Physics 211 Lecture 18 Today’s Agenda by fsm30412

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									                 Physics 211: Lecture 18

                         Today’s Agenda

   More about rolling

   Direction and the right hand rule

   Rotational dynamics and torque

   Work and energy with example




                                          Physics 211: Lecture 18, Pg 1
   Rotational v.s. Linear Kinematics

   Angular                                 Linear

 = constant                            a  constan t

 = 0 + t                              v = v 0 + at

               1 2                                       1 2
 = 0 + 0 t + t                       x = x0 + v 0 t + at
               2                                         2

And for a point at a distance R from the rotation axis:

           x = Rv = Ra = R




                                             Physics 211: Lecture 18, Pg 2
                                                        Roll objects
                            Rolling Motion              down ramp

       Objects of different I rolling down an inclined plane:


                 v=0             K = - U = Mgh
         R       = 0
          M      K=0
                                                         1 2 1
h                                                     K = I  + Mv 2
                                                         2     2

                                                                     v = R




                                                     Physics 211: Lecture 18, Pg 3
                                  Rolling...

      If there is no slipping:




                                                               v
                                  2v         v
                         v
                                                            Where v = R


In the lab reference frame             In the CM reference frame



                                                      Physics 211: Lecture 18, Pg 4
                               Rolling...

     1 2 1                                             hoop:       c=1
  K = I  + Mv 2       Use v = R and I = cMR2 .
     2     2                                           disk:       c = 1/2
             1            1      1                     sphere: c = 2/5
        K=     c MR 2 2 + Mv 2 = ( c + 1)Mv 2
             2            2      2
                                                       etc...

So:     1                                               1
          ( c + 1)Mv 2 = Mgh                v = 2 gh
                                                       c +1
        2


 The rolling speed is always lower than in the case of simple sliding
 since the kinetic energy is shared between CM motion and rotation.

         We will study rolling more in the next lecture!

                                                   Physics 211: Lecture 18, Pg 5
                       Direction of Rotation:
   In general, the rotation variables are vectors (have direction)
   If the plane of rotation is in the x-y plane, then the convention is

      CCW   rotation is in
       the + z direction                 y

                                                 x

                                    z

      CW   rotation is in
       the - z direction                 y

                                                  x

                                     z

                                                      Physics 211: Lecture 18, Pg 6
                   Direction of Rotation:
                   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 211: Lecture 18, Pg 7
                                Example:

   A flywheel spins with an initial angular velocity 0 = 500 rad/s.
    At t = 0 it starts to slow down at a rate of 0.5 rad/s2. How long
    does it take to stop?

   Realize that  = - 0.5 rad/s2.
                                                                   
   Use      0  t     to find when  = 0 :
                                                                         
                           0
                    t 
                           
                           500 rad / s
   So in this case t                  2
                                           1000 s  16.7 min
                            0.5 rad / s


                                                     Physics 211: Lecture 18, Pg 8
                    Lecture 18, Act 1
                       Rotations
   A ball rolls across the floor, and then starts up a ramp as
    shown below. In what direction does the angular acceleration
    vector point when the ball is on the ramp?

    (a) down the ramp
    (b) into the page
    (c) out of the page




                                              Physics 211: Lecture 18, Pg 9
                     Lecture 18, Act 1
                         Solution
   When the ball is on the ramp, the linear
    acceleration a is always down the ramp (gravity).
   The angular acceleration is therefore counter-clockwise.

   Using your right hand rule,  is out of the page!




                                                  
                                 a




                                                 Physics 211: Lecture 18, Pg 10
                      Rotational Dynamics:
                      What makes it spin?

       Suppose a force acts on a mass constrained to move in a
        circle. Consider its acceleration in the ^ direction at some
                                                 
        instant:                                                     ^
          a = r                                             ^     r
                                                               
                                                                 F
                                       
        Now use Newton’s 2nd Law in the ^              F
        direction:
                                                            a
          F = ma = mr                                            m

   Multiply by r :                                    r
         rF = mr2                                



                                                   Physics 211: Lecture 18, Pg 11
                       Rotational Dynamics:
                       What makes it spin?

    rF = mr2   use     I   = mr 2
        = I
   Define torque:  = rF.
                                                                     ^
                                                            ^        r
       is the tangential force F                         
       times the lever arm r.
                                                            F
                                                  F
            = I                                      a
                                                                m
   Torque has a direction:
      + z if it tries to make the system         r
       spin CCW.
      - z if it tries to make the system
       spin CW.
                                              



                                              Physics 211: Lecture 18, Pg 12
                    Rotational Dynamics:
                    What makes it spin?

