Torque
•  Tutorial homework due Thursday/Friday.
•  Covering torque today.
1
Torque
Force causes linear acceleration while torque causes angular
acceleration
Force applies
Force applies
small torque
large torque
Force applies
no torque
Use Greek letter tau for torque: τ
Axis of
rotation
τ = rF sin φ
where r is the distance between the force
application and the axis, F is the force magnitude, and φ
is the angle between r and F.
2
Calculating torque 3 ways
τ = rF sin φ
distance r times force times
sine of angle between them.
τ = r ( F sin φ ) = rF⊥
τ = rF sin φ
F
φ
F
φ
r
F sin φ
distance r times the
perpendicular force
τ = F ( r sin φ ) = Fl
F
φ
Force times the perpendicular
distance. Also called lever arm or moment arm.
r
l = r sin φ
r
3
Clicker question 1
Set frequency to BA
Three forces labeled A, B, C are applied to a rod which pivots
on an axis through its center. Which force causes the largest
magnitude torque? sin 45° = cos 45° = 1/ 2 = 1/ 1.414
2F
A.  A
L
B.  B
C
L/2
C.  C
45°
D.  Two are tied for largest
A B
L/4
F
F
E.  All three are the same
L
LF
LF
L
LF
F
sin
90°
=
LF
sin
45°
=
2F
sin
90°
=
C:
B:
A:
2
2
1.4
4
2
A generates the largest magnitude torque
4
Newton’s 2nd law for angular motion
Consider a force pushing on mass m perpendicular
to the rod R. At the moment it is into the screen but
in general, the force is always tangent to the motion.
Newton’s 2nd law for this is Ftan = matan
R
m
Ftan
But we know that atan = Rα so Ftan = mRα
2
Multiplying both sides by R: RFtan = mR α
2
But RFtan is the torque applied and mR is the
moment of inertia for this situation so τ = Iα
This is Newton’s 2nd
law for angular motion
A force causes linear acceleration; a
torque causes angular acceleration
Moment of inertia resists angular acceleration
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like mass resists linear acceleration
Summary of torque so far
A force F applied a distance r from an axis provides a torque
of
!
τ = rF sin φ , where φ is the angle between the r! and F vectors.
SI units of torque are newton·meter (N·m). Not joules.
!
!
Newton’s 2nd law Fnet = ma can be rewritten as
!
!
τ net = Iα for rotational motion.
α is the angular acceleration. It is also related to the tangential
acceleration: atan = rα
The addition of torques requires a modification to the free body
diagram. We need to show where the force acts on the object.
Special note: Gravity always acts at the center-of-mass
of the object (also called the center-of-gravity).
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Generally use M for a rotating mass and m for a point-like mass
Clicker question 2
Set frequency to BA
A mass is hanging from the end of a horizontal bar which pivots
about an axis through its center and is initially stationary. The
bar is released and begins rotating. As the bar rotates from
horizontal to vertical, the magnitude of the torque on the bar…
A.  increases
B.  decreases
C.  remains constant
τ = rF sin φ
r and F stay the same but φ decreases
from 90° when horizontal to 0° when vertical
7
Massive pulleys
A m=2 kg weight is attached to the end of a rope coiled
around a pulley with mass of M=6 kg and radius of 0.1 m.
T
What is the acceleration of the weight?
Weight free body diagram: T − mg = ma
Fg = mg
For the pulley, rope (and therefore the tension
force) comes off at 90° relative to R so τ = RT
(the other forces produce no torque).
Disk moment of inertia is
1
2
T
R
2
kg
MR 2 so τ = Iα becomes RT = 12 MR 2α
Acceleration on outside of pulley equals acceleration of 2 kg
weight because connected by rope. However, signs need to
match so −a = Rα
1
2 a
RT = − MR
2
R
1
becomes T = − 2 Ma
8
Massive pulleys
A 2 kg weight is attached to the end of a rope coiled
around a pulley with mass of 6 kg and radius of 0.1 m.
What is the acceleration of the weight?
1
T
=
−
Ma
We have two equations T = mg + ma and
2
2
kg
mg + ma = − 12 Ma rearranges to −2mg = Ma + 2ma
2
−2mg
−2
⋅
2
kg
⋅10
m/s
−40 N
Solving a =
=
=
= −4 m/s2
M + 2m
6 kg + 2 ⋅ 2 kg
10 kg
gives
Since we defined up as positive, the
acceleration is 4 m/s2 downward.
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Clicker question 3
Set frequency to BA
For the 2 kg weight attached to 6 kg pulley with radius
of 0.10 m, what is the kinetic energy of the weight
after dropping for 1 second. Remember a=4 m/s2 and
K=½mv2, K=½Iω2,and Idisk=½MR2.
A.  2 J
B.  4 J
C.  8 J
D.  16 J
E.  32 J
After 1 second
v = v0 + at = 0 + 4 m/s2 ⋅1 s = 4 m/s
2
kg
Kinetic energy for the weight is
2
2
1
1
K = 2 mv = 2 2 kg ⋅ ( 4 m/s) = 16 J
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