Exam#1 (chapter 1-8)

time: Wednesday 03/24 8:30 am- 9:20 am

 Location: physics building room 112

If you have special needs, e.g. exam time extension, and has not contact me
before, please bring me the letter from the Office of the Dean of Students before
03/17.

AOB
• multiple choice problems.
• Prepare your own scratch paper, pencils, erasers, etc.
• Use only pencil for the answer sheet
• Bring your own calculators
• No cell phones, no text messaging, no computers
• No crib sheet of any kind is allowed. Equation sheet will be provided.

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 The center of gravity
is the point about which the weight of the object itself
exerts no torque.
We can locate the center of gravity by finding the point
where it balances on a fulcrum.

What’s the center of gravity of a disk?

A. any point on the edge of the disk.
B. Center of the disk
C. Any point half way between the center and the edge.

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 How far can
the child walk
without
tipping the
plank?

• For a uniform plank, its center of gravity is at its geometric center.

• The pivot point will be the edge of the supporting platform.
• The plank will not tip as long as the counterclockwise torque from the
weight of the plank is larger than the clockwise torque from the
weight of the child.
• The plank will verge on tipping when the magnitude of the torque of
the child equals that of the plank.

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An 80-N plank is placed on a dock as shown. The
plank is uniform in density so the center of
gravity of the plank is located at the center of the
plank. A 150-N boy standing on the plank walks
out slowly from the edge of the dock. What is the
torque exerted by the weight of the plank about
the pivot point at the edge of the dock?
a) +80 N·m
b) -80 N·m
c) +160 N·m
d) -160 N·m
e) +240 N·m
f) -240 N·m
1 m  80 N = +80 N·m
(counterclockwise)

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 Quiz: An 80-N plank is placed on a dock as shown. The plank is uniform in
density so the center of gravity of the plank is located at the center of the plank.
A 150-N boy standing on the plank walks out slowly from the edge of the dock.
How far from the edge of the dock can the 150-N boy walk until the plank is just
on the verge of tipping?

a) 0.12 m
b) 0.23 m
c) 0.53m
d) 1.20 m

80 N·m / 150 N = 0.53 m

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Demos: 1Q-09
Demonstration
Gyroscopes
 Rotational Inertia and Newton’s Second
Law

• In linear motion, net force and mass determine the

acceleration of an object.
• For rotational motion, torque determines the rotational
acceleration.
• The rotational counterpart to mass is rotational inertia or
moment of inertia.
– Just as mass represents the resistance to a change in linear
motion, rotational inertia is the resistance of an object to change
in its rotational motion.
– Rotational inertia is related to the mass of the object.
– It also depends on how the mass is distributed about the axis of
rotation.
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 Rotational Inertia and Newton’s Second
Law
• Newton’s second law for linear motion:
Fnet = ma
• Newton’s second law for rotational motion: Massless bar


net = I
– The rotational acceleration produced is equal to the torque
divided by the rotational inertia.

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 Rotational Inertia and Newton’s Second
Law

• For an object with its mass concentrated at a point:

– Rotational inertia = mass x square of distance from axis
– I = mr2

• The total rotational inertia of an object like a merry-go-

round can be found by adding the contributions of all the
different parts of the object.

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Two 0.2-kg masses are located at either end of
a 1-m long, very light and rigid rod as shown.
What is the rotational inertia of this system
about an axis through the center of the rod?

a) 0.02 kg·m2
b) 0.05 kg·m2
c) 0.10 kg·m2
d) 0.40 kg·m2

I = mr2
= (0.2 kg)(0.5m)2 x 2
= 0.10 kg·m2

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Rotational
inertias for
more complex
shapes:

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 Angular Momentum

• Linear momentum is mass (inertia) times linear velocity:

p = mv
• Angular momentum is rotational inertia times rotational
velocity:
L = I
– Angular momentum may also be called rotational momentum.
– A bowling ball spinning slowly might have the same angular
momentum as a baseball spinning much more rapidly, because of
the larger rotational inertia I of the bowling ball.
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 Angular momentum is a vector
• The direction of the rotational-velocity vector is given by
the right-hand rule.
• The direction of the angular-momentum vector is the same
as the rotational velocity.
Inertia I, rotational velocity 
Angular momentum : L I

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 Conservation of Angular Momentum
net = I =

i.e. the direction of the angular

momentum change is the same as that
of the net toque.

When net = 0, = 0, i.e. L = const.

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 Conservation of Angular Momentum
Inertia m : Fnet ma
p mv
If Fnet 0, p constant

Inertia I :  net I

L I
If  net 0, L constant
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Demos: 1Q-09
Demonstration
Gyroscopes
Quiz: Some new 2-wheel motocycles can stand
by itself without support when not moving.

A. Not possible.
B. Possible. Angular Momentum conservation is
in action.

https://www.youtube.com/watch?v=1KmhTfhaWG8

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Kinetic Energy

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 1Q- 23 Conservation of angular momentum
Changing the moment of inertia of a skater

How does conservation

of angular momentum
manifest itself ?

When pulling in
both arms, the
person will spin:
A. Faster
B. Slower
C. Same speed
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Fall 2009
 1Q-32 Stability Under Rotation
Example of Gyroscopic Stability: Swinging a spinning Record

Why does the Record not

“flop around” once it is L
set spinning ? L

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Fall 2009
1Q-30 Bicycle Wheel Gyroscope
Gyroscopic action and precession

L
What happens
to the wheel, F
does it fall
down?

mg
F = mg

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 1Q-21 Conservation of angular momentum
Conservation of angular momentum using a spinning wheel

When the wheel is

inverted, the stool
A. Rotate
B. Stay as is.

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 A student sits on a stool holding a bicycle wheel with a
rotational velocity of 5 rad./s about a vertical axis. The
rotational inertia of the wheel is 2 kg·m2 about its center and
the rotational inertia of the student and wheel and platform
about the rotational axis of the platform is 6 kg·m2. What is
the initial angular momentum of the system?

a) 10 kg·m2/s upward
b) 25 kg·m2/s
downward
c) 25 kg·m2/s upward
d) 50 kg·m2/s
downward
L = I = (2 kg·m2)(5 rad./s)
= 10 kg·m2/s
upward from plane of wheel

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Quiz: A student sits on a stool holding a bicycle wheel with a
rotational velocity of 5 rad./s about a vertical axis. The rotational
inertia of the wheel is 2 kg·m2 about its center and the rotational
inertia of the student and wheel and platform about the rotational
axis of the platform is 6 kg·m2. If the student flips the axis of the
wheel, reversing the direction of its angular-momentum vector, what
is the rotational velocity of the student and stool about their axis
after the wheel is flipped?

a) 1.67 rad/s
b) 3.33 rad/s
c) 60 rad/s
d) 120 rad/s

 = L / I = (20 kg·m2/s) / (6 kg·m2)

= 3.33 rad/s
in direction of original angular velocity

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