Introduction
In this topic, we will study the angular motion with angular kinematics and rotation dynamics. We will investigate the relationship between, q the angular displacement, w the angular velocity and a the angular acceleration.
4.1 Angular Velocity (halaju sudut)
>Defination of angular displacement,q
q = r = radius , s = linear displacement/perimeter
unit : radian
> Definition for angular velocity
The rate of change of angular displacement for a circular motion.
Unit: Radians per second
> Definition of period, T
The time taken for the particle to complete a circle of the circular motion. For a complete cycle, where q = 2p radians
Other information
360° º 2p
Example 1
What angle in degree has a car travelled around a circular track if the track has a radius of 100 m and distance by the car is 300 m ?
Solution
Example 2
For a point on the circumference (radius = 10 cm) which move through(from beginning to the end) the circle in 60 s. calculate:
(a) angular velocity of the circumference.
(b) linear speed in m/s.
4.2 Centripetal Acceleration (pecutan memusat)
> Definition
Centripetal acceleration is the acceleration where the direction of the velocity change Dv is, perpendicular to the direction of v and is directed towards the centre of the circle.
> Formula
Centripetal Acceleration, a = vw
= rw^{2}
becouse of Centripetal Acceleration, there will be centripetal force
F_{centripetal} = ma = m = mw^{2}r
Example 3
A 900 g object is whirled in a circle at the end of a string. If the string is 0.3 m long and the force in the string is 6.4 N, what is the speed of the object?
Solution
The force acting on an object in circular motion is:
example 4
A rope is wound automatically by a machine at a constant angular velocity. The radius of the roller increases at a steady rate.
Sketch a graph to show:
(a) The variation of speed v of rope with radius,r.
(b) Centripetal force F on the rope at difference distance r from the centre rotation.
solution
Example 5
What is the maximum speed that a car can cross a humpbacked bridge and just remain in contact with it? The radius of the bridge is 20 m.
Solution
Push of bridge on car wheels = P
Weight of car = W
The resultant force F = W – P
This resultant force is the centripetal force.
4.3 Centripetal Force
> Definition
The force on a body moving in a circle and always directed towards the centre O of the circle.
Ø Formula F= ma
Horizontal circle
v = speed of the particle
r = radius of the circle
•Other information
When the speed increases gradually, the tension in the string increases until the braking tension is attained.
– Then the object continues the motion in straight line with uniform speed at the direction tangential to the circle. (First Newton’s Law)
Vertical Circle
Tension maximum

Tension minimum

When q = 0°,
T is maximum

When q = p
T is minimum

•To maintain the movement at vertical circle:.
T ≥ 0
v ≥ The minimum velocity of the object at the highest point.
· Other information
The object remains without dropping because the centripetal force on the object is greater than the weight of the object.
Example 6
Plane flying in a uniform horizontal circle at constant speed and height,
(A) has no resultant force acting on it.
(B) experiences a resultant force acting away from the centre of the circle.
(C) experiences a resultant force acting towards the centre of the circle.
(D) experiences an increasing force acting towards the centre of the circle.
Solution
Objects moving with uniform circular motion experience a resultant force acting towards the centre of the circle.It is called the centripetal force.
Example 7
An object is whirled in uniform circular motion in a vertical circle. If the weight of the object is W the tension in the string is T. what is the centripetal Force. F, at the moment that the object passes through the bottom point of the circle.
Solution
Value of the centripetal force must remain constant at all points in the circle if the motion of the object is uniform circular motion.
At the bottom of the circle the weight of the object is acting down and the tension in the string I acting up. The resultant of these two forces must be the centripetal force. Add the two forces (which are vectors, of course) and you get:
F_{c}= TW
This means that the tension in the string has to be large enough to overcome the weight of the object and also provide the centripetal force.
Note that at the top of the circle both the tension and the weight act down. Because the two forces act in the same direction, the tension must be smaller than it is at the bottom if the resultant they produce (the centripetal force) is to be the same.
Example 8
Which of the following statements is correct for an object moving with uniform horizontal circular motion?
(A) The speed of the object varies but the velocity is constant.
(B) The kinetic energy of the object varies but its momentum is constant. 1
(C) The momentum of the object varies but its speed is constant. 1
(D) The linear velocity of the object varies but its momentum is constant. I
Solution
The only answer to satisfy these conditions is C.
It an object is travelling in a circle, its direction is always changing, so no vector quantities can remain constant e.g. velocity or momentum. Scalar quantities can remain constant, however e.g. 1 speed and kinetic energy.
Note that although the value of acceleration remains the same and it is always towards the centre of the circle, the direction of the acceleration is always changing.
Example 9
An object of mass 2 kg is attached to the end of a string length 1 m and whirled in a vertical
circle at a linear speed of 4 ms^{1}. Find the tension in the string at the top and the bottom of the circle.
Solution
Example 10
A pilot of mass 60 kg flies in a vertical circle of radius 1 km at a speed of 150 m/s.
What is the push of the seat on the pilot at the top of his flight?
Conical Pendulum
Horizontal component

Vertical component

= mrw^{2 }
= m(l) sin q w^{2}
= mlw^{2}

T cos q = mg
ml w^{2} cos q = mg

This method can be used to calculate the value of g but is not suitable because:
(i) It is difficult to measure accurately the angle q.
(ii) It is hard to fix q at a constant value when measuring the period T.
Motion of a motorcyclist round a curved track
The force F produces a moment in the clockwise direction. So, to prevent from toppling over, the person has to bend inwards so that the weight mg and the reaction R form a couple to balance the moment produced by F.

Clockwise moment = anticlockwise moment
F x h = mga
The angle of q increases if:
(a) The speed v increases. (b) The radius, r of the track is small.
The centripetal force is provided by the lateral frictional forces F_{1} and F_{2} between the inner and outer pair of tyres.
There is no acceleration in vertical direction, the normal reactions at the inner and the outer pairs of tyres
R_{1} +R_{2} = mg
The clockwise moments = Total anticlockwise moments
(F_{1 }+ F_{2} )h + R_{1}a = R_{2}a
Solving the equation simultaneously
For the normal reaction R_{1} at the inner pair of tyres is less than R_{2} the normal reaction at the outer pair of tyres.
The car over turn when
v is the maximum speed of the car to round the curved track without overturning. The speed can be increased when:
(a) The radius r of the circular track is bigger.
(b) The centre of gravity G of the car is lower, h is smaller.
(c) The wheel base is bigger.
Example 11
A bicycle and rider of total mass 120kg are rounding a bend of radius 20m. The maximum sideways frictional force available is 600N. Find the greatest linear speed of the bike before it will skid out of the bend.
Solution
F =ma
V = 10 ms^{1}
Example 12
Pendulum moves in a horizontal circle at an angle of 60° to the horizontal as shown. Calculate the speed at which it must move to maintain this angle.
Solution
Ttension in the rope W – the weight of the mass
Resolve the tension vertically and horizontally
T sin 60°
T cos 60°
(i ) Since the pendulum is not moving up or down the vertical forces must be equal.
T cos 60° = mg ……………1
(ii)The centripetal force is given by the force acting towards the centre of the circle. 2
Its all rather neat divide ………2 by……..1
v=2.94 ms 1
Hence speed of the pendulum is 2.9 ms^{1}.
Example 13
An aircraft is flying at a constant speed of 100ms^{1} in a horizontal circle of radius 25 km. A plumb line, attached to the roof of the cabin is inclined at an angle q to the vertical while the aircraft is turning.
(a) What is the centripetal acceleration of the aircraft?
(b) Calculate the angle q.
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