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Chapter 1
UNIFORM CIRCULAR MOTION
Introduction
There are many cases of objects moving in a curve about some fixed point. The earth and moon
revolve continuously around the sun. The tip of the balance wheel of a mechanical watch moves
to and fro in a circular path about the fixed axis of the wheel e.t.c.
In these cases, motion of an object moving in a circle with uniform speed round a fixed point will
be considered.
Radian measure
In circular motion, it is more convenient to measure angles in 𝑟𝑎𝑑𝑖𝑎𝑛𝑠. Consider a circle with
centre 𝑶 below. A particle moves from 𝑨 to 𝑩 and the distance moved is 𝒔 and at the same time,
there is a change in angle 𝜽 with respect to the centre.
The angle 𝜽 is known as angular displacement. The ratio of the arc length 𝒔 to the radius 𝒓 of the
circle, is the angle size in radians.
∴ Angle in radians, 𝜃 =
𝑎𝑟𝑐 𝑙𝑒𝑛𝑔𝑡ℎ, 𝒔
………………………………………………………………………(i)
For a complete revolution, the angle covered in radians is 2 𝝅 radians.
∴ 360° = 2 𝜋 radians.
𝑨
𝑩
𝑶
𝜽
Angular velocity, 𝝎
𝐼𝑡 𝑖𝑠 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑎𝑠 𝑡ℎ𝑒 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡. Thus, if 𝒕 is the time taken by
a particle to move from 𝑨 to 𝑩 in the diagram above, then,
Angular velocity, 𝜔 =
𝜃
𝑡
𝜽
𝒕
………………………………………………………………. (ii)
Angular velocity is measured in radians per second (rads
) also, from equation (i) above,
Dividing by time, 𝒕, taken by the particle to cover 𝒔,
𝑠
𝑡
𝜃𝑟
𝑡
, but
𝑠
𝑡
is the linear velocity, 𝑣 of the particle and
𝜃
𝑡
is its angular velocity, 𝜔.
∴ 𝒗 = 𝝎𝒓 ………………..……………………………………………………. (iii)
Worked examples
Example 1
A wheel of radius 50 cm is moved through a quarter turn. Calculate:
(i) the angle rotated in radians.
(ii) distance moved through the circumference.
Solution
(i) Angle in degrees =
1
4
× 360 = 90°
∴ 360° = 2 𝜋 radians
90 × 2 𝜋
360
= 0. 5 𝜋 rads
(ii) Linear distance, 𝑠 = 𝜃𝑟
= 50 × 1. 571
Centripetal force
𝑇ℎ𝑖𝑠 𝑖𝑠 𝑡ℎ𝑒 𝑓𝑜𝑟𝑐𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑡𝑜 𝑘𝑒𝑒𝑝 𝑎𝑛 𝑜𝑏𝑗𝑒𝑐𝑡 𝑚𝑜𝑣𝑖𝑛𝑔 𝑖𝑛 𝑎 𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟 𝑡𝑟𝑎𝑐𝑘. The force
acts towards the centre of the circular track. The force depends on:
(i) 𝑀𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑜𝑏𝑗𝑒𝑐𝑡 𝒎. The larger the mass, the greater is the force required.
(ii) 𝐴𝑛𝑔𝑢𝑙𝑎𝑟 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦, 𝝎. Increase in angular velocity, results into an increase in the force
required.
2
(iii) 𝑅𝑎𝑑𝑖𝑢𝑠, 𝒓, of circular track; the larger the radius, the greater is the force required
Combining the three factors above,
2
r
2
𝑟; but experiments show that 𝑘 = 1.
2
Since 𝜔 =
𝑣
𝑟
, then,
2
𝟐
𝑽
𝑨
𝑽
𝑩
𝑨
𝑩
𝑶
𝜽
By Newton’s second law of motion, 𝐹 = 𝑚𝑎.
Then, 𝐹 = 𝑚 (
𝑣
2
𝑟
) or 𝐹 = 𝑚(𝑤
2
Therefore, centripetal acceleration, 𝒂 =
𝟐
or 𝝎
𝟐
Whirling an object in a vertical plane
Consider the three positions 𝑨, 𝑩 and C of the object in the diagram below.
