Lecture 18
Goals:
• Define and analyze torque
• Introduce the cross product
• Relate rotational dynamics to torque
• Discuss work and work energy theorem with
respect to rotational motion
• Specify rolling motion (center of mass velocity to
angular velocity
• So what causes rotation?
Physics 201: Lecture 18, Pg 1
So what forces make things rotate?

Which of these scenarios will cause the bar to spin?
Fixed rotation axis.
A
F
F
B
F
C
F
F
D
F
F
F
E
Physics 201: Lecture 18, Pg 2
Net external torques cause objects to spin

An external force (or forces) properly placed induced changes
in the angular velocity. This action is defined to be a “torque”

A force vector or a component of a force vector whose “line of
action” passes through the axis of rotation provides no torque

A force vector or the component of a force vector whose “line
of action” does NOT pass through the axis of rotation
provides torque

The exact position where the force is applied matters.
Always make sure the force vector’s line of action contacts the
point at which the force is applied.
Physics 201: Lecture 18, Pg 3
Net external torques cause objects to spin



Torque increases proportionately with increasing force
Only components perpendicular to r vector yields torque
Torque increases proportionately with increasing distance
from the axis of rotation.
 | r || F | sin 


This is the magnitude of the torque
 is the angle between the radius and
F

r
the force vector

Use Right Hand Rule for sign
Physics 201: Lecture 18, Pg 4
Force vector line of action must pass through contact point

Force vector cannot be moved anywhere

Just along line of action
F
r
r
line of action


F
Physics 201: Lecture 18, Pg 5
Resolving force vector into components is also valid

Key point: Vector line of action must pass through
contact point (point to which force is applied)
F = Fr + Ft
line of action
r
Ft

Fr
Fr
F
F t = F sin 
Physics 201: Lecture 18, Pg 6
Exercise Torque


In which of the cases shown below is the torque provided by
the applied force about the rotation axis biggest? In both
cases the magnitude and direction of the applied force is the
same.
Remember torque requires F, r and sin 
or the tangential force component times perpendicular distance
L
F
A.
B.
C.
Case 1
Case 2
Same
F
L
axis
case 1
case 2
Physics 201: Lecture 18, Pg 7
Torque is constant along the line of action
Even though case 2 has a much larger radius vector the
torque remains constant.
 Case 1 :  = L F
 Case 2 :  = r F sin  = F r sin 
 Notice that the sin  = sin (p-) = - cos p sin  = sin 
 Case 2 :  = r F sin  = F r sin  = F r sin (p-) = F L
Here  = 135° and |r| = 2½ L so r sin  = 2-1/2 (2½ L ) = L


p-
L
F
F
L
p-
axis
case 1
case 2
Physics 201: Lecture 18, Pg 8
Torque, like w, is a vector quantity


Magnitude is given by (1)
|r| |F| sin 
(2)
|Ftangential | |r|
(3)
|F| |rperpendicular to line of action |
Direction is parallel to the axis of rotation with respect to the
“right hand rule”
r sin 
F
tangential
line of action
r

F
F
F
Fradial
 Torque
is the rotational equivalent of force
Torque has units of kg m2/s2 = (kg m/s2) m = N m
Physics 201: Lecture 18, Pg 9
Torque can also be calculated with the vector cross product
 
  r F


The vector cross product is just a definition

r  xˆi  yˆj  zkˆ
ˆi
 
rF  x
ˆj
y
Fx
Fy

F  Fx ˆi  Fy ˆj  Fz kˆ
kˆ
z  ( yFz - zFy )ˆi 
Fz
( zF - xF )ˆj 
x
z
( xFy - yFx )kˆ 
Physics 201: Lecture 18, Pg 10
Torque and Newton’s 2nd Law



F  ma
FTangential  maTangential
Ft r  mat r
at 2
2
Ft r  m r  mr   I
r
Ft r    I
Applying Newton’s second law
Physics 201: Lecture 18, Pg 11
Example: Wheel And Rope

A solid 16.0 kg wheel with radius r = 0.50 m rotates
freely about a fixed axle. There is a rope wound
around the wheel.
Starting from rest, the rope is pulled such that it
has a constant tangential force of F = 8 N.
How many revolutions has the wheel made after
10 seconds?
F
r
Physics 201: Lecture 18, Pg 12
Example: Wheel And Rope
m = 16.0 kg radius r = 0.50 m
 F = 8 N for 10 seconds
Constant F  Constant   constant 
Isolid disk = ½ mr2 = 2 kg m2
I  =  = r F   = 4 Nm/ 2 kg m2 = 2 rad/s2
  0  w0 Dt + ½  Dt2   - 0  w0 Dt + ½  Dt2
Rev = ( - 0) / 2p ( 0 + ½  Dt2 )/ 2p
Rev = (0.5 x 2 x 100) / 6.28 = 16

F
r
Physics 201: Lecture 18, Pg 13
Work

Consider the work done by a force F acting on an
object constrained to move around a fixed axis. For
an angular displacement d then ds = r d
F
dW = FTangential dr = Ft ds
dW = (Ft r) d
axis of
R
rotation
d

dr =Rd
dW =  d (and with a constant torque)
 We can integrate this to find:
W =     (f - i)
Physics 201: Lecture 18, Pg 14
Rotation & Kinetic Energy...

