Clicker Question
Room Frequency BA
angle of rotation (rads)

v 2

 2 f
r


atan
r
angular velocity (rad/s)
angular acceleration (rad/s2)
Recall that centripetal acceleration is expressed in terms of
tangential velocity as: ar = v2/r. How is it expressed in terms of
angular velocity ω?
A) ar = ω2/r
C) ar = rω2
B) ar = rω
D) ar = r2ω2
ar = v2/r = (rω)2/r = rω2
Remember: This Week
• Read Ch. 8: And read Dr. Dubson’s excellent notes on Ch. 8.
• CAPA assignment #10 is due this Friday at 10 PM.
• Recitation: Review 1
• Lab Make-up (Labs 1-3): arrange with your TA.
Rotational Kinematics
angle of rotation (rads)
Today:

v 2

 2 f
r


atan
r
angular velocity (rad/s)
angular acceleration (rad/s2)
torque (N m)
moment of inertia (kg m2)
Newton’s 2nd Law
KE  PE  constant
(Conservation
KEtrans  KErot  PE  constant
IC
of Mechanical
Energy)
Rotational Kinematics
angle of rotation (rads)
Constant acceleration a:

v 2

 2 f
r


atan
r
angular velocity (rad/s)
angular acceleration (rad/s2)
Constant angular acceleration α:
v  v0  at
   0  t
1 constant
x  x0  vot  at 2
2
v 2  v02  2ax
1 2
   0   ot   t
2
 2   02  2
Clicker Question
Room Frequency BA
A space ship is initially rotating on its axis with an
angular velocity of ω0 = 0.5 rad/sec. The Captain
creates a maneuver that produces a constant angular
acceleration of α = 0.1 rad/sec2 for 10 sec.
1) What is its angular velocity at the end of this
maneuver?
A) 1 rad/sec
C) 0.6 rad/sec
   0  t
v 2
 
 2 f
r

1 2
   0   ot   t
2
 2   02  2
B) 1.5 rad/sec
D) 5 rad/sec
   0   t  0.5 rad/sec (0.1 rad / sec2 )(10sec)  1.5 rad/sec
Clicker Question
Room Frequency BA
A space ship is initially rotating on its axis with an
angular velocity of ω0 = 0.5 rad/sec. The Captain
creates a maneuver that produces a constant angular
acceleration of α = 0.1 rad/sec2 for 10 sec.
1) What is its angular velocity at the end of this
maneuver?   1.5 rad/sec
   0  t
v 2
 
 2 f
r

1 2
   0   ot   t
2
 2   02  2
2) At how many rpms is it rotating after the maneuver?
A) 10 rpm
B) (30/π) rpm
C) (45/π) rpm
D) π rpm
 revols 1.5 revols  60 sec  45 revols 45
f 


rpm

 
2 sec
2 sec
min
 min

Clicker Question
Room Frequency BA
A space ship is initially rotating on its axis with an
angular velocity of ω0 = 0.5 rad/sec. The Captain
creates a maneuver that produces a constant angular
acceleration of α = 0.1 rad/sec2 for 10 sec.
1) What is its angular velocity at the end of this
maneuver?   1.5 rad/sec
   0  t
v 2
 
 2 f
r

1 2
   0   ot   t
2
 2   02  2
2) At how many rpms is it rotating after the maneuver? f 
45
rpm

3) How far (in radians) did the space ship rotate while the Captain was
applying the angular acceleration?
A) 1 rad
B) 5 rad
C) 10 rad
D) 45 rad
1 2
1
   0   ot   t = (0.5 rad/sec)(10 sec) + (0.1 rad/sec 2 )(10 sec)2
2
2
= 5 rad +5 rad = 10 rad
Clicker Question
Room Frequency BA
A space ship is initially rotating on its axis with an
angular velocity of ω0 = 0.5 rad/sec. The Captain
creates a maneuver that produces a constant angular
acceleration of α = 0.1 rad/sec2 for 10 sec.
1) What is its angular velocity at the end of this
maneuver?   1.5 rad/sec
   0  t
v 2
 
 2 f
r

1 2
   0   ot   t
2
 2   02  2
2) At how many rpms is it rotating after the maneuver? f 
45
rpm

3) How far (in radians) did the space ship rotate while the Captain was
applying the angular acceleration?  = 10 rad
4) How many degrees is this?
A) 180°
B) 1800°
C) (1800/π) degrees
D) 360°
 180 deg 
10rad 
  rad 

1800

deg
Rotational Dynamics
angle of rotation (rads)
Today:
KE  PE  constant

v 2

 2 f
r


atan
r
angular velocity (rad/s)
angular acceleration (rad/s2)
torque (N m)
moment of inertia (kg m2)
Newton’s 2nd Law
KEtrans  KErot  PE  constant
IC
Torque
Rotational Dynamics
F
r
s
θ
r
F
Consider the work done by a
constant force F in moving
an object from point A to
point B on a circle.
A
W  (Component of force along
displacement) x displacement
= F s  F (r ) = (rF )  
Torque
  rF
Units: N m
Torque
 Torques are related to F
 The amount of work done by
F increases linearly with r
and F .
Torque:
  rF
= (lever arm) x (Component of force perpendicular to lever arm)
Torque
  rF
Therefore, to easily rotate an object about an axis, you want
a large lever arm r and a large perpendicular force F :
Torque
  rF
Therefore, to easily rotate an object about an axis, you want
a large lever arm r and a large perpendicular force F :
Clicker Question
Room Frequency BA
  rF
Torque
You pull on a door handle a distance r = 1m from the hinge with a
force of magnitude 20 N at an angle θ =30° from the plane of the
door.
What’s the torque you exerted on the door? (Note: sin 30° =1/2)
A) 8 Nm
B) 10 Nm
C) 12 Nm
D) 20 Nm
  rF  rF sin   (1m)(20N )(1 / 2)  10Nm
Clicker Question
Room Frequency BA
  rF
Torque
Three forces labeled A, B, and C are applied to a rod which pivots
on an axis through its center.
Which force causes the largest size torque?
sin 45  cos 45  1 / 2  0.707
|  A | LF sin 45  0.707LF
L
|  B | F  0.5LF
2
L
|  C | 2F   0.5LF
4
Clicker Question
Room Frequency BA
  rF
Torque
Three forces labeled A, B, and C are applied to a rod which pivots
on an axis through its center.
What is the net torque on the rod?
A) 0.707 LF
E) Zero
B) -0.707 LF
 A  LF sin 45  0.707LF
L
 B  F  0.5LF
2
L
 C  2F   0.5LF
4
 net  (0.707  0.5  0.5)LF
 1.707LF
C) 1.707 LF
D) -1.707 LF
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.
Torque and Moment of Inertia
r
r
m
  rF  I
Consider a point mass m to which
we apply a constant tangential force
F .
atan  r
F  matan  mr
  rF  mr   I
2
I  mr 2
Thus, the moment of inertia for a
point mass is equal to mr2.
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.
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
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