At position 𝑨, resultant force towards the centre = 𝑇 1
1
𝑚𝑣
2
𝑟
𝟏
𝟐
At position 𝑩, resultant force towards the centre = 𝑇
2
𝟐
𝟐
At position 𝑪, resultant force towards the centre = 𝑇
3
3
2
𝑨
𝑩
𝑪
𝒎𝒈
𝒎𝒈
𝒎𝒈
𝑻
𝟐
𝑻
𝟑
𝑻
𝟏
Exercise
1. A stone of mass 2 kg attached to the end of a string 2.5 m long is whirled horizontally in space
at 5 rev/s. Calculate:
(i) its speed in m/s. (𝟕𝟖. 𝟓𝟒 𝒎/𝒔)
(ii) force exerted by the string. (𝟒𝟗𝟑𝟓 𝑵)
2. A string of length 60 cm is used to whirl a 0.5 kg stone in a vertical circle. If the tension of the
string at the bottom of the circle is 28 N, calculate: (i) the speed of the stone in m/s (ii) the
minimum speed required to keep the stone in circular motion. (𝟓. 𝟐𝟓𝟒 𝒎𝒔
−𝟏
, 𝟐. 𝟒𝟒𝟗 𝒎𝒔
−𝟏
)
3. An object of mass 0.5 kg is rotated in a horizontal circle by a string 1m long. The maximum
tension in the string before it breaks is 50 N. What is the greatest number of revolutions per
second of the object? (𝟏. 𝟓𝟏𝟎 𝒓𝒆𝒗/𝒔)
4. What is the force necessary to keep a mass of 0.8 kg revolving in a horizontal circle of radius
0.7 m with a period of 0.5 s? (𝟖𝟖. 𝟒𝟑 𝑵)
Case examples of circular motion
(a) 𝑪𝒐𝒏𝒊𝒄𝒂𝒍 𝒑𝒆𝒏𝒅𝒖𝒍𝒖𝒎
Suppose a small object 𝑨 of mass 𝒎 is tied to a string 𝑂𝐴 of length 𝑙 and then whirled round in a
horizontal circle of radius 𝑟, with 𝑶 fixed directly above the centre 𝑩 of the circle. If the circular
speed of 𝑨 is constant, the string turns at a constant angle 𝜃 to the vertical. This is called a
The horizontal component, 𝑻 𝒔𝒊𝒏 𝜽, of the tension 𝑻, in the string acts towards the centre 𝑩 and is
equal to centripetal,
2
∴ 𝑇 sin 𝜃 =
𝑚𝑣
2
𝑟
………………………………………………………..…. (i)
Also, the weight 𝒎𝒈 , of the object is balanced by the vertical component, 𝑻 𝒄𝒐𝒔 𝜽, of the tension
So, 𝑇 cos 𝜃 = 𝑚𝑔………………………………………………………...(ii)
Dividing equation (i) by equation (ii),
𝑇 sin 𝜃
𝑇 cos 𝜃
𝑚𝑣
2
𝑟
÷ 𝑚𝑔
sin 𝜃
cos 𝜃
𝑣
2
NB:
𝟐
(b) 𝑴𝒐𝒕𝒊𝒐𝒏 𝒐𝒇 𝒄𝒂𝒓 / 𝒕𝒓𝒂𝒊𝒏 𝒓𝒐𝒖𝒏𝒅 𝒃𝒂𝒏𝒌𝒆𝒅 𝒕𝒓𝒂𝒄𝒌
Let’s consider a car or train moving around a banked track in a circular of horizontal radius 𝒓. If
the only forces at the wheels is the normal reactions 𝑹 and there is no side-slip or strain at the
wheels, the force towards the centre of the track is 𝑹𝒔𝒊𝒏 𝜽, where 𝜽 is the angle of inclination of
the plane to the horizontal.
𝑚 = Mass of bicycle and rider.
𝐹 = Frictional force acting on the wheel by the ground.
𝑁 = Normal reaction on the wheel by the ground.
𝑣 = Speed of the bicycle.
Taking moments about 𝑐. 𝑜. 𝑔.
𝑭 × ℎ = 𝑵 × 𝑎 ……………………………………….….. ( 1 )
Resolving horizontally, 𝐹 =
𝑚𝑣
2
𝑟
Resolving vertically, 𝑁 = 𝑚𝑔………………………………………………..…. ( 3 )
Therefore,
𝟐
2
𝟐
2
2
𝑀𝑚
𝑟
, where 𝜇 is the coefficient of friction between the wheel and the road.
2
Therefore, 𝜇 ≥ tan 𝜃 is the condition to avoid skidding.
