Applications of torque
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Long Answer Exam 2
Note 25%
of students
got all of
the points
possible
Exam 2 Total Points
Sum of Exams 1+2
This is 30%
of the points
for
determining
the final
grade.
Grading
3 midterm exams, 15% each:
45%
Final exam:
23%
CAPA homework
15%
Tutorial participation + homeworks
10%
In-lecture “clicker” participation:
4%
“SmartPhysics” Prelecture participation 3%
---------------------------------------
---------
Total
100%
Summary of torque so far
Torque is the cross product (or vector product) of the
vector which is the distance from the axis to the force
application and the force vector :
In many cases we only care about the magnitude
where is the angle between the and
vectors
SI units of torque are newton·meter (N·m). This has the
same dimensions as energy but only use Joules for energy
Newton’s 2nd law
can be rewritten as
for cases of rotational motion.
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?
Weight free body diagram:
For the pulley, rope (and therefore the tension
force) comes off at 90° relative to R so
.
Disk moment of inertia is
so
2
kg
becomes
Acceleration along outside of pulley equals acceleration of 2
kg weight because connected by rope so
becomes
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?
We have two equations
rearranges to
Solving
gives
and
2
kg
Clicker question 1
Set frequency to BA
For 2 kg weight attached to 6 kg pulley with R=0.10 m,
what is the kinetic energy of the pulley after dropping for
1 second. Remember v=4 m/s for the weight after 1
second and K=½mv2 and K=½Iω2 and Idisk=½MR2.
A.  4 J
B.  8 J
C.  16 J
D.  24 J
E.  48 J
2
kg
Clicker question 1
Set frequency to BA
For 2 kg weight attached to 6 kg pulley with R=0.10 m,
what is the kinetic energy of the pulley after dropping for
1 second. Remember v=4 m/s for the weight after 1
second and K=½mv2 and K=½Iω2 and Idisk=½MR2.
A.  4 J
B.  8 J
C.  16 J
D.  24 J
E.  48 J
The weight velocity and the outside
pulley velocity are the same since
they are connected by the rope.
Can get angular velocity from rim velocity
and then find kinetic energy
2
kg
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.
The weight acceleration is found to be 4 m/s2.
After 1 second the total kinetic energy was found to be
After 1 second the weight will descend
losing gravitational potential energy
2
kg
Conservation of
energy still holds
Possible pitfalls
Sometimes more than one R in the problem.
R is found in several places (choose correctly)
1. Torque calculation
2. Moment of inertia calculation
3. Conversion between rope linear velocity
and pulley angular velocity
2 kg
If pulley has a hole in the center could
add another R which affects moment of
inertia
2 kg
Notes on gravity
The potential energy due to gravity for an extended
body is measured from the center of mass.
When calculating torque due to gravity, the location of the
force of gravity can be assumed to be the center of mass.
Note the torque will change
as the rod falls due to the
angle changing
A few more tips
If torques and forces aren’t helping, try
conservation of energy (and vice versa)
Everything you have already learned continues to work
Newton’s 2nd law is always true
(for inertial reference frames)
Work and conservation of energy still apply.
Rotation & Translation
Problems so far had a stationary axis.
Other problems have a moving axis like
a boomerang, bowling ball, or yo-yo.
This is a combination of translational
motion and rotational motion
Split motion into two parts
1. Translational motion of the center of mass
2. Rotational motion around the center of mass
Separate kinetic energy terms
We can separate the total kinetic energy into two parts,
translational kinetic energy and rotational kinetic energy.
Sometimes center of mass velocity & angular velocity are related.
Rolling without slipping:
translational
rotational
total
Clicker question 2
Set frequency to BA
A mass m hangs from string wrapped around a pulley of
radius R. The pulley has a moment of inertia I and its pivot is
frictionless. Because of gravity the mass falls and the pulley
rotates. The magnitude of the torque on the pulley is…
A.  greater than mgR
B.  less than mgR
C.  equal to mgR
R
m
Clicker question 2
Set frequency to BA
A mass m hangs from string wrapped around a pulley of
radius R. The pulley has a moment of inertia I and its pivot is
frictionless. Because of gravity the mass falls and the pulley
rotates. The magnitude of the torque on the pulley is…
A.  greater than mgR
B.  less than mgR
C.  equal to mgR
R
(Hint: Is the tension in the string = mg?)
Free body diagram of mass gives T=mg-ma
when a is positive so T<mg so
m
Yo-yo
A yo-yo of mass M and radius R starts from rest
and is dropped, attached by an unrolling string.
What is the center of mass velocity after falling a height h?
We know that falling a height loses gravitational
potential energy which in this case is
This energy goes into kinetic energy:
Since this is rolling without slipping
Setting
Other aspects of translational + rotational motion
Sometimes conservation of energy doesn’t tell us everything
we want to know. In these cases, you probably need both
versions of Newton’s 2nd law
and
You can read the book to see how the previous
yo-yo example works when investigating the
forces and torques (Section 11.5 on page 281).
We are going to try a slightly harder yo-yo problem.
Pulling the yo-yo
Yo-yo with inner radius R1 and outer radius R2 is
on a table. String is pulled with force F as shown.
There is enough static friction to prevent the yo-yo
from sliding. Which way does the yo-yo move?
Rolling without slipping:
and
but you need to make sure you get the signs right!
so
Substituting
Rearranging:
gives