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Year 13 Physics
Rotation & Circular Motion
Rotation
When either a rigid body or a particle
rotates about some fixed point, we can
describe the motion in a similar fashion
to an object moving in a straight line.
v
r
θ
Consider a particle moving in a circle,
Radius r, speed v. In time t, the radius
Sweeps out an angle θ.
Some important definitions.
Angular Displacement (angle)
An angle θ (measured in radians) is
defined as the ration of length of arc
d, to the radius, r.
d

r
 d  r
Units: radians (rad)
d
Angular Velocity
We define angular velocity, ω (omega) as:
change in angular displacement
time


t
Units: rad.s-1
Angular Acceleration
If an particle moving in a circle speeds up, its angular velocity also increases.
We define angular acceleration, α (alpha) as:
change in angular velocity
time


t
Units: rad.s-2
Relationship between angular and linear quantities
As d
= rθ we can write:
v
d r


 r.
 r
t
t
t
Similarly, as:
a
i.e.
v r


 r.
 r
t
t
t
v  r
a  r
Equations of Rotational Motion
Consider the particle shown moving in
a circle radius r.
At time = 0 it has angular velocity ωi.
It undergoes constant angular
acceleration α to reach ωf in time t.
In this time the radius line rotates
through an angle θ.
As for the linear situation it is possible
to derive the following equations.
The only difference being that:
ω replaces v
α replaces a
θ replaces i
ωf
r
θ
α
d
ωi
Rotational
 f  i   t

(i   f )t
2
Linear
v f  vi  at
d
(vi  v f )t
2
  i t  12 t 2
d  vi t  12 at 2
 2f  i2  2
v 2f  vi2  2ad
Angular Momentum
An object moving in a straight line has linear momentum given by: p = mv.
We define angular momentum of a moving particle about some
fixed point P as:
L  mvr
m
r
Note: r is the distance measured from
P to the ‘line of action’ of the velocity
at right angles as shown.
For a particle moving in a circle this
would always be the radius of the circle.
P
v
Torque
A torque is a ‘turning’ force.
In order to cause an object to turn or rotate, the must be some pivot point P,
about which it turns and the force must act some distance away from that
point.
F
d
P
NB: The distance d is measured from the pivot at right angles to the ‘line
of action’ of the force F.
We define: Torque = Force x distance.
  Fd
Units are: Newton.metres (N.m)
Rotational Dynamics
For the particle, mass m moving
in a circle, it has linear energy given by:
E  12 mv 2
but v = rω
so this becomes:
E  12 mr 2 2
The particle has angular momentum.
L = mvr
Using the same substitution, this becomes:
L  mr 2
v
r
θ
m
s
If a force acts in the direction of motion, the particle will accelerate.
F=ma
Using the substitutions:
  Fr
a  r
This becomes:
  mr 2
Rotational Inertia
Each of the last three equations contain the term mr2.
We define this quantity for a PARTICLE as Rotational Inertia I.
I  mr 2
NB. For more complex shapes we must use:
I   r 2 dm
These three equations become:
ER  12 I 2
L  I
  I
NB. These are the rotational analogies to their linear counterparts:
I replaces m
ω replaces v
α replaces a
τ replaces F
L replaces p
ER replaces Ek
This is called Rotational kinetic energy)
Example: A particle, mass 2.0kg moves in a circle, radius 5.0m,
with constant speed. It completes 1revolution every 4.0s. Find:
2r 2  5

 7.9m.s 1
T
4
(a) Its speed.
v
(b) Its angular velocity.

(c) Its rotational inertia.
The object is a particle so:
v 7.9

 1.6rad .s 1
r
5
I  mr 2  2  52  50kg.m2
(d) Its rotational energy.
ER  12 I 2  0.5  50 1.6 2  62 J
Rotational Inertia for ‘Real’ Objects
The value of I for rotating bodies other than a particle is usually found
by considering energy considerations or using the conservation of
angular momentum.
Solid cylinder, mass M, radius R
I  12 MR 2
Solid sphere, mass M, radius R
I  52 MR 2
Conservation of Angular Momentum
When the nett torque on a system is zero, angular momentum is
always conserved.
I11  I 22
This is best shown with an example.
An ice skater goes into a spin with arms extended. In this position, she
has a rotational inertia I = 6.8kg.m2.
In this position she turns 8 revolutions in 5s.
Find her angular velocity and angular momentum in this position.
8  2
 10rad .s 1
t
5
L  I  6.8 10  68kg.m 2 .s 1



Next, she pulls her arms close to her body. What effect will this have on
her rotational inertia, and why?
What will happen to her angular velocity, and why?
It will reduce the value of I as the mass is closer to the axis of rotation.
As angular momentum, Iω is conserved, if I decreases, angular velocity ω
will increase.
In this new position, it is found that she completes12 revolutions in the next
5s. Find her new angular velocity, hence her new rotational inertia.
12  2
 15rad .s 1
t
5
I 22  68kg.m 2 .s 1

I2 


68
 4.5kg.m 2
15
Find her energy before and energy after she tucks her arms in.
Ei  12 I112  0.5  6.8 10 2  340 J
E f  12 I 222  0.5  4.5 152  506J
How has this increase in energy come about?
In order to bring her arms in towards the axis of rotation, she must do work.
The work done is equal to the change in energy.
Circular Motion
When a particle moves in a circle with
constant speed it accelerates.
This is because its velocity (a vector)
is continually changing, as its direction
is always changing.
v
m
r ar
This centripetal (or radial) acceleration ar
is directed towards the centre of the circle.
v2
ar 
r
Centripetal (radial) Force
This is the force required to hold an object in a circle. Fr = mar.
mv2
Fr 
r
Frequency and Period
The time taken for a rotating object to complete one revolution
is the PERIOD,
T.
The number of revolutions in 1 second is the FREQUENCY, f.
These two quantities are related :
1
T
f
Which leads to:
f 
1
T
In one revolution:
v
circumference
period

1revolution
period
v
2r
 2rf
T

2
 2f
T
Example: The Conical Pendulum
θ
Consider a mass on a string suspended
as shown moving in a horizontal circle.
l
The forces acting on the mass are:
Tension
v
r
m
W
These forces can be
shown on a vector
diagram.
FR is the resultant
force acting towards
the centre.
W
Tension
FR
Find an expression for the period of rotation.
Note that:
r  l sin 
FR  mg tan 
v
2r
T
tan  
Substitute into
Hence:
sin 
cos 
mv 2
FR 
r
T  2
l cos 
g