8.4 Torque
Torque
Rotational Dynamics
Problem-Solving
We began this course with chapters on kinematics, the
description of motion without asking about its causes.
We then found that forces cause motion, and used
Newton’s laws to study dynamics, the study of forces and
motion.
In chapter 8 we have a deceptively brief section on
angular kinematics. It is brief because we already
learned how to solve such problems in chapters 2 and 3.
What’s next?
kinematics ⇒ dynamics
rotational kinematics ⇒ rotational dynamics
The rotational analog of force is torque.
Consider two equal and opposite forces acting at the
center of mass of a stationary meter stick.
F
Consider two equal and opposite forces acting on a
stationary meter stick.
F
F
Does the meter stick move?
ΣFext = macm = 0 so acm = 0.
F
Does the meter stick move?
ΣFext = macm = 0 so acm = 0.
The center of mass of the meter stick does not
accelerate, so it does not undergo translational motion.
However, the meter stick would begin to rotate about its
center of mass.
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Let’s apply a force to a rod
and see how we get a torque.
A torque is produced by a force
acting on an extended (not
pointlike) object.
F
The torque depends on how strong
the force is, and where it acts on
the object.
O
You must always specify your reference axis for
calculation of torque. By convention, we indicate that
axis with the letter “O” and a dot.
Torques cause changes in rotational motion.
Torque is a vector. It is not a force,* but is related to
force.
First apply the force.
You need to choose an axis of
rotation. Usually there will be a
“smart” choice. Label it with a
point (or line) and an “O.”
+
O
Choose the direction of rotation
that you want to correspond to
positive torque.*
F
Label the positive direction with a
curved arrow and a “+” sign! Do this
around the point labeled “O.”
*So never set a force equal to a torque!
*Traditionally, the counterclockwise direction is chosen to be
positive. You are free to choose otherwise.
Draw a vector from the origin
to the tail of the force vector.
Give it a name (typically R).
Important: θ is the angle
between R and F.
Label the angle between R
and F (you may have to
slide the vectors around
to see this angle).
The symbol for torque is the
Greek letter tau (τ). The
magnitude of the torque due
to F is RF sinθ.
+
“Slide” R and F around until
their tails touch. θ is the
angle between them.
O
R
θ
F
+
O
R
R
θ
θ
F
Look at your diagram and determine if the torque would
cause a + or a – rotation (according to your choice of +).
In this case, the rotation would be -, so τz=-RF sinθ.
“Between” means go from R to F.
Don’t be fooled by a problem which
gives an angle θ not “between” the
vectors! (Example coming soon.)
F
This θ is the correct
“angle between.”
Watch out for diagrams
containing some other
angle!
2
In this diagram, which is the angle between R and F?
Often it is easy to visualize F⊥,
the component of F which is
perpendicular to R.
F
θ?
R
NO!
Because sin(θ)=-sin(-θ), either choice will give you the
correct answer (switch direction of + rotation and
switch sign on sine gives no net switch in sign).
Summarizing:
+
R⊥ is called the lever arm or
The z axis passes through the
point O and is perpendicular to
the plane of the paper.
R⊥
R⊥
O
R
F⊥
θ
θ
Sometimes it is easier to
visualize R⊥, the component of R
which is perpendicular to F.
F
The magnitude of the torque due to F is R⊥F, and in this
case τz=-R⊥F. (Note R⊥=R sinθ.)
Important reminder: label the point O about which your
torques are calculated and draw a curved arrow around it
with a + sign to show what you have chosen for a
positive sense of direction.
τz = RF⊥ = R⊥F = RF sinθ
moment arm. The line along
which F is directed is its line of
action.
+
The magnitude of the torque
due to F is RF⊥, and in this case
τz=-RF⊥. (Note F⊥=F sinθ.)
There are two choices for the angle between R and F.
OSE:
There are other ways to find the
torque.
O
Draw the curved arrow around
the point O, not somewhere else!
R
F⊥
θ
A torque producing a + rotation
is +. A torque producing a rotation is -.
R
O
F
+
F
To find the direction of the torque, curl your fingers
around the direction of rotation from R into F. The thumb
of your right hand points in the direction of the torque.
You don’t need to know this for the exam!
