# Physics 111 Lecture 19 Todayâ€™s Agenda by tlc69476

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```									                Physics 111: Lecture 19

Today’s 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
   Atwood’s Machine with a massive pulley

Physics 111: Lecture 19, Pg 1
Review: Direction & The Right Hand Rule
y
   To figure out in which direction the rotation
vector points, curl the fingers of your right
hand the same way the object turns, and                                    x
your thumb will point in the direction of the
rotation vector!                                         z
y
   We normally pick the z-axis to be the
rotation axis as shown.
= z                                                                  x
= z                                                  z
= z

   For simplicity we omit the subscripts
unless explicitly needed.
Physics 111: Lecture 19, Pg 2
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.

Physics 111: Lecture 19, Pg 3
Lecture 19, Act 1
Rotations
   Two wheels can rotate freely about fixed axles through their
centers. The wheels have the same mass, but one has twice
   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                             F2
F1
(c) 4

Physics 111: Lecture 19, Pg 4
Lecture 19, Act 1
Solution

We know          I
but      FR      and        I  mR 2

so FR  mR 2             F2 mR2 R2
    
F  mR             F1 mR1 R1

F2
Since R2 = 2 R1               2
F1
F2
F1

Physics 111: Lecture 19, Pg 5
Review: Work & Energy

   The work done by a torque  acting through a displacement
 is given by:

W  

   The power provided by a constant torque is therefore given
by:

dW     d
P            
dt    dt

Physics 111: Lecture 19, Pg 6
Falling weight & pulley

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

a        mg
L

Physics 111: Lecture 19, Pg 7
Falling weight & pulley...

   For the hanging mass use F = ma
           I
mg - T = ma
R
   For the pulley + flywheel use  = I
 = TR = I
a
   Realize that a = R         TR  I                  T
R
m
   Now solve for a using the above
equations.                                  a        mg
L
 mR   2
a   2
g
 mR  I 

Physics 111: Lecture 19, Pg 8
Flywheel
Falling weight & pulley...         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
1 2                   2L
L  at              t
2                     a                            T

 mR 2                            m
where     a   2
g
 mR  I                     a        mg
L

Physics 111: Lecture 19, Pg 9
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

Physics 111: Lecture 19, Pg 10
Rotation around a moving axis...
F
   The CM moves according to F = MA                A
M
1 2     F 2
   The distance moved by the CM is thus D        At     t
2      2M
     RF     2F
   The disk will rotate about           =     =        =
its CM according to  = I                I 1          MR
MR 2
2
1 2     F 2
   So the angular displacement is       t     t
2      MR

M      A
1         
I  MR 2           R
2                                 F
Physics 111: Lecture 19, Pg 11
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:
F 2                  F 2
D      t   (a)            t   (b)
2M                   MR
 2                    The length of string
Divide (b) by (a):           R  2 D
D R                    pulled out is L = R:

L  2D


F                                         F

D                    L

Physics 111: Lecture 19, Pg 12

   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!

M        A

R
F
Physics 111: Lecture 19, Pg 13
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 last
problem)...
I
M     R



Physics 111: Lecture 19, Pg 14
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
M
   Now consider rotation about the CM                                f
and use  = I realizing that                            R
 = Rf and A = R
        Mg
A              A
Rf  I           f I
R              R2

Physics 111: Lecture 19, Pg 15
Rolling...

Mg sin  - f = ma                   A
   We have two equations:                                   f I
R2

   We can combine these to eliminate f:

MR 2 sin 
A=g
MR 2 + I
I
A          M    R
For a sphere:

MR 2 sin          5
A=g                    = gsin 
2             7                        
2
MR + MR 2
5

Physics 111: Lecture 19, Pg 16
Lecture 19, Act 2
Rotations
   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

Physics 111: Lecture 19, Pg 17
Lecture 19, Act 2
Solution
    Consider one of them. Say it has radius R, mass M and falls
a height H.
1        1
Energy conservation: - DU = DK            MgH      I  2  MV 2
2        2
1                      V
but   I  MR 2    and      
2                      R
2
1 1 2 V   1
MgH   MR  2  MV 2
2 2  R    2

1       1      3
MgH      MV 2  MV 2  MV 2
4       2      4
H

Physics 111: Lecture 19, Pg 18
Lecture 19, Act 2
Solution

3                      3
So:   MgH      MV 2             gH  V 2
4                      4
4
V      gH
3

So, (c) does not depend on size,
as long as the shape is the same!!

H

Physics 111: Lecture 19, Pg 19
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
Physics 111: Lecture 19, Pg 20
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:
2                       5g                              5 g
 = MgR = MR 2               =                  = 0 + t =         t (b)
5                       2R                               2R


x                 v f= R                   v0
f = Mg

D
Physics 111: Lecture 19, Pg 21
Sliding to Rolling...
5g
   We have two equations:     v  v 0  gt (a)      =     t (b)
2R
2 R
   Using (b) we can solve for t as a function of           t
5 g

   Plugging this into (a) and using vf = R (the condition for
rolling without slipping):
5
vf  v0         Doesn’t depend
7            on , M, g!!


x                  vf= R                v0
f = Mg

D
Physics 111: Lecture 19, Pg 22
Lecture 19, Act 3
Rotations
   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?

1                 2                  1
(a)               (b)                (c)
5                 5                  2

Recall that I  2 MR 2 for a solid sphere about
5
an axis through its CM:

Physics 111: Lecture 19, Pg 23
Lecture 19, Act 3
Solution
   The total kinetic energy is partly due to rotation and partly
due to translation (CM motion).

1        1
K=      I  2  MV 2
2        2

rotational    translational
K               K

Physics 111: Lecture 19, Pg 24
Lecture 19, Act 3
Solution
1        1                                                           V
K=     I  2  MV 2            Since it rolls without slipping:  
2        2                                                           R
rotational   Translational
K                K

2    2 V
2
1 2
I     MR  2
       R  2
K ROT                5
 2
KTRANS     1    2      MV 2     5
MV
2

Physics 111: Lecture 19, Pg 25
Atwoods Machine with Massive Pulley:

   A pair of masses are hung over a                           y
massive disk-shaped pulley as shown.
Find the acceleration of the blocks.                             x
M
   For the hanging masses use F = ma           
 -m1g + T1 = -m1a                                       R
 -m2g + T2 = m2a

a                   T1           T2
I                                           a
   For the pulley use  = I
R
a 1                           m1             m2
 T1R - T2R I  MRa
R 2
a
1                                           m2g
(Since I      MR 2 for a disk)        m1g
2
Physics 111: Lecture 19, Pg 26
Large and small pulleys
Atwoods Machine with Massive Pulley...

   We have three equations and three                         y
unknowns (T1, T2, a). Solve for a.
x
-m1g + T1 = -m1a      (1)                           M

-m2g + T2 = m2a       (2)                                R

1
T1 - T2      Ma   (3)                       T1           T2
2                                                      a

m1             m2
    m1  m 2    
a                g
 m1  m 2  M 2                               a
m2g
m1g

Physics 111: Lecture 19, Pg 27
Recap of today’s lecture

   Review                                       (Text: 9-1 to 9-6)
   Many body dynamics
   Weight and massive pulley                    (Text: 9-4)
   Rolling and sliding examples                 (Text: 9-6)
   Rotation around a moving axis: Puck on ice   (Text: 9-4)
   Rolling down an incline                      (Text: 9-6)
   Bowling ball: sliding to rolling
   Atwood’s Machine with a massive pulley       (Text: 9-4)

   Look at textbook problems Chapter 9: # 53, 89, 92, 113, 125

Physics 111: Lecture 19, Pg 28

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