Racquetball Striking a Wall
Rotational
Dynamics
Stewart Hall
General Physics I
Mechanics
Physics 140
Mt. Etna
Torque
•
•
•
•
•
The “rotating effect” of a force is greater, if the
force is greater
The “rotating effect” of a force is greater, if the
force is applied further from the axis of rotation
The “rotating effect” of the force also depends on
the direction of the force: it is maximized by a
perpendicular force; no effect of a parallel force
These three properties lead to the following



definition of the torque:   r  F
Vector r is directed from the axis to the point F
where the force is applied:
axis
r
Magnitude of torque
•
Magnitude of torque:
  r F
A rotating bar
• A mass is hanging from the end of a horizontal bar
which pivots about an axis through it center, but is
being held stationary. The bar is released and
begins to rotate. As the bar rotates from horizontal
to vertical, the magnitude of the torque on the bar…
A. Decreases
B. Increases
C. Remains the same
A rotating bar
• A mass is hanging from the end of a horizontal bar
which pivots about an axis through it center, but is
being held stationary. The bar is released and
begins to rotate. As the bar rotates from horizontal
to vertical, the magnitude of the torque on the bar…
A. Decreases
B. Increases
C. Remains the same
r
φ
F
  rF sin 
and φ decreases
A rotating bar
• A mass is hanging from the end of a horizontal bar
which pivots about an axis through it center, but is
being held stationary. The bar is released and
begins to rotate. As the bar rotates from horizontal
to vertical, the magnitude of the torque on the bar…
A. Decreases
B. Increases
C. Remains the same
r
F
φ
Lever arm decreases
F
  rF sin 
and φ decreases
Torque
•
Depends on the axis
Torque
•
•
Depends on the axis
Depends on where the force is applied
Torque
•
•
•
Depends on the axis
Depends on where the force is applied
Specifically, gravitational force (weight) can be
viewed as being applied at the center of mass
Torque
•
•
•
•
Depends on the axis
Depends on where the force is applied
Specifically, gravitational force (weight) can be
viewed as being applied at the center of mass
What about the direction?
Torque
•
•
•
•
•
Depends on the axis
Depends on where the force is applied
Specifically, gravitational force (weight) can be
viewed as being applied at the center of mass
What about the direction?
As usual for any vector product, direction of the

torque is given by the right-hand rule:
F

r
  
  r F
 is into page

Torque
•
•
•
•
•
•
Depends on the axis
Depends on where the force is applied
Specifically, gravitational force (weight) can be
viewed as being applied at the center of mass
What about the direction?
As usual for any vector product, direction of the

torque is given by the right-hand rule:
F

r
  
  r F
 is into page

In practice, we will be speaking about clockwise
and counterclockwise torques
The Right Hand Rule:
Calculating a Torque
r rotates CCW into F
F
 points out of page
  r F
r

points
out of
page
r
Axis

points
into
page
r
r
F
Axis
r rotates CW into F
 points into page
Problem 10.2
A.
B.
C.
D.
E.
16 N m
19 N m
28 N m
52 N m
64 N m
Problem 10.2
A.
B.
C.
D.
E.
16 N m
19 N m
28 N m
52 N m
64 N m
|τ1| = 40 N m
|τ2| = 12 N m
The two torques are in
the opposite directions!
Rotational Dynamics
•
No total torque = no angular acceleration
•
If the total torque exerted by all forces on a rigid body
around some axis is not zero, then the angular acceleration
of the body is given by


1
1
   tot   i
I
I
1
1
 z   tot, z   i , z
I
I
where I is the moment of inertia around the same axis
•
Analogous to the Newton’s 2nd law for translational motion
 1 
a  Ftot
m
•
For rotations, both the moment of inertia and the torques
depend on the axis (which must be same!)
Accelerating two wheels
•
A.
B.
C.
D.
E.
Two wheels with fixed axles, each have the same
mass M, but wheel 2 has twice the radius of wheel
1. Each is accelerated from rest with a force
applied as shown. In order to impart identical
angular accelerations to both wheels, how should
F2 compare to F1? Assume that all the mass of
the wheels is concentrated in the rims.
F2
F1
F2 = ½F1
F2 = F1
R
F2 = 2F1
2R
Wheel 1, radius R, mass M
F2 = 4F1
None of the above
Wheel 2, radius 2R, mass M
Accelerating two wheels
•
A.
B.
C.
D.
E.
Two wheels with fixed axles, each have the same
mass M, but wheel 2 has twice the radius of wheel
1. Each is accelerated from rest with a force
applied as shown. In order to impart identical
angular accelerations to both wheels, how should
F2 compare to F1? Assume that all the mass of
the wheels is concentrated in the rims.
F2
F1

