Motion in a Circle 1
Kinematics of uniform circular motion
a) define the radian and express angular displacement in radians
b) understand and use the concept of angular speed to solve problems
c) recall and use v = rω to solve problems
1. Uniform circular motion
v
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. Degrees and radians
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.
arc length s
θ= s=r θ
radius r
2. For a full circle (360°), s = circumference of the circle = 2 π r ,
2 πr
θ= =2 π rad .
r
360
360 °=2 π rad . , 1 rad= =57.3 °
2π
3. To change degrees to radians:
θ( degrees)
s ( rad )= x2 π
360
To change radians to degrees:
Motion in a Circle 2
s (radians)
θ ( degrees )= x 360
2π
3. 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.
θ
ω=
t
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).
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),
angle turned
time for one revolution T =
angular velocity
2π
T=
ω
1
2. Frequency (f) is the number of revolutions per second, f = ,
T
2π
hence ω= =2 π f
T
3. For one revolution: speed,
distance moved 2 πr
v= =
timetaken T
2π
but ω= ,
T
Hence v=ω r
4. Linear speed v increases with the radius r of the circle, but the angular velocity ω is constant.
5. Centripetal force and acceleration
Centripetal acceleration and centripetal force
a) describe qualitatively motion in a curved path due to a perpendicular force, and
understand the centripetal acceleration in the case of uniform motion in a circle
b) recall and use centripetal acceleration equations a = rω 2 and a = v 2/ r
c) recall and use centripetal force equations F = mrω2 and F = mv 2 /r
v
object
direction of a and F
centre
Motion in a Circle 3
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
F=m a ,
the acceleration is in the direction of the resultant or centripetal force, and is called the centripetal
acceleration. It can be shown that
2
v
the centripetal acceleration , a=
r
2
¿ a=r ω
6. The centripetal force is
2
v
centripetal force F =m
r
2
¿ F=m r ω
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.
https://www.youtube.com/watch?v=ID0R43My4Co
https://www.youtube.com/watch?v=Zulw5bQ18Kk
Worked Example
Paper 41, June 2010, Q1
(a) Define the radian.
Motion in a Circle 4
(b) A stone of weight 3.0 N is fixed, using glue, to one end P of a rigid rod CP, as shown in Fig.
The rod is rotated about end C so that the stone moves in a vertical circle of radius 85 cm. The
angular speed ω of the rod and stone is gradually increased from zero until the glue snaps. The
glue fixing the stone snaps when the tension in it is 18 N. For the position of the stone at which the
glue snaps,
(i) on the dotted circle of Fig. 1.1, mark with the letter S the position of the stone,
(ii) calculate the angular speed ω of the stone
Solution
6. Examples of Circular Motion
(i) Conical pendulum
https://www.youtube.com/watch?v=jnkgydYz1y0
θ
T cos θ
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
Motion in a Circle 5
2
mv
Dividing: T sin θ r
tanθ= =
T cos θ mg
2
v
tanθ=
gr
The horizontal component of the tension provides the centripetal force.
2. Banking
https://www.youtube.com/watch?v=eGZWVwcaq0U
N
θ
θ
F
θ mg
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:
2
mv
horizontally: N sin θ+ F cos θ=
r
vertically: N cos θ=mg
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
2
N sinθ v
=tanθ=
N cos θ rg
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
https://www.youtube.com/watch?v=NH72nlsr_D4
reaction R
W = mg
(radius of bridge = r)
Motion in a Circle 6
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.
2
mv
mg – R=
r
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:
2
mv
=mg
r
v=√ rg
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.
2
mv
At the bottom: T B−m g=
r
2
mv
At the top: T T +m g=
r
2
mv
At the side: T S=
r
(b) Looping the loop
Motion in a Circle 7
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.
2
mv
2. The resultant of these two forces is the centripetal force , which is constant for a given
r
radius r and speed v.
2
mv
At the bottom: R B−mg=
r
2
mv
At the top: RT + mg=
r
At the sides: The resultant of RS and mg
2
mv
gives
r
3. As the speed of rotation decreases,
2
mv
decreases and when RT becomes zero, the man in the ride will just make the loop.
r
Any speed slower than this makes
2
mv
mg > ,
r
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
2
mv
mg < .
r
