Rotational Dynamics
angle of rotation (rads)

v 2

 2 f
r

angular velocity (rad/s)

atan
r
angular acceleration (rad/s2)
torque (N m)
moment of inertia (kg m2)
Newton’s 2nd Law
Kinetic Energy (joules J)
KE  PE  constant
KEtrans  KErot  PE  constant
(Conservation of Mechanical
Energy)
IC
ptot   mi vi  constant
i
Tod
Ltot 
 Ii i  constant
i
Torque and Moment of Inertia
  rF
Rotational Analog to Newton’s Second Law (F = ma):
Units of τ: Nm
Units of I: kg m2
Units of α: rad/s2
  I
Torque = (moment of inertia) x (rotational acceleration)
Moment of inertia I depends on the distribution of mass and,
therefore, on the shape of an object.
I point  mr
2
Moment of inertia for a point mass
is the mass times the square of the
distance of the mass from the axis
of rotation.
Clicker Question
Room Frequency BA
  rF  I
up
I point  mr 2
down
A light rod of length 2L has two heavy masses (each with mass m)
attached at the end and middle. The axis of rotation is at one end.
1) What is the moment of inertia about the axis?
A) mL2
B) 2mL2
C) 4mL2
D) 5mL2
I   mi ri2 = mL2  m(2L)2  5mL2
E) 9mL2
Clicker Question
Room Frequency BA
  rF  I
up
I point  mr 2
down
A light rod of length 2L has two heavy masses (each with mass m)
attached at the end and middle. The axis of rotation is at one end.
1) What is the moment of inertia about the axis? I  5mL2
2) What is the net torque due to gravity when it’s released?
A) 2mgL
B) -2mgL
C) 3mgL
D) -3mgL
E) 4mgL
   1   2  -r1F  r2 F  L(mg)  2L(mg)  3Lmg
   |  | =  3mgL because all torques are CW.
Clicker Question
Room Frequency BA
  rF  I
I point  mr 2
up
down
A light rod of length 2L has two heavy masses (each with mass m)
attached at the end and middle. The axis of rotation is at one end.
1) What is the moment of inertia about the axis? I  5mL2
2) What is the net torque due to gravity when it’s released?   3mgL
3) If the bar’s released from rest, what’s the magnitude of its angular
acceleration?
 3mgL
3g
 

2
I
5mL
5L
3g
5g
7L
3L
A)
B)
C)
D)
5L
3L
3g
5g |  |  3g
5L
Falling Chimney Demonstration
Clicker Question
Room Frequency BA
A rod of length L and mass M makes
an angle θ with a horizontal table.
L
1) What is the magnitude of the
torque τ exerted on the rod by
gravity?
A) ML
B) Mg sin θ
Mg
θ
C) MgL
τ = Fg, perp (L/2) = Mg cosθ (L/2)
D) 0.5 MgL cos θ
Clicker Question
Room Frequency BA
A rod of length L and mass M makes
an angle θ with a horizontal table.
1) What is the magnitude of the
torque τ exerted on the rod by
gravity? 0.5 MgL cos θ
L
Mg
θ
2) What is the angular acceleration α of the rod when it is released?
(Note: the moment of inertia is I = M L2/ 3)
A) L(cosθ)/3g
B) (3g/2L) cos θ
C) 3g/(Lcos θ)
α = τ/I = (3/2) MgL cos θ/ ML2 = (3g/2L) cos θ
D) gL/3
Clicker Question
Room Frequency BA
A rod of length L and mass M makes
an angle θ with a horizontal table.
1) What is the magnitude of the
torque τ exerted on the rod by
gravity? 0.5 MgL cos θ
L
Mg
θ
2) What is the angular acceleration α of the rod when it is released?
(Note: the moment of inertia is I = M L2/ 3) (3g/2L) cos θ
3) What is the tangential acceleration a of the far end of the rod?
A) (3/2) g/L2 cos θ
B) 2g/3 sin θ
a = L α = (3g/2) cos θ
C) 1.5 g cos θ D) gL/3
Note: ay= a cos θ = 1.5 g cos2θ > g
if cos2 θ > 2/3 !
Falling Chimney Demonstration
  rF  I
Torque and Moment of Inertia
I point  mr 2
Consider the moment of inertia of a hoop of total mass M and radius R:
ri
I
Ipoint = mr2
Ihoop = MR2
Clicker Question
Room Frequency BA
Torque and Moment of Inertia
  rF  I
I
m r
2
i i
i
A force F is applied to a hoop of mass M
and radius R. What’s the resulting
magnitude of the angular acceleration?
A) RF/M
B) F/MR2
C) MR2
D) F/MR
  I hoop  RF
RF
RF
F



2
I hoop MR
MR
I hoop  MR
2
Clicker Question
Room Frequency BA
  rF  I
Torque and Moment of Inertia
I
m r
2
i i
i
Two wheels have the same radius R and total
mass M. They are rotating about their fixed
axes. Which has the larger moment of inertia?
A) Hoop
B) Disk
C) Same
The hoop’s mass is concentrated at its rim,
while the disk’s is distributed from its center
to its rim. So, the hoop will have the larger
moment of inertia.
I hoop  MR
2
Torque and Moment of Inertia
I
2
m
r
 ii
i
(i) Hollow Sphere: (2/3)MR2
needed for CAPA 11
(disk)
Clicker Question
Room Frequency BA
Torque and Moment of Inertia
I
2
m
r
 ii
i
Consider a uniform rod with an axis of rotation through is center and
an identical rod with an axis of rotation through on end. Which has a
larger moment of inertia?
A) IC > IE
B) IC < IE
C) IC = IE
If more mass is further from the axis, the moment of inertia increases.