Rotation
• So far we have looked at motion in a straight
line or curved line- translational motion.
• We will now consider and describe rotational
motion – where an object turns about an axis.
• So far we have looked at motion in a straight
line or curved line- translational motion.
• We will now consider and describe rotational
motion – where an object turns about an axis.
• We will start by mentioning some keywords.
1.
2.
3.
4.
Angular position
Angular displacement
Angular velocity
Angular acceleration
• These terms are analogous to their linear
equivalents.
Angular position
• The angular position is an angle measured
between a reference line and a fixed direction
taken as zero.
Reference line
Angular position
• The angular position θ is: 

s
(radians)
r
s: arc length
Reference line
Angular displacement
• The angular displacement gives the change in
angular position of a rotating body.
   2  1
t2
Δθ
θ2
t1
θ1
Counter-clockwise rotation is positive
displacement.
Angular velocity
t2
Δθ
θ2
t1
θ1
• If an object moves through the angle an
angular position of θ1 to θ2 the average
angular velocity is:
 

t

 2  1
t 2  t1
(average angular velocity)
Angular velocity
t2
Δθ
θ2
t1
θ1
• Thus, 
 lim
Units: rads/s
t  0

t

d
dt
(instantaneous angular velocity)
Angular acceleration
• Similarly we can definite the angular
acceleration as,
 avg 
  lim

t  0
t

t
Units: rads/s2

 2  1

t 2  t1
d
dt
(average angular acceleration)
(instantaneous angular acceleration)
Example
• The rotation of a wheel about its central axis is
given by   t 3  27 t  4 . Calculate the angular
velocity and acceleration.
Example
• The rotation of a wheel about its central axis is
given by   t 3  27 t  4 . Calculate the angular
velocity and acceleration.
• Sol:
 
d
 3 t  27
2
dt
 
d
dt
 6t
Equations of linear and angular motion
Motion
Linear
Angular
Formula
Missing variable
v=v0 + at
x-x0
x-x0=v0t + 1/2at2
v
v2=v02 + 2a(x-x0)
t
x-x0=1/2(v + v0)t
a
x-x0=vt - 1/2at2
v0
ω=ω0 + αt
θ-θ0
θ-θ0=ω0t + 1/2αt2
ω
ω2=ω02 +2a(θ-θ0)
t
θ-θ0=1/2(ω + ω0)t
α
θ-θ0=ωt - 1/2αt2
ω0
Relating linear and angular
• Recall that linear and angular variables can be
related as follows:
• Position:
s  r
• Speed:
v
ds
r
dt
 v  r
d
dt
(linear velocity)
Relating linear and angular
• Time:
T 
2 r

2
v
• Acceleration:
a
dv
dt


d
dt
r
Relating linear and angular
• Time:
T 
2 r

2
v
• Acceleration:
dv
a


d
dt
r
dt
• Thus:
a t   r (tangential acceleration)
ar 
v
2
r
  r (radial acceleration)
2
Example
• A grindstone having a constant angular
acceleration of 0.35rad/s2 starts from rest with
an arbitrary reference line horizontal at
angular position   0 . What is the angular
displacement of the reference line at t=18s?
0
• Sol: What equation do we use?
• Sol:
   0   0t 
1
2
t 2
• Sol:
   0   0t 
1
2
t 2
0  0
0  0
(starts from rest)
  0  0 
1
2
0 . 35 18 2
   0  56 . 7 rad  3200   9 rev
• What is the wheel’s angular velocity at t=18s?
• Sol:
  0  t
   0  0 . 35  18  6 . 3 rad / s
The Kinetic Energy of Rotation
• It is clear that a rotating body has kinetic
energy.
• It is clear that a rotating body has kinetic
energy.
• However it is not clear how to calculate the KE
of the body since (1) the particles making the
body move at different velocities, (2) the KE of
the body as a whole is zero since the com has
a velocity of zero.
• The kinetic energy is found by summing the KE
of the particles of the body and writing the
velocity in terms of the angular velocity
(which is the same for all particles).
K 
1
2
K 
m 1 v1 
2

 K 
K 
1
2
1
2

1
2
m 2 v 2  ... 
2
2
m ivi
1
2
m i  ri 
 m r 
2
i i
2
2
1
2
2
m nvn
• The kinetic energy is found by summing the KE
of the particles of the body and writing the
velocity in terms of the angular velocity
(which is the same for all particles).
K 
1
2
K 
m 1 v1 
2

 K 
K 
1
2
1
2

1
2
m 2 v 2  ... 
2
1
2
2
m ivi
1
2
m i  ri 
 m r 
2
2
m nvn
2
2
i i
Moment of inertia
• The kinetic energy is found by summing the KE
of the particles of the body and writing the
velocity in terms of the angular velocity
(which is the same for all particles).
K 
1
2
K 
m 1 v1 
2

