Torque
Units: m N
2
Lever Arm
is the vector from the point of application
of a force to the axis of rotation (not
necessarily the radius!).
F
F
F
Convention:
CCW is positive (thumb is up)
CW is negative (thumb is down)
Which of the forces pictured as acting upon the
rod will produce a torque about an axis
perpendicular to the plane of the diagram at the
left end of the rod?
a)
b)
c)
d)
F1
F2
Both.
Neither.
F2 will produce a torque
about an axis at the left
end of the rod. F1 has no
lever arm with respect to
the given axis.
The two forces in the diagram have the same
magnitude. Which orientation will produce the
greater torque on the wheel?
a)
b)
c)
d)
F1
F2
Both.
Neither.
F1 provides the larger torque. F2
has a smaller component
perpendicular to the radius.
A 50-N force is applied at the end of a wrench
handle that is 24 cm long. The force is applied in a
direction perpendicular to the handle as shown.
What is the torque applied to the nut by the
wrench?
a)
b)
c)
d)
6 N·m
12 N·m
26 N·m
120 N·m
0.24 m  50 N = 12 N·m
What would the torque be if the force were applied
half way up the handle instead of at the end?
a)
b)
c)
d)
6 N·m
12 N·m
26 N·m
120 N·m
0.12 m  50 N = 6 N·m
What is the torque about the axle of the merry-goround due to the 50-N force?
a)
b)
c)
d)
+60 N·m
-60 N·m
+120 N·m
-120 N·m
-(1.2 m  50 N) = -60 N·m
(clockwise)
What is the net torque acting on the merry-goround?
a)
b)
c)
d)
e)
+36 N·m
-36 N·m
+96 N·m
-60 N·m
+126 N·m
96 N·m (counterclockwise)
- 60 N·m (clockwise)
= +36 N·m (counterclockwise)
Equilibrium
We want to balance a 3-N weight against a
5-N weight on a beam. The 5-N weight is
placed 20 cm to the right of a fulcrum.
What is the torque produced by the 5-N
weight?
a)
b)
c)
d)
+1 N·m
-1 N·m
+4 N·m
+4 N·m
F=5N
 = - Fl
l = 20 cm = 0.2 m
= - (5 N)(0.2 m)
= -1 N·m
How far do we have to place the 3-N
weight from the fulcrum to balance the
system?
a)
b)
c)
d)
2 cm
27 cm
33 cm
53 cm
F=3N
 = +1 N·m
l=/F
= (+1 N·m) / (3 N)
= 0.33 m = 33 cm
Quick Quiz
A constant net torque is applied to an
object. Which one of the following will not
be constant?
1.
2.
3.
4.
angular acceleration
angular velocity
moment of inertia
center of gravity
Center of Mass (CM)
You have learned have that:
perpendicular
  F  r
r or l is the distance to the center of
the rotation!!!!!!
But torque is also equal to:
  I 
Is called MOMENT OF INERTIA (I), and is
measured in kgm2
I  m r
2
a 
Where   t 
r
t
and is measured in rad/s2
MOMENT OF INERTIA (I), measured in kgm2
I  m r
Objects with different shapes,
will have different I:
R
I  MR
L
2
ML2
I
12
2
Objects with the same shape, but
spinning around different axis,
will have different I:
ML2
I
3
ML2
I
12
The only cases when you can actually calculate
the MOMENT OF INERTIA (I) will be:
Case A:
When the object is very small, punctual (not an
extended object), located at a distance r from the
center of the rotation.
Ex. a small ball, a piece of putty, etc.
I  mr
2
The only cases where you can actually calculate
the MOMENT OF INERTIA (I) will be:
Case B:
Only if you can ignore the mass of
the connecting rod:
I  ml  ml  2ml
2
2
2
Case C:
Only if you can ignore the
mass of the connecting rod:
I  m1r  m2 r  m3r  m4 r
2
2
2
r = 0.50m for all masses!
2
Case D:
Only if you can ignore the mass of
the connecting rod:
3
2
4
1
I  m r  m2 r2  m3r3  m4 r4
2
1 1
r1  r3  0
r2  r4  0.50m
2
2
2
Otherwise, consult the table in your book.
Newton’s 2nd Law for Rotation
•
•
•
•
Draw free body diagrams of each object
Only the cylinder is rotating, so apply S = I 
The bucket is falling, but not rotating, so apply SF = ma
Remember that a =  r
Quick Quiz
The two rigid objects shown in the figure below have
the same mass, radius, and angular speed. If the
same braking torque is applied to each, which takes
longer to stop?
1. A
2. B
3. more information is needed
Rotational Kinetic Energy (KEr)
I 
KEr 
2
2
Unit: joule (J)
From now on, when bodies are TRANSLATING (moving along a
straight line) AND ALSO ROTATING (such as a ball rolling down an
incline), the Total Mechanic Energy (TME) will be:
TME  KEr  KEt  PE
I
mv
TME 

 mgh
2
2
2
2
I
mv
TME 

 mgh
2
2
2
2
Total KE
Work  KEt  KEr  PE
Work done by
external force F
If Work by external force is zero, then:
(KEt  KEr  PEg )i  (KEt  KEr  PEg )f
Quick Quiz
Two spheres, one hollow and one solid, are rotating
with the same angular speed around an axis through
their centers. Both spheres have the same mass and
radius. Which sphere, if either, has the higher
rotational kinetic energy?
1. the hollow sphere
2. the solid sphere
3. They have the same kinetic energy.
Angular Momentum (L)
L  I 
L is measured in kgm2/s or also Js
And, just like we’ve seen for linear momentum (p), the relation
between L and torque is:
 I   L
  I   I 


t
t
t
Angular momentum (L) is a vector.
L has the direction of the change in torque.
L is in the same axis as torque (parallel or antiparallel).
L is in the same direction as angular velocity.
L is positive is
counterclockwise.
L is negative if
clockwise.
Conservation of Angular
Momentum (L)
If
 net  0
Then L is conserved:
Linitial  L final
I i  i  I f   f
Conservation of Angular Momentum
• Linear momentum is conserved if the net external force
acting on the system is zero.
• Angular momentum (L) is conserved if the net external
torque (τ) acting on the system is zero.
Fnet = ma
p = mv
= 0,
p = constant
1 2
KE = mv
2
Inertia m :
Inertia I :
If Fnet
If t net = 0,
t net = Ia
L = Iw
L = constant
1 2
KE = Iw
2
Conservation of Angular Momentum
How do
spinning
skaters or
divers
change
their
rotational
velocities?
Conservation of Angular Momentum
Change the
rotational inertia (I)
in order to
change the
rotational velocity (ω)
Larger I (because R is larger)
results in
smaller ω, in order to keep
L = I ω constant.
Smaller I (because R is smaller)
results in
larger ω, in order to keep
L = I ω constant.
Conservation of Angular Momentum
Change the
rotational inertia (I)
in order to
change the
rotational velocity (ω)
Quick Quiz
A horizontal disk with moment of inertia I1 rotates with angular
speed ω1 about a vertical frictionless axle. A second horizontal
disk, with moment of inertia I2 drops onto the first, initially not
rotating but sharing the same axis as the first disk. Because
their surfaces are rough, the two disks eventually reach the
same angular speed ω. The ratio ω/ω1 is equal to
1.
2.
3.
4.
I1 / I2
I2 / I1
I1 / (I1 + I2)
I2 / (I1 + I2)