                                                  ri Fi ,   mi ri  i
   So for a collection of many particles                             2
    arranged in a rigid configuration:           i              
                                                              i 
                                                      i         I

   Since the particles are connected rigidly,
    they all have the same .
            i  I 
           i
                                    m4                                        F1
                               F4                                r1 m1
       NET  I                     r4

                                    m3
                                          r3       r2
                                                        m2
                                                        F2
                                     F3
                                                    Physics 211: Lecture 18, Pg 13
                            Rotational Dynamics:
                            What makes it spin?

          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.
   Torque has units of kg m2/s2 = (kg m/s2) m = Nm.




                                                     Physics 211: Lecture 18, Pg 14
                              Torque

   Recall the definition of torque:

      = rF

       = r F sin                           Fr  F
       = r sin  F
                                                         
                                               F
      = r pF
                                                         
     rp = “distance of closest approach”       r

   Equivalent definitions!
                                                    rp


                                           Physics 211: Lecture 18, Pg 15
                               Torque

     = r Fsin                                    F

   So if  = 0o, then  = 0                  r




   And if  = 90o, then  = maximum
                                        F
                                               r




                                            Physics 211: Lecture 18, Pg 16
                     Lecture 18, Act 2
                         Torque
   In which of the cases shown below is the torque provided by the
    applied force about the rotation axis biggest? In both cases the
    magnitude and direction of the applied force is the same.



    (a) case 1
                                                        L
    (b) case 2
    (c) same                            F                              F
                           L
                                        axis

                               case 1          case 2

                                                  Physics 211: Lecture 18, Pg 17
                       Lecture 18, Act 2
                           Solution
   Torque = F x (distance of closest approach)
      The applied force is the same.
      The distance of closest approach is the same.



           Torque is the same!



                            F                                  F
               L                       L



                   case 1              case 2

                                                Physics 211: Lecture 18, Pg 18
                      Torque and the
                     Right Hand Rule:
   The right hand rule can tell you the direction of torque:
      Point your hand along the direction from the axis to the
       point where the force is applied.
      Curl your fingers in the direction of the force.
      Your thumb will point in the direction
       of the torque.
                                                           F
                                      y

                                                     r
                                             x
                                  
                              z
                                                 Physics 211: Lecture 18, Pg 19
                    The Cross Product

   We can describe the vectorial nature of torque in a
    compact form by introducing the “cross product”.
      The cross product of two vectors is a third vector:


                  AXB=C                                B

   The length of C is given by:                            
                  C = AB sin                                       A

   The direction of C is perpendicular to         C
    the plane defined by A and B, and in
    the direction defined by the right hand
    rule.



                                                Physics 211: Lecture 18, Pg 20
                   The Cross Product

   Cartesian components of the cross product:

         C=AXB
                                                     B
       CX = AY BZ - BY AZ

       CY = AZ BX - BZ AX                                        A

       CZ = AX BY - BX AY                        C


                   Note: B X A = - A X B


                                             Physics 211: Lecture 18, Pg 21
             Torque & the Cross Product:

   So we can define torque as:

     =rXF
      = rF sin 


    X = rY FZ - FY rZ = y FZ - FY z               F
    Y = rZ FX - FZ rX = z FX - FZ x
    Z = rX FY - FX rY = x FY - FX y     
                                                  r
                                                         y

                                                                x
                                                         z

                                       Physics 211: Lecture 18, Pg 22
                     Comment on  = I

   When we write  = I we are really talking about the z
    component of a more general vector equation. (Recall that
    we normally choose the z-axis to be the the rotation axis.)