The kinetic energy of a rotating system looks
similar to that of a point particle:
Point Particle
K  mv
1
2
Rotating System
2
v is “linear” velocity
m is the mass.
K  Iw
1
2
2
w is angular velocity
I is the moment of inertia
about the rotation axis.
I   mi ri
2
i
Physics 201: Lecture 18, Pg 15
Work & Kinetic Energy:

Recall the Work Kinetic-Energy Theorem:
DK = WNET or WEXT

This is true in general, and hence applies to
rotational motion as well as linear motion.

So for an object that rotates about a fixed axis:
DK  I (w - w )  WNET
1
2
2
f
2
i
Physics 201: Lecture 18, Pg 16
Example: Wheel And Rope

A solid 16 kg wheel with radius r = 0.50 m rotates
freely about a fixed axle. There is a rope wound
around the wheel.
Starting from rest, the rope is pulled such that it has
a constant tangential force of F = 8 N.
What is the angular velocity after 16 revolutions ?
F
r
Physics 201: Lecture 18, Pg 17
Example: Wheel And Rope
Mass 16 kg radius r = 0.50 m Isolid disk=½mr2= 2 kg m2
 Constant tangential force of F = 8 N.
 Angular velocity after you pull for 32p rad?
W = F (xf - xi) = r F D =0.5 x 8.0 x 32p J = 402 J
DK = (Kf - Ki) = Kf = ½ I w2 = 402 J
w = 20 rad/s

F
r
Physics 201: Lecture 18, Pg 18
Exercise Work & Energy

Strings are wrapped around the circumference of two solid
disks and pulled with identical forces for the same linear
distance.

Disk 1, on the left, has a bigger radius, but both have the
same mass. Both disks rotate freely around axes though
their centers, and start at rest.
 Which disk has the biggest angular velocity after the
pull?
w
w1
2
W =    F d = ½ I w2
Smaller I bigger w
(A) Disk 1
start
(B) Disk 2
(C) Same
finish
F
F
d
Physics 201: Lecture 18, Pg 19
Home Example: Rotating Rod

A uniform rod of length L=0.5 m and mass m=1 kg is free to
rotate on a frictionless pin passing through one end as in
the Figure. The rod is released from rest in the horizontal
position. What is
(A) its angular speed when it reaches the lowest point ?
(B) its initial angular acceleration ?
(C) initial linear acceleration of its free end ?
L
m
Physics 201: Lecture 18, Pg 20
Example: Rotating Rod

A uniform rod of length L=0.5 m and mass m=1 kg is free to rotate
on a frictionless hinge passing through one end as shown. The rod
is released from rest in the horizontal position. What is
(B) its initial angular acceleration ?
1. For forces you need to locate the Center of Mass
CM is at L/2 ( halfway ) and put in the Force on a FBD
2. The hinge changes everything!
L
m
mg
S F = 0 occurs only at the hinge
but z = I z = r F sin 90°
at the center of mass and
(ICM + m(L/2)2) z = (L/2) mg
and solve for z
Physics 201: Lecture 18, Pg 21
Example: Rotating Rod

A uniform rod of length L=0.5 m and mass m=1 kg is free to rotate
on a frictionless hinge passing through one end as shown. The rod
is released from rest in the horizontal position. What is
(C) initial linear acceleration of its free end ?
1. For forces you need to locate the Center of Mass
CM is at L/2 ( halfway ) and put in the Force on a FBD
2. The hinge changes everything!
L
m
a=L
mg
Physics 201: Lecture 18, Pg 22
Example: Rotating Rod

A uniform rod of length L=0.5 m and mass m=1 kg is free to rotate
on a frictionless hinge passing through one end as shown. The rod
is released from rest in the horizontal position. What is
(A) its angular speed when it reaches the lowest point ?
1. For forces you need to locate the Center of Mass
CM is at L/2 ( halfway ) and use the Work-Energy Theorem
2. The hinge changes everything!
L
m
mg
L/2
W = mgh = ½ I w2
W = mgL/2 = ½ (ICM + m (L/2)2) w2
and solve for w
mg
Physics 201: Lecture 18, Pg 23
Connection with CM motion

If an object of mass M is moving linearly at velocity
VCM without rotating then its kinetic energy is
2
K T  12 MVCM

If an object of moment of inertia ICM is rotating in place
about its center of mass at angular velocity w then its
kinetic energy is
K R  I CMw
1
2

2
What if the object is both moving linearly and rotating?
K  I CMw  MV
1
2
2
1
2
2
CM
Physics 201: Lecture 18, Pg 24
Rolling Motion

Now consider a cylinder rolling at a constant speed.
VCM
CM
The cylinder is rotating about CM and its CM is moving at
constant speed (VCM). Thus its total kinetic energy is
given by :
K TOT  I CMw  MV
1
2
2
1
2
2
CM
Physics 201: Lecture 18, Pg 26
Rolling Motion

Again consider a cylinder rolling at a constant
speed.
2VCM
VCM
CM
Physics 201: Lecture 18, Pg 27
Rolling Motion

Now consider a cylinder rolling at a constant speed.
VCM
CM
The cylinder is rotating about CM and its CM is moving at
constant speed (VCM). Thus its total kinetic energy is
given by :
K TOT  I CMw  MV
1
2
2
1
2
2
CM
Physics 201: Lecture 18, Pg 28
Motion

Again consider a cylinder rolling at a constant speed.
Rotation only
vt = wR
CM
Both with
|vt| = |vCM |
2VCM
VCM
CM
Sliding only
VCM
CM
Physics 201: Lecture 18, Pg 29
For Tuesday

Read though 11.3
Physics 201: Lecture 18, Pg 30