −𝑢)/� = a
� � =�
v=ωr
� � =�
Linear Angular
v
Velocity
r v =ωr ω=
a
Acceleration a =αr α=
r
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Thus, the motor has a retardation of 21 rad s-2, and makes 375 revolutions in coming to
rest.
Example 7
A large gear-wheel rotating at 40 rpm has an angular retardation of 1/60 rad s-2.
Calculate its angular velocity after 30 seconds, and the number of revolutions it makes (a) in
40 seconds, and (b) in coming to rest.
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using ω1 = ω0+αt
1
= 4.19 - (30)
60
= 3.69 rad s-1
60
= 3.69 ×
2
= 35.2 rpm
Thus, the velocity after 30 s is 35.2 rpm. In 30 s the wheel makes 24.6 revolutions; and in
coming to rest, 84 revolutions.
Centripetal Acceleration
Consider a particle which is moving on a circular path of radius r, with a constant angular
velocity ω.
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When the body is at A it possesses an instantaneous linear velocity v, tangential to the circle.
Suppose that during the small interval of time δt, the body moves from A to B, the arc AB
subtending the small angle δθ at the centre of the circle O.
Now although the body is moving with a uniform speed in a circular path the direction of the
instantaneous linear velocity is continually changing and the body is therefore subject to
acceleration in the direction of the instantaneous change in linear velocity.
Referring to the velocity vector diagram, vector Oa represents the velocity at A while vector
Ob represents the velocity at B.
It will be seen that there is a vector change in velocity δv during the motion from A to B of
magnitude ab, acting in a radial direction.
The direction of this instantaneous acceleration ω2 r is towards the centre of the circle O and
is given the particular name centripetal acceleration. For uniform speed of rotation the
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Example 8
A spin drier of diameter 0.2 m is initially at rest and is started with a constant angular
acceleration of 2.5 rad s-2.
(a) Find the angular velocity in rpm after 4 s starting from rest.
While it runs at a steady 720 rpm,
(b) Find the tangential velocity for a point on the surface of the drum
(c) Find the centripetal acceleration of clothes.
Solution
(a) Given ω0 = 0, α = 2.5 rad s-2
Using ω1 = ω0+αt
ω1 = 0 + (2.5) (4)
= 10 rad s-1
60
= 10 ×
2x
= 95.5 rpm
Thus, if the spin drier is initially at rest, after 4 s, the angular velocity of the drum is 95.5
rpm. At constant angular velocity of 720 rpm, the tangential velocity is 75.4 m s -1, and the
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