 K 
K 
1
2
1
2

m 2 v 2  ... 
2
1
2
1
2
2
m ivi
1
2
m i  ri 
 m r 
2
i i
2
2

1
2
I
2
m nvn
2
Moment of inertia
• The momentum of inertia (rotational inertia)
about some rotational point is the measure of
the resistance to a change in the angular
acceleration due to the action of a torque.
I 

i
m i ri
2
(Moment of Inertia)
• The rotational inertia depends not only on the
mass of the object but how it is distributed
wrt the rotational axis.
Axis of rotation
(1)
(2)
Consider the two rods which have identical total mass . Both rods balance
at the centre. However rod 2 rotates more freely than rod 1. Rod 1 has a
larger moment of inertia than rod 2.
Moment of Inertia of some objects
• The moment of inertia for a continuous body
is:
I 

2
r dm
• The moment of inertia for a continuous body
is:
I 

2
r dm
• If the rotational inertia of a body is known
about any axis which passes through its com
then the parallel-axis theorem can be used to
find the moment of inertia about any parallel
axis.
I  I cm  Mh
2
(parallel-axis theorem)
Example
Example
• Find the rotational inertia about the com of
the rigid body consisting of two particles of
mass m connected by rod (with negligible
mass)of length L.
m
m
cm
L
L
2
2
Example
• Find the rotational inertia about the com of
the rigid body consisting of two particles of
mass m connected by rod (with negligible
mass)of length L.
m
I cm 

i
m
cm
L
L
2
2
m i ri  m 1 r12  m 2 r22  m  12 L   m  12 L 
2
2

1
2
mL
2
2
Example
• Calculate the rotational inertia about the end
of the body.
m
I  I cm  Mh 
2
 mL
2
m
cm
1
2
L
L
2
2
mL  2 m  12 L 
2
2
Example
• Consider a thin, uniform rod of mass M and
length L. What is the rotational inertia of the
rod through its com?
L
L
2
2
Example
• Consider a thin, uniform rod of mass M and
length L. What is the rotational inertia of the
rod through its com?
dx
dm
x
I 

2
r dm
L
L
2
2
Consider an element dm of
width dx. The mass of dm is
the mass per unit length
times the length of dm
Example
• Consider a thin, uniform rod of mass M and
length L. What is the rotational inertia of the
rod through its com?
dx
dm
x
I 

2
r dm
 M 
dm  
 dx
 L 
L
L
2
2
Consider an element dm of
width dx. The mass of dm is
the mass per unit length
times the length of dm
Example
• Consider a thin, uniform rod of mass M and
length L. What is the rotational inertia of the
rod through its com?
dx
dm
x
L
2
 I 
L
L
2
2
2 M 
x
  L  dx
L
2
Consider an element dm of
width dx. The mass of dm is
the mass per unit length
times the length of dm
Example
• Consider a thin, uniform rod of mass M and
length L. What is the rotational inertia of the
rod through its com?
dx
dm
x
L
L
L
2
2
L
3

2
M
x
M


2
 I   x 
 dx 


L
3
 L 

 L
L
2
2
2

1
12
ML
2
Consider an element dm of
width dx. The mass of dm is
the mass per unit length
times the length of dm
Torque
• A simple definition of torque is an influence
which tends to change the rotational motion
of an object.
• The Torque = Force applied X perpendicular
distance from the axis or point of rotation to
the line of action of the force.
  r  F sin  
 
Alt:   r  F
• Newton’s 2nd law can be rewritten for a
rotating body as:
Ft  ma t
(at is the tangential acceleration)
• Newton’s 2nd law can be rewritten for a
rotating body as:
Ft  ma t (at is the tangential acceleration)
   rF t  ma t r  m  r r  m  r 2
• Hence we can write that,
  I
Example
• A uniform disk of mass M=2.5kg and radius
R=20cm is mounted on a fixed horizontal axle.
A block of mass m=1.2kg hangs from a
massless cord that is wrapped around the rim
of the disk. Find the acceleration of the falling
block, the angular acceleration of the disk and
the tension in the cord, assuming that the
cord does not slip and there is no friction at
the axle.
• Sol: Diagram
• Sol: Free body diagrams
T
Mg
T
• Considering the block:
T
T  mg  ma
Mg
• Considering the block:
T
T  mg  ma
Mg
• Considering the disk:
  r F  I 
T
• Considering the block:
T
T  mg  ma
Mg
• Considering the disk:
  r F  I 
  RT 
1
2
MR 
2
T
where the torque is negative because it causes a
clockwise rotation
• Considering the block:
T
T  mg  ma
Mg
• Considering the disk:
  r F  I 
  RT 
  RT 
1
2
1
2
 
MR 
2 a
1

T


Ma
MR
2
R
2
T
a
R
• Substituting for T we get,
a  g
2m
M  2m
  4 . 8 ms
• Hence,
T
2
Mg
T   12 Ma  6 . 0 N
• Finally,
 
a
R
  24 rad / s
2
T