1
Circular Motion
Radians
One revolution is equivalent to 3600 which is also equivalent to 2𝜋 radians. Therefore we can
𝜋
say that 360° = 2𝜋 𝑟𝑎𝑑𝑖𝑎𝑛𝑠 , 180° = 𝜋 𝑟𝑎𝑑𝑖𝑎𝑛𝑠, 90° = 2 𝑟𝑎𝑑𝑖𝑎𝑛𝑠.
360°
Hence 1 𝑟𝑎𝑑𝑖𝑎𝑛 = 2𝜋
Conversions
2𝜋 𝜋
Rule 1: Converting from degrees to radians you multiply by which is equivalent to
360° 180°
360° 180°
Rule 2: Converting from radians to degrees you multiply by which is equivalent to
2𝜋 𝜋
Example 1
Convert:
(a) 450 to radians
(b) 2400 to radians
3𝜋
(c) radians to degrees
5
2𝜋
(d) radians to degrees
3
Angular Displacement (Ѳ)
An object moving along a circular path passes through an angle Ѳ measured at the centre of the
circle as shown in the diagram below. The angle Ѳ is called the angular displacement and it is
measured in radians.
𝑠
We can calculate this angular displacement Ѳ by using: 𝜃 = 𝑟
2
Angular Velocity (ω)
When an object moves along the arc at a constant speed, we can calculate the angular velocity by
using the formula:
𝜃
𝜔= 𝑡
where 𝜃 is the angle in radians
𝜔 is the angular velocity in radians per seconds
t is the time in seconds
Period (T)
The period is the time taken for the object to complete one revolution (through 2𝜋 𝑟𝑎𝑑𝑖𝑎𝑛𝑠 ).
We can calculate the period by using the equation:
2𝜋
𝑇=
𝜔
Instantaneous Angular Velocity
If the angular velocity is not constant, we can work it out by using the formula below:
𝑣
𝜔=
𝑟
We can derive this formula by first calculating the linear velocity which is the rate of change of
distance along the circumference (s) with time (t).
𝑠
𝑣= 𝑡
We know that:
𝑠
𝜃= ∴ 𝑠 = 𝜃𝑟
𝑟
Hence
𝜃𝑟
𝑣= eq (i)
𝑡
But we are given that
𝜃
𝜔= 𝑡
Therefore we can rewrite eq (i) as 𝑣 = 𝜔𝑟 and hence
𝑣
𝜔=
𝑟
3
Centripetal Force
If a body is moving in a circular path, a force must be acting on it or it will move off in a straight
line (according to Newton’s First Law).
If the body is also moving at a steady speed, this force cannot have a component which is the
direction of motion of the body (else the body will increase and decrease in speed). The force
therefore acts perpendicularly to the direction of the motion of the body and is directed towards
the centre of the circular path. This force is known as the centripetal force.
Example: If a brick is spun in a circle on the end of a piece of rope, the tension in the rope
provides the centripetal force. If the rope breaks, the brick will fly off at a tangent.
Centripetal Acceleration
An object moving in a circular path with a constant speed has a changing velocity. This occurs
because the direction of the velocity changes even though its magnitude remains the same. Such
an object must have an acceleration, which is acting in the same direction as the force, towards
the centre of the circle. This acceleration is known as the centripetal acceleration.
We can use the formula below to calculate the centripetal acceleration of an object
𝑎 = 𝜔2 𝑟
If we are given the linear speed of the object we can also the following equation
𝑣2
𝑎 =
𝑟
If we use Newton’s Second Law which calculates the resultant force by using the formula
(F = ma), we will get the following equations.
𝐹 = 𝑚 𝜔2 𝑟
𝑣2
𝐹=𝑚
𝑟
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𝒗𝟐
Deriving that the centripetal acceleration (𝒂 = )
𝒓
Let us consider a particle moving at a constant speed (v) along an arc NOP (as shown in the
diagram above).
The component of the velocity of the particle has the same value at P as at N. Therefore the
acceleration of the x- component is zero.
𝑠𝑝𝑒𝑒𝑑
We know that 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 = 𝑡𝑖𝑚𝑒
Hence for the x-component, the acceleration (ax) can be calculated by:
(𝑣 cos 𝜃 − 𝑣 cos 𝜃)
𝑎𝑥 =
𝑡
∴ 𝑎𝑥 = 0
The y – component of acceleration (ay) can be given by:
𝑣 sin 𝜃 − (−𝑣 sin 𝜃)
𝑎𝑦 =
𝑡
𝑣 sin 𝜃 + 𝑣 sin 𝜃
𝑎𝑦 =
𝑡
2𝑣 sin 𝜃
∴ 𝑎𝑦 = … 𝑒𝑞 𝑖
𝑡
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We can also calculate the speed of the particle along the arc by using the formula:
𝑎𝑟𝑐 𝑙𝑒𝑛𝑔𝑡ℎ (𝑁𝑂𝑃)
𝑠𝑝𝑒𝑒𝑑 (𝑣) =
𝑡
𝑎𝑟𝑐 𝑙𝑒𝑛𝑔𝑡ℎ (𝑁𝑂𝑃)
∴ 𝑡 =
𝑠𝑝𝑒𝑒𝑑 (𝑣)
We can find the arch length (OP) by using 𝑠 = 𝜃𝑟
Therefore the arc length (NOP) would be 2𝑠 = 2𝜃𝑟
Hence we can then say that
2𝜃𝑟
𝑡 = … 𝑒𝑞 𝑖𝑖
𝑠𝑝𝑒𝑒𝑑 (𝑣)
If we rearrange equation i to get
2𝑣 sin 𝜃
𝑡 =
𝑎𝑦
and then equate to equation ii, we get
2𝑣 sin 𝜃 2𝜃𝑟
=
𝑎𝑦 𝑣
Hence
𝑣 2 sin 𝜃
𝑎𝑦 =
𝜃𝑟
If 𝜃 is very small then sin 𝜃 ≈ 𝜃
Then we can say that
𝑣2
𝑎𝑦 =
𝑟
6
Banked Track
Consider a vehicle of mass (m) kg, travelling on a banked circular track. The speed of the vehicle
is constant. Suppose the track is inclined at the angle Ѳ as shown in the diagram below. We can
make the following statements.