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Example 8-8. The biceps muscle
exerts a vertical force of 700 N on the
lower arm, as shown in the figure.
Calculate the torque about the axis of
rotation through the elbow joint.
F
30°
There is no new litany for torques.
You should adapt the litany for force
problems.
When you work with torques, the
first thing you need to do is draw
an extended free-body diagram.
Before that, we need to have a
diagram of the “thing” we are
investigating.
r=5 cm
We have our diagram. Now we
must do a free-body diagram. For
rotational motion, we must do an
extended free-body diagram, which
shows where the forces are
applied.
F
F
O +
r
θ
r
θ
We are not interested in the upper arm!
From the extended free-body
diagram, I see that the angle
between r and F is 90+θ, so
τz=RF sin(90+θ) would work.
I think it is better to look for r⊥ or
F⊥. In this case, r⊥ is easy to see.
F
r
θ
Label the rotation axis.
Choose a + direction for
rotation.
How about this for an OSE?
τz = RF sinθ
No! No! No!
F
r
O + ⊥
θ
r
From the diagram r⊥=r cosθ.
OSE: τz = R⊥F = RF cosθ. Done! (Except for plugging in
numbers.)
“That was a lot of work for something that took 2
lines in the text!”
If more than one torque acts on an object, the net
torque is the algebraic sum of the two torques
(“algebraic” means there may be signs involved).
Example 8-10. Calculate the net
torque on the compound wheel shown
in the drawing.
F2
θ
The diagram will serve as an extended
free-body diagram. No need for a
separate one.
τz,net = ∑τz = τz,F1 + τz,F2
τz,net = +r1F1 + r2(-F2cosθ)
r2
r1
F1
τz,net = r1F1 - r2F2cosθ
No. I showed you a general approach to torque
problems. The text just solved one simple problem.
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8.5 Rotational Dynamics;
Torque and Rotational Inertia
We saw in our study of dynamics that forces cause
acceleration:
ΣF = ma.
Torques produce angular acceleration, and the rotational
equivalent of mass is the moment of inertia, I:
OSE:
Στz = Iαz.
This is really a vector equation, but our problems will all
have a unique axis of rotation, which is “like” a onedimensional problem, so that the only vestiges of the
vector nature of τz will be the sign.
What is this moment of inertia, I?
It is the rotational analog of mass.
I depends on the mass of the object. It also depends on
how the mass is distributed relative to the axis of
rotation.*
Figure 8-20 gives I for various objects of uniform
composition. You will be given this figure (or its
equivalent, or appropriate portions of it) on an exam or
quiz.
Solid cylinder, mass M, radius R
I=½MR2
It doesn’t matter how thick the cylinder is!
*This means a single object can have different I’s for
different axes of rotation!
8.6 Solving Problems in Rotational Dynamics
Example 8-13 (modified). A cord of negligible mass is
wrapped around a frictionless pulley of mass M and
radius R. The pulley rotates about a fixed axle which
passes through its center. A bucket of mass m hangs
from the cord. Calculate the angular acceleration of the
pulley and the linear acceleration of the bucket.
The litany for force problems still works!
Step 1. Draw a basic sketch.
R
M
m
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Step 2. Draw free-body diagrams. For objects that
rotate, the free-body diagram must be extended; it must
show the actual points of application of forces.
α
T
a
R
x
w=mg
P (force due to axle)
+
O
T
x
W=Mg
bucket
pulley
Step 3. Label each vector (done).
Step 4. Draw axes (done).
To avoid extraneous
minus signs, make sure
your a and α have the
same “sense.”
Step 5. Draw projections of forces not along axes (done).
Step 6. OSE
bucket:
ΣFx = max
pulley:
Στz = Iαz
(We don’t need the sum of
forces equation for the
pulley in this example.)
α
T
a
R
x
w=mg
bucket
P (force due to axle)
+
O
T
x
W=Mg
pulley
Step 7. Write out sum of forces/torques equation
explicitly, then replace generic quantities with labeled
quantities.
bucket
Tx + Wx = max
-T + mg = ma
pulley
τT,z + τW,z + τP,z = Iαz
+RT + 0 + 0 = Iα
Step 8. Solve. You need to use the OSE a = Rα to
connect a and α.
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