Torque
Torque and golden rule of mechanics
r1
F2
r2
r1
d2
d1
r2
F1
W1 =W2
F1||
r1
Definition of torque
F2
  F  r  Fr sin   Fr 

F1
F1 r1 = F2 r2
F1 d1 = F2 d2
r2
r1 /d1 = r2 /d2
F1 
 
  rF

r

F
Torque and moment of inertia
F
r
  F  r  ma tan r  m  r r  mr 
2
  I
Analog of
Newton’s Law
for rotation
I  mr
2
I 
Analog of mass
for rotation

m i ri
2
Example: Two blocks of masses m1 = 15 kg and m2 =20 kg are connected
through a string that goes through the pulley of mass M = 2.00 kg and
radiuss R = 25.0 cm. What is the angular acceleration of the pulley?
R
M
I 
T2
T1
 
1
2
MR
T 2 R  T1 R 
a
R
Ma
 m 1  g  m 2  m 1 
1
2
M a
T 2  T1 
m 2

1
2
m 2 g  T2  m 2 a
m1g
 
MR
T1  m 1 g  m 1 a
m2g
a
1
2
T 2  T1 
R
m2
m1
  I
2
m 2
m 2
 m1 
 m1 
1
2
M
g
R
1
2
Ma
 5 . 45 rad / s
2
2
a
R
Example: What is the angular acceleration of a tree as it falls down?
Model of a tree as uniform rod of length L.
  mg  12 L  cos 
I 
L
1
3
 
mL

2

mg  12 L  cos 
1
3
I

a  L 
3
2
mL
2

3 g cos 
2L
g cos 
mg
The top of the tree will have an acceleration larger than g when:
3
2
g cos   g

cos  
2
3

  48 . 2


v2

v cm

v cm

v cm

v cm

v cm
Rolling


v 3  2 v cm


v 3  v cm
+

v4


v1   v cm

v2
=
45°

v cm
45°

v4

v1  0
Rolling without slipping: v cm   R
Example: The rear wheel on a clown’s bicycle has twice the radius of
the front wheel. Is the linear speed at the very top of the rear wheel
greater than, less than, or the same as that of the front wheel?
A. Twice greater than
B. Twice less than
C. Four times greater than
D. Four times less than
E. The same
F. Non of these
Example of Rolling: roller race
Which of these get to the bottom of the ramp first?
KE 
1
2
KE 
1
2
Mv
2
CM
Mv
2
CM

1
2
I CM 
(1  k )
2

1
2
k 
2
CM
Mv

1
2
 v CM 
kMR 

 R 
2
2
I
MR
2
Mgh 
1
2
Mv CM (1  k )
2
Example: Cylinder 1 is released on an incline and rolls down w/o slipping.
Cylinder 2 has an initial angular speed at the bottom of the ramp and
starts rolling up w/o slipping. What is the direction of the static friction
force in each case?
A.
fs
1
Rolling down
B.
Vrelative
Rolling up
fs
2
C.
Vrelative
Friction must oppose relative motion. In the absence of friction:
cylinder 1 would slide down (no rotation),
cylinder 2 would keep rotating at the base of the ramp without going up.
In both cases, relative velocity (at the point of contact) is down the incline,
but for motion w/o slipping it should be zero, and fS must point up.
The weight and the normal force produce zero torque about the CM.
To produce “clockwise” angular acceleration and the appropriate torque
direction (into the page), fS must point up in both cases.
Example: A cylinder of mass M and radius R rolls down an incline of
angle θ with the horizontal. If the cylinder rolls without slipping, what is
its acceleration?
Newton's 2nd law for rotation:
f s R  I CM 
Rolling without slipping:
v CM  R   a CM  R 
f s  I CM 
1
R


1
2
MR
2
I CM 
1
2
MR
fs 
1
2
Ma CM
 a CM  1


 R R

2
Mg sin   f s  Ma CM
Newton's 2nd law for translation of the CM:
N
Mg sin  
fS
a CM 
2
3
3
2
Ma CM
g sin 
compare to g sinθ, the results for
an object sliding without friction
mg
θ
Example: A disk of radius R and mass M that mounted on a massless shaft of
radius r << R and rolling down an incline with a groove. What is its acceleration?
M
N
fS
R
m~0
r
I CM 
1
2
MR
2
Mg
M g sin   fS  M a C M
a CM 
fS r  I C M 
a CM  r 
a CM 
M g sin 
M 
MR
2r
2
2
1

1
R
2r
2
2
M g sin 
I CM
M  2
r
g sin 
Very small if R >> r !