Physics 106P: Lecture 23 Notes

1
Average 76
2
What country would you come up in if you drilled
a hole straight through the earth from Buenos
Aires?

• Canada
• United States
• Russia
• China
• Finland

3
Physics 211 Lecture 19Todays Agenda

• Review
• Many body dynamics
• Weight and massive pulley
• Rolling and sliding examples
• Rotation around a moving axis Puck on ice
• Rolling down an incline
• Bowling ball sliding to rolling
• Atwoods Machine with a massive pulley

4
Review Direction The Right Hand Rule

• To figure out in which direction the rotation
  vector points, curl the fingers of your right
  hand the same way the object turns, and your
  thumb will point in the direction of the rotation
  vector!
• We normally pick the z-axis to be the rotation
  axis as shown.
• ??? ?z
• ?? ?z
• ?? ?z
• For simplicity we omit the subscripts unless
  explicitly needed.

5
Review Torque and Angular Acceleration

• ?????????????????? ????NET
  I???????????
• This is the rotational analogue of FNET ma
• Torque is the rotational analogue of force
• The amount of twist provided by a force.
• Moment of inertia I is the rotational analogue of
  mass
• If I is big, more torque is required to achieve a
  given angular acceleration.

6
Lecture 19, Act 1Rotations

• Two wheels can rotate freely about fixed axles
  through their centers. The wheels have the same
  mass, but one has twice the radius of the other.
• Forces F1 and F2 are applied as shown. What is
  F2 / F1 if the angular acceleration of the wheels
  is the same?

(a) 1 (b) 2 (c) 4
7
Lecture 19, Act 1Solution
We know
8
Review Work Energy

• The work done by a torque ? acting through a
  displacement ? is given by
• The power provided by a constant torque is
  therefore given by

9
Falling weight pulley

• A mass m is hung by a string that is wrapped
  around a pulley of radius R attached to a heavy
  flywheel. The moment of inertia of the pulley
  flywheel is I. The string does not slip on the
  pulley.
• Starting at rest, how long does it take for the
  mass to fall a distance L.

I
?
R
T
m
mg
a
L
10
Falling weight pulley...

• For the hanging mass use F ma
• mg - T ma
• For the pulley flywheel use ? I?
• ? TR I?
• Realize that a ?R
• Now solve for a using the above equations.

I
?
R
T
m
mg
a
L
11
Falling weight pulley...
Flywheel w/ weight

• Using 1-D kinematics (Lecture 1) we can solve for
  the time required for the weight to fall a
  distance L

I
?
R
T
m
where
mg
a
L
12
Rotation around a moving axis.

• A string is wound around a puck (disk) of mass M
  and radius R. The puck is initially lying at
  rest on a frictionless horizontal surface. The
  string is pulled with a force F and does not slip
  as it unwinds.
• What length of string L has unwound after the
  puck has moved a distance D?

M
R
F
Top view
13
Rotation around a moving axis...

• The CM moves according to F MA

• The disk will rotate about its CM according to ?
  I?

M
A
?
R
F
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Rotation around a moving axis...

• So we know both the distance moved by the CM and
  the angle of rotation about the CM as a function
  of time

(a)
(b)
The length of string pulled out is L R?
Divide (b) by (a)
?
F
F
D
L
15
Comments on CM acceleration

• We just used ? I? for rotation about an axis
  through the CM even though the CM was
  accelerating!
• The CM is not an inertial reference frame! Is
  this OK??(After all, we can only use F ma in
  an inertial reference frame).
• YES! We can always write ? I? for an axis
  through the CM.
• This is true even if the CM is accelerating.
• We will prove this when we discuss angular
  momentum!

16
Rolling

• An object with mass M, radius R, and moment of
  inertia I rolls without slipping down a plane
  inclined at an angle ? with respect to
  horizontal. What is its acceleration?
• Consider CM motion and rotation about the CM
  separately when solving this problem (like we
  did with the lastproblem)...

I
R
M
?
17
Rolling...

• Static friction f causes rolling. It is an
  unknown, so we must solve for it.
• First consider the free body diagram of the
  object and use FNET MACM
• In the x direction Mg sin ? - f MA
• Now consider rotation about the CMand use ? I?
  realizing that
• ? Rf and A ?R

M
f
R
Mg
?
18
Rolling...

• We have two equations Mg sin ? - f MA
• We can combine these to eliminate f

I
A
R
M
For a sphere
?
19
Lecture 19, Act 2Rotations

• Two uniform cylinders are machined out of solid
  aluminum. One has twice the radius of the other.
• If both are placed at the top of the same ramp
  and released, which is moving faster at the
  bottom?

(a) bigger one (b) smaller one (c) same

20
Lecture 19, Act 2Solution

• Consider one of them. Say it has radius R, mass
  M and falls a height H.

H
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Lecture 19, Act 2 Solution
So
So, (c) does not depend on size, as long as the
shape is the same!!
H
22
Sliding to Rolling
Roll bowling ball

• A bowling ball of mass M and radius R is thrown
  with initial velocity v0. It is initially not
  rotating. After sliding with kinetic friction
  along the lane for a distance D it finally rolls
  without slipping and has a new velocity vf. The
  coefficient of kinetic friction between the ball
  and the lane is ?.
• What is the final velocity, vf, of the ball?

?
vf ?R
v0
f ?Mg
D
23
Sliding to Rolling...

• While sliding, the force of friction will
  accelerate the ball in the -x direction F
  -?Mg Ma so a -?g
• The speed of the ball is therefore v v0 - ?gt
  (a)
• Friction also provides a torque about the CM of
  the ball.Using ? I? and remembering that I
  2/5MR2 for a solid sphere about an axis through
  its CM

?
v f ?R
x
v0
f ?Mg
D
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Sliding to Rolling...
(a)
(b)

• We have two equations
• Using (b) we can solve for t as a function of ??
• Plugging this into (a) and using vf ?R (the
  condition for rolling without slipping)

Doesnt depend on ?, M, g!!
?
x
vf ?R
v0
f ?Mg
D
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Lecture 19, Act 3Rotations

• A bowling ball (uniform solid sphere) rolls along
  the floor without slipping.
• What is the ratio of its rotational kinetic
  energy to its translational kinetic energy?

Recall that for a solid sphere about an
axis through its CM
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Lecture 19, Act 3Solution

• The total kinetic energy is partly due to
  rotation and partly due to translation (CM
  motion).

rotational K
translational K
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Lecture 19, Act 3 Solution
rotational K
Translational K
28
Atwoods Machine with Massive Pulley
y

• A pair of masses are hung over a massive
  disk-shaped pulley as shown.
• Find the acceleration of the blocks.

x
M

• For the hanging masses use F ma
• -m1g T1 -m1a
• -m2g T2 m2a

?
R
T2
T1
a

• For the pulley use ? I?
• T1R - T2R

m2
m1
a
m2g
(Since for a disk)
m1g
29
Atwoods Machine with Massive Pulley...
Large and small pulleys
y

• We have three equations and three unknowns (T1,
  T2, a). Solve for a.
• -m1g T1 -m1a (1)
• -m2g T2 m2a (2)
• T1 - T2 (3)

x
M
?
R
T2
T1
a
m2
m2
m1
m1
a
m2g
m1g
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Recap of todays lecture

• Review
• Many body dynamics
• Weight and massive pulley
• Rolling and sliding examples
• Rotation around a moving axis Puck on ice
• Rolling down an incline
• Bowling ball sliding to rolling
• Atwoods Machine with a massive pulley