(d) 𝑴𝒐𝒕𝒊𝒐𝒏 𝒊𝒏 𝒗𝒆𝒓𝒕𝒊𝒄𝒂𝒍 𝒄𝒊𝒓𝒄𝒍𝒆 𝒂𝒏𝒅 𝒄𝒓𝒊𝒕𝒊𝒄𝒂𝒍 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚, 𝒗
𝒄
The minimum velocity is normally calculated when the object is at the top of the circular track.
The minimum velocity is also known as the critical velocity, 𝒗 𝒄
. At the top of the circle, if
𝑚𝑣
𝑐
2
𝑟
= 𝑚𝑔, the tension will become zero, hence there will be no circular motion.
Therefore,
𝑐
2
∴ Critical velocity, 𝒗 𝒄
An acrobat in a circus rides a motorcycle and makes a loop along a circular track in a vertical plane
if the critical velocity is equal to √
i.e. 𝒗
𝒄
A bucket of water can be swung round in a vertical circle without spilling the water. The water in
the bucket will stay in, if the centripetal force is greater than or equal to the total weight of water
and the bucket.
A car can travel over a limp back bridge without ‘flying’ off the ground if the reaction on the
wheels does not reduce to zero.
From the diagram above, for no ‘flying’, Resultant Force towards the centre = 𝑚𝑔 – 𝑅.
𝑚𝑣 𝑐
2
𝑟
𝑚𝑣
𝑐
2
𝑟
𝑐
2
𝑐
𝑹
Periodic time, 𝑻 =
𝟐𝝅
𝝎
2 × 3. 142
Example 3
A car travels over a lump back bridge of radius of curvature 40 m. Calculate the maximum velocity
of the car if its wheels are to stay in contact with the bridge.
Solution
Force towards centre = 𝑚𝑔 – 𝑅 =
𝑚𝑣
2
𝑟
For no flying, 𝑅 = 0
𝑐
2
𝑐
10 × 40
𝒄
Example 4
A basket containing eggs is rotated in a vertical circle of radius 40 cm. What should be its minimum
speed so that the eggs do not fall out of the basket?
Solution
At the uppermost point, weight of eggs should be equal to centripetal force, so that there is no
unbalanced force to push the eggs downwards.
𝑐
2
𝑐
2
𝒄
𝑐
= √ 10 × 0. 4 = √ 4
𝒄
Applications of uniform circular motion
(a) 𝑺𝒂𝒕𝒆𝒍𝒍𝒊𝒕𝒆
Consider a body projected horizontally from the top of the tower with some velocity. The body
will follow a parabolic path under the effect of gravity and strikes the earth’s surface at a point. If
the body is projected with the velocity greater than the initial velocity, then the body will strike
the surface at another point which is farther from the previous point. If the velocity is gradually
increased, the horizontal range will also increase and finally a stage will come when the body will
not strike the earth’s surface but will always be in a state of free fall under gravity to fall to the
earth. Then the body will describe a stable circular path around the earth and becomes a satellite
of the earth. Satellites are mainly used in weather focusing and in telecommunications.
(b) 𝑪𝒆𝒏𝒕𝒓𝒊𝒇𝒖𝒈𝒆
It consists of a small metal or plastic container tubes which can be electrically or manually rotated
in a circle. If we consider two particles of different masses 𝒎
𝟏
and 𝒎
𝟐
, each of them requires a
centripetal force to keep it in circular motion. The more massive particle requires a greater force
𝟎
Chapter 2
FLOATING AND SINKING
A heavy object is surprisingly lighter and much easier to lift when under water than when it is in
air. A Greek scientist Archimede was the first person to realise that there is an upward force on an
object placed in a fluid which comes from the fluid itself and makes the object appear to lose
weight.
An overflow (Eureka) can is placed on the bench with a beaker below its spout. Water is poured
into the can until it starts to flow out through the spout. When the water has stopped dripping, the
beaker is replaced with another dry beaker whose weight has been determined.
A suitable solid body is tied securely with a thread and is suspended from the lower end of the
spring balance to determine its weight in air. The body is then lowered into the water slowly while
still attached to the spring balance. When completely submerged, its weight is read off again from
the spring balance. The weight of the beaker with the collected water is measured.
(a) Weight of object in air (b) Weight of object in water
Overflow can
Beaker
Displaced water
Spring balance
Solid
1
2
3
4
Upthrust on the solid = (𝑊
1
2
Weight of displaced water = (𝑊 4
3
The upthrust on the solid immersed in water is equal to the water displaced.