Statements:
The weight of the vehicle (W) acts vertically downwards
The Normal (R) acts perpendicular to the track.
The Normal (R) can be separated into its vertical and horizontal components (as shown
in the diagram above)
The centripetal force required to make the vehicle go around the bend is provided by the force
exerted on the tyres by the road (the Normal, R). Since the vehicle of mass (m) is moving with a
constant speed (v) around the bend of radius (r), the centripetal force will provide an acceleration
which can be calculated by using the equation:
𝑣2
𝑎𝑦 =
𝑟
If we use Newton’s Second Law (𝐹 = 𝑚𝑎), we get that
𝑚𝑣 2
𝐹=
𝑟
Since the vehicle acquires a horizontal component as a result of banking we can rewrite the
equation to state
𝑚𝑣 2
𝑅 sin 𝜃 = … 𝑒𝑞 𝑖
𝑟
Note that there is no vertical acceleration but we can make the statement that
𝑅 cos 𝜃 = 𝑚𝑔 … 𝑒𝑞 𝑖𝑖
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If we divide equation i by equation ii, we get that
𝑚𝑣 2
𝑅 sin 𝜃
= 𝑟
𝑅 cos 𝜃 𝑚𝑔
𝑚𝑣 2
tan 𝜃 = ÷ 𝑚𝑔
𝑟
𝑚𝑣 2 1
tan 𝜃 = ×
𝑟 𝑚𝑔
𝑣2
∴ tan 𝜃 =
𝑔𝑟
This equation can be applied to railway trains travelling around a bend on level tracks and to
airplanes banking in order to make a turn (See Muncaster Page 73).
Motion in a Vertical Circle
If we use a small mass which is threaded on wire of length (r) and moving in a circle in the
vertical plane, we can calculate the tension in the wire by using the following equation.
𝑚𝑣 2
𝑇 − 𝑚𝑔 cos 𝜃 =
𝑟
𝑚𝑣 2
∴ 𝑇 = + 𝑚𝑔 cos 𝜃
𝑟
𝑚𝑣 2
When small mass at BOTTOM: 𝜃 = 0° ∴ 𝑇 = + 𝑚𝑔
𝑟
𝑚𝑣 2
When small mass at TOP: 𝜃 = 180° ∴ 𝑇 = – 𝑚𝑔
𝑟
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The Conical Pendulum
Consider a mass (m) is attached to a string of length (L). Suppose the mass travels in a horizontal
circular path so that the radius (r) and the string is at angle Ѳ to the vertical, as shown in the the
diagram below.
Note:
There are two forces acting on the pendulum bob
The weight of the bob (𝑊 = 𝑚𝑔)
The tension in the string (F)
The horizontal component of the tension provides the centripetal force and can be calculated by
using the equation
𝑚𝑣 2
𝑇 sin 𝜃 = … 𝑒𝑞 𝑖
𝑟
Since there is no vertical acceleration, the vertical component can be calculated by
𝑇 cos 𝜃 = 𝑚𝑔 … 𝑒𝑞 𝑖𝑖
If we divide equation i by equation ii, we get that
𝑣2
∴ tan 𝜃 = … 𝑒𝑞 𝑖𝑖𝑖
𝑔𝑟
(NB: same as the equation used in objects that are banking)
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We know that 𝑣 = 𝜔𝑟 … 𝑒𝑞 𝑖𝑣
If we substitute equation iv into equation iii, we get that
𝑣2
tan 𝜃 =
𝑔𝑟
(𝜔𝑟)2
tan 𝜃 =
𝑔𝑟
𝜔2 𝑟 2
tan 𝜃 =
𝑔𝑟
𝜔2 𝑟
∴ tan 𝜃 =
𝑔
From the diagram above we see that
𝑟 = 𝐿 sin 𝜃
Hence
𝜔2 𝐿 sin 𝜃
tan 𝜃 =
𝑔
But
sin 𝜃
tan 𝜃 =
cos 𝜃
Therefore
sin 𝜃 𝜔2 𝐿 sin 𝜃
=
cos 𝜃 𝑔
Making 𝜔 the subject, we can write that
sin 𝜃 ×𝑔
𝜔2 =
cos 𝜃 ×𝐿 sin 𝜃
But
2𝜋
𝜔=
𝑇
Hence
2𝜋 2 𝑔
( ) =
𝑇 Lcos 𝜃
10
4𝜋 2 𝑔
=
𝑇2 Lcos 𝜃
2
4𝜋 2 Lcos 𝜃
𝑇 =
𝑔
Square rooting both sides we get that
4𝜋 2 Lcos 𝜃
𝑇= √
𝑔
Lcos 𝜃
∴ 𝑇 = 2𝜋√
𝑔