rF
F2 = ½F1
 
2
I
Mr
F2 = F1
R
F2 = 2F1
2R
F  Mr
Wheel 1, radius R, mass M
F2 = 4F1
None of the above
Wheel 2, radius 2R, mass M
Example
• A block of mass m is attached to a rope wound
around the outer rim of a disk of radius R and
moment of inertia I, which is free to rotate around an
axle passing through its center of mass. The rope
does not slip on the disk. Because of gravity the
block m falls and the pulley rotates. The magnitude
of the torque on the pulley is…
R
A. less than mgR
I
B. equal to mgR
C. greater than mgR
m
D. impossible to predict
Example
• A block of mass m is attached to a rope wound
around the outer rim of a disk of radius R and
moment of inertia I, which is free to rotate around an
axle passing through its center of mass. The rope
does not slip on the disk. Because of gravity the
block m falls and the pulley rotates. The magnitude
of the torque on the pulley is… = TR
R
A. less than mgR
T
I
B. equal to mgR
C. greater than mgR
m
D. impossible to predict
Example
• A block of mass m is attached to a rope wound
around the outer rim of a disk of radius R and
moment of inertia I, which is free to rotate around an
axle passing through its center of mass. The rope
does not slip on the disk. Because of gravity the
block m falls and the pulley rotates. The magnitude
of the torque on the pulley is… = TR
R
A. less than mgR - because T < mg!
I
B. equal to mgR
T
C. greater than mgR
m
a
D. impossible to predict
mg
Example
•
A block of mass m is attached to a rope wound
around the outer rim of a disk of radius R and
moment of inertia I, which is free to rotate around
an axle passing through its center of mass. The
rope does not slip. What are the acceleration of
the block and the tension in the string?
Hanging block:
R α
mg  T  ma
Pulley:
T
I
T
a
F 
m
mg
  TR  I
Constraint:
a  R
 F  mg  T  ma
  TR  I
 T  mg  ma
a  R
 (mg  ma)R  Ia / R
 a
g
I


 T  mg 
2 
 I  mR 
I
1
mR 2
R
T
I
T
a
 (mg  ma)R  I
m
mg
α
Another example: Yo-yo
•
Work done by gravity goes
not only to translational
motion, but also to rotation:
GPE = KEtrans + KErot
and KErot >> KEtrans
ICM
Another example: Yo-yo
•
•
Work done by gravity goes
not only to translational
motion, but also to rotation:
GPE = KEtrans + KErot
and KErot >> KEtrans
Alternatively, the torque is
proportional to r (small),
while the moment of inertia
is proportional to R2 (large),
resulting in a very small
angular acceleration
ICM
Another example: Yo-yo
•
•
•
Work done by gravity goes
not only to translational
motion, but also to rotation:
GPE = KEtrans + KErot
and KErot >> KEtrans
Alternatively, the torque is
proportional to r (small),
while the moment of inertia
is proportional to R2 (large),
resulting in a very small
angular acceleration
2
mr
Acceleration: a
g
CM
mr  I CM
2
ICM
Translational motion:
Dynamics of a yoyo
mg  T  ma
Rotational motion:
a
  Tr  I CM  I CM
r
I CM
T  2 a
r
T
Solving for acceleration:
r
M, I
mg
a
I CM
mg  2 a  ma
r
m
mr 2
a
g
g
2
I
mr  I CM
m  CM
r2
Dynamics of a yoyo
mr 2
a
g
2
mr  I CM
Approximate the yo-yo as a uniform
disk of radius R:
I CM
1
 mR 2
2
Then:
T
R
r
M, I
mg
mr 2
2r 2
a
g 2
g
2
1
R  2r
mr 2  mR 2
2

2
g
rR

3


2

r
 

2

  g r  0
R
