Physics 111 Lecture 19 Today’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
    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                             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
             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!


                            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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