A2 notes on Circular Motion

1. Motion in a Circle 1
1. Uniform circular motion
v
m
ω
r O
1. Consider an object of mass m moving around the circumference in a circle of
radius r at a constant speed.
2. The object is in uniform circular motion. Examples of circular motion include:
(i) A stone being whirled round on a string
(ii) A CD or record spinning on its turntable
(iii) Satellites moving in orbits around the Earth.
2. Linear and angular velocity
1. The linear velocity v is the velocity the object would move off at, along a tangent to the circle, if
there was no centripetal force.
2. Angular velocity ω (or angular frequency) is the rate of change of its angular displacement
about O.
=
where angular displacement (θ) is the angle (in radians) turned in a given direction, e.g. clockwise,
and t is the time taken. Angular speed is the angle (in radians) swept out by the radius per
second. The unit of angular speed and angular velocity is radian per second (rad s-1
).
3. Degrees and radians
1.
r s
θ
r
The radian is the SI unit of angle.
The radian is the angle subtended at the centre of a circle by an arc length s equal to the radius r.
= =
2. For a full circle (360°), s = circumference of the circle = 2 ,
= = 2 .
360 ° = 2 . , 1 =
"#$
= 57.3°
3. To change degrees to radians:
' ( =
' (
"#$
) 2

2. Motion in a Circle 2
To change radians to degrees:
' *+ ** ( =
' (
) 360
4. Period, frequency, linear and angular velocity
1. The period (or time period) T is the time taken to complete one revolution (i.e. to turn through
2π radians),
,-.* /0 01* *2034,-01 5 =
1+3* ,4 1*
1+43 2*306-,7
5 =
2
2. Frequency (f) is the number of revolutions per second, / =
8
9
,
hence ω =
9
= 2 /
3. For one revolution: speed,
2 =
;<=
; >
=
9
but ω = 9
,
Hence 2 =
4. Linear speed v increases with the radius r of the circle, but the angular velocity ω is constant.
5. Centripetal force and acceleration
v
object
direction of a and F
centre
1. A resultant (or unbalanced) force is needed to make an object move in a circle, otherwise it
would continue to move in a straight line, according to Newton’s first law. 2. Since the speed is
constant, the force acting on it must be perpendicular to the direction of motion otherwise it would
increase or decrease the speed, if it had a component in the direction of motion. The resultant
force is the centripetal force, (centre seeking) because it pulls the object towards the centre.
3. The centripetal force acts at right angles to the direction of motion, so no work is done on the
object because there is no movement in the direction of the force. The kinetic energy and speed
of the object is unchanged.
4. The velocity continually changes direction.
Acceleration is the rate of change of velocity, so the object is accelerating.
5. Applying Newton’s second law
? = . ,
the acceleration is in the direction of the resultant or centripetal force, and is called the centripetal
acceleration. It can be shown that
,ℎ* 6*1, -A*, 3 66*3* ,-01, =
=B
0 =
6. The centripetal force is
6*1, -A*, 3 /0 6* ? = .
=B
or ? = .

3. Motion in a Circle 3
7. The centripetal force may be due to a single force, or to a combination of several forces. For
example
(i) for an object whirling round the end of a string, the tension in the string is the centripetal force,
(ii) for a satellite moving round the earth, the force of gravity between the satellite and the Earth is
the centripetal force.
(iii) For a capsule on the London Eye, the centripetal force is the resultant of the support on the
capsule and the force of gravity on it.
6. Proof of =
=B
vB
B δθ
vA vB
O δθ vA
A δv
1. The object turns through an angle δθ when it moves from A to B at constant speed. The
distance AB along the circle is
E = 2 E,.
2. The velocity direction of the object turns through the same angle δθ.
The change in velocity,
E2 = 2*306-,7 , F – 2*306-,7 , H,
as shown in the triangle of vectors.
The change in velocity δv is towards O, i.e. the acceleration is towards the centre of the circle.
3. The triangles ABO and the velocity vector triangles are similar, since they have the same
shape with the same angle between the sides and with two sides of the same length.
4. Since δθ is small,
E =
I
and E =
I=
=
Eliminating δθ gives
I
=
I=
=
But E = 2 E,,
= I
=
I=
=
So
I=
I
=
=B

4. Motion in a Circle 4
5. The acceleration is
I=
I
,
we get centripetal acceleration
=
=B
4. Using 2 = gives 6*1, -A*, 3 66*3* ,-01 =
=B
=
7. Examples
(i) Conical pendulum
θ
T cos θ
T
r
mg
The forces acting on the mass are the tension in the thread and the weight. The tension is
resolved into horizontal and vertical components:
Vertically: T cos θ = m g
Horizontally: T sin θ = mv2
/r
Dividing: tan =
9
9 <
=
MNB
O
;
tan =
2
+
The horizontal component of the tension provides the centripetal force.
2. Banking
N
θ
θ
F
θ mg

5. Motion in a Circle 5
On a banked track the normal reaction and friction both have a component towards the centre of
the track. These combine to give the resultant centripetal force to make the car go round the
track. Resolving:
horizontally: P -1 + ? 60 =
; =B
vertically: P 60 = .+
At one particular speed and angle of banking, the frictional force F = 0.
The centripetal force is then N sin θ.
Dividing the two equations gives
R STU
R VWS
= tan =
=B
The equation shows that for a given radius of bend, the angle of banking is only correct for one
speed.
3. Car travelling over a circular bridge (humpback) bridge
reaction R
v
W = mg
(radius of bridge = r)
At the top of the bridge, the force from the road on the car is R upwards and is in the opposite
direction to the weight mg.
The resultant or net inward force is the centripetal force.
.+ – X =
; =B
The weight mg is a constant, so as v increases, R decreases.
When R = 0, the wheels are just about to leave the ground.
The maximum speed at which this occurs is:
; =B
= .+
2 = Y +

6. Motion in a Circle 6
4. Motion in a vertical circle
(a) Maximum and minimum tension
TT
mg TS
TB
mg
mg
An object is whirled in a vertical circle at a constant speed. The net or resultant force is the
centripetal force. For a given speed, the centripetal force is constant. The weight is constant.
The tension in the string varies.
At the bottom: 5Z − . + =
; =B
At the top: 59 + . + =
; =B
At the side: 5 =
. 22
(b) Looping the loop
mg
RT RS
RB
mg
mg
1. It is possible to loop the loop on a roller coaster ride without falling out.
The forces acting on the rider are
(i) His weight, mg, which does not change.
(ii) The contact force R that the seat exerts on the rider. This varies in size as the car goes round
the circle.

7. Motion in a Circle 7
2. The resultant of these two forces is the centripetal force
; =B
, which is constant for a given
radius r and speed v.
At the bottom: XZ − .+ =
. 22
At the top: X9 + .+ =
. 22
At the sides: The resultant of RS and mg
gives
; =B
3. As the speed of rotation decreases,
; =B
decreases and when RT becomes zero, the man in the ride will just make the loop.
Any speed slower than this makes
mg >
; =B
,
and the man will fall out of his seat.
(c) Water in a bucket
Water can be rotated in a vertical circle without the water falling out at the top, provided that its
weight
mg <
; =B
.