Archimedes’ principle
It states that, 𝑤ℎ𝑒𝑛 𝑎 𝑠𝑜𝑙𝑖𝑑 𝑖𝑠 𝑝𝑎𝑟𝑡𝑖𝑎𝑙𝑙𝑦 𝑜𝑟 𝑡𝑜𝑡𝑎𝑙𝑙𝑦 𝑖𝑚𝑚𝑒𝑟𝑠𝑒𝑑 𝑖𝑛 𝑎 𝑓𝑙𝑢𝑖𝑑, 𝑡ℎ𝑒 𝑢𝑝𝑡ℎ𝑟𝑢𝑠𝑡
Deduction of Archimedes’ principle from laws of liquid pressure
Consider a block of length, 𝒍 and cross-sectional area, 𝒂 placed in a liquid of density 𝝆 as shown
below. The downward force, 𝐹 1
1
𝑔)𝑎 and
the upward force, 𝐹 2
2
Resultant upward force = 𝐹
2
1
2
1
Weight of solid (sinker) in water = 𝑊
1
Weight of floater + sinker in water = W
2
Weight of sinker in water + cork in air = 𝑊
3
∴ Weight of floater in air = 𝑊 3
1
Weight of floater in water = 𝑊
2
1
∴ Upthrust = loss of weight for the floater = (𝑊
3
1
2
1
3
2
∴ R.D. =
𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑓𝑙𝑜𝑎𝑡𝑒𝑟 𝑖𝑛 𝑎𝑖𝑟
𝑈𝑝𝑡ℎ𝑟𝑢𝑠𝑡
R.D. =
𝑊
3
−𝑊
1
(𝑊
3
−𝑊
2
)
(c) 𝑹𝒆𝒍𝒂𝒕𝒊𝒗𝒆 𝒅𝒆𝒏𝒔𝒊𝒕𝒚 𝒐𝒇 𝒂 𝒍𝒊𝒒𝒖𝒊𝒅
Weight of a solid in air = 𝑊 1
Weight of same solid in water = 𝑊 2
Weight of same solid in liquid = 𝑊 3
R.D. =
𝑈𝑝𝑡ℎ𝑟𝑢𝑠𝑡 𝑜𝑛 𝑠𝑜𝑙𝑖𝑑 𝑖𝑛 𝑙𝑖𝑞𝑢𝑖𝑑
𝑈𝑝𝑡ℎ𝑟𝑢𝑠𝑡 𝑜𝑓 𝑠𝑜𝑙𝑖𝑑 𝑖𝑛 𝑤𝑎𝑡𝑒𝑟
∴ R.D. =
𝑊
1
−𝑊
3
𝑊
1
−𝑊
2
(d) 𝑨𝒓𝒄𝒉𝒊𝒎𝒆𝒅𝒆𝒔’ 𝒑𝒓𝒊𝒏𝒄𝒊𝒑𝒍𝒆 𝒂𝒏𝒅 𝒑𝒓𝒊𝒏𝒄𝒊𝒑𝒍𝒆 𝒐𝒇 𝒎𝒐𝒎𝒆𝒏𝒕𝒔
A solid is tied at the extreme end of the metre rule and then balanced at 𝐴 with a mass, 𝒎 about
its centre of gravity, as shown below. The distance 𝒙
𝟏
is recorded.
String
𝟏
𝟏
The solid is then submerged completely into water and a new length 𝒙
𝟐
found for the rule to be at
equilibrium.
By principle of moments,
1
1
and, 𝑚𝑔𝑥
2
2
Dividing equation (𝑖𝑖) by (𝑖),
𝑚𝑔𝑥
2
𝑚𝑔𝑥 1
𝑊
2
𝑦
𝑊
1
𝑦
𝑥 2
𝑥 1
𝑊
2
𝑊
1
Subtracting both sides from 1,
𝑥
2
𝑥
1
𝑊
2
𝑊
1
1
2
1
𝑊
1
−𝑊
2
𝑊
1
Or,
𝑥
1
𝑥
1
−𝑥
2
𝑊
1
𝑊
1
− 𝑊
2
Since relative density =
𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑜𝑙𝑖𝑑 𝑖𝑛 𝑎𝑖𝑟
𝑈𝑝𝑡ℎ𝑟𝑢𝑠𝑡
R.D. =
𝑊
1
𝑊
1
− 𝑊
2
String
𝟐
𝟐