                           z = Izz
                                                               z

                                                  Iz
                                         z

   We usually omit the
    z subscript for simplicity.
                                                  z


                                               Physics 211: Lecture 18, Pg 23
                           Example

   To loosen a stuck nut, a (stupid) man pulls at an angle of
    45o on the end of a 50 cm wrench with a force of 200 N.
      What is the magnitude of the torque on the nut?
      If the nut suddenly turns freely, what is the angular
       acceleration of the wrench? (The wrench
       has a mass of 3 kg, and its shape
       is that of a thin rod).                             45o
                                               F = 200 N


                                                           L = 0.5 m




                                               Physics 211: Lecture 18, Pg 24
                           Example
                                                  Wrench w/ bolts

   Torque  = LFsin  = (0.5 m)(200 N)(sin 45)    = 70.7 Nm

   If the nut turns freely,  = I
       We know  and we want , so we need to figure out I.


   1     1                                                  45o
I  ML2  3 kg 0.5 m 2  0.25 kgm2
   3     3                                   F = 200 N


                                                          L = 0.5m
    So = / I = (70.7 Nm) / (0.25 kgm2)

                                                      
                 = 283 rad/s2

                                              Physics 211: Lecture 18, Pg 25
                             Work

   Consider the work done by a force F acting on an object
    constrained to move around a fixed axis. For an
    infinitesimal angular displacement d:

      dW      .
           = F dr = FR d cos()                                    F
           = FR d cos(90-)                                     
           = FR d sin()                          R
                                                       d        dr = R d
           = FR sin() d
                                           axis
      dW =  d

   We can integrate this to find: W = 
   Analogue of W = F •r
   W will be negative if  and  have opposite signs!



                                               Physics 211: Lecture 18, Pg 26
                  Work & Kinetic Energy:

   Recall the Work/Kinetic Energy Theorem:       K = WNET

   This is true in general, and hence applies to rotational
    motion as well as linear motion.

   So for an object that rotates about a fixed axis:


                 K      
                        1 2
                        2
                                   
                          I f   i2  WNET




                                                 Physics 211: Lecture 18, Pg 27
                  Example: Disk & String

   A massless string is wrapped 10 times around a disk of
    mass M = 40 g and radius R = 10 cm. The disk is
    constrained to rotate without friction about a fixed axis
    though its center. The string is pulled with a force F = 10 N
    until it has unwound. (Assume the string does not slip, and
    that the disk is initially not spinning).

      How fast is the disk spinning after the string has
       unwound?


                                                                   M
                                                             R


                                                             F


                                                Physics 211: Lecture 18, Pg 28
                      Disk & String...

   The work done is W = 
      The torque is = RF (since  = 90o)
      The angular displacement  is
       2 rad/rev x 10 rev.
                                                                 M
                                                            R
   So W = (.1 m)(10 N)(20rad) = 62.8 J
                     
              
                                                        F




                                              Physics 211: Lecture 18, Pg 29
                      Disk & String...
                                           Flywheel, pulley,
                             1 2                 & mass
WNET = W = 62.8 J = K        I
                             2
Recall thatIfor a disk about
its central axis is given by:
       1
                                                              M
                                                         R
    I  MR 2
       2
                                                              
  So K   MR 2   2  W
         1 1
                
         2 2    

          4W         4 62 .8 J 
                                      = 792.5 rad/s
          MR 2
                   .04 kg .1    2




                                           Physics 211: Lecture 18, Pg 30
                       Lecture 18, Act 3
                        Work & Energy
   Strings are wrapped around the circumference of two solid disks
    and pulled with identical forces for the same distance.
    Disk 1 has a bigger radius, but both have the same moment of
    inertia. Both disks rotate freely around axes though their
    centers, and start at rest.
      Which disk has the biggest angular velocity after the pull ?



                                          1                    2
    (a) disk 1
    (b) disk 2
    (c) same
                                               F                     F



                                                   Physics 211: Lecture 18, Pg 31
                       Lecture 18, Act 3
                           Solution
    The work done on both disks is the same!
      W = Fd
    The change in kinetic energy of each will
     therefore also be the same since W = K.
                       1
    But we know K      I 2
                       2                1                        2
    So since I1 = I2

            1 = 2
                                                 F                     F

                                                          d

                                                     Physics 211: Lecture 18, Pg 32
                   Spinning Disk Demo:
                                                                  I
   We can test this with our big flywheel.

                                    negligible
                    1 2 1
         W  K      I  mv 2     in this case
                    2     2                                 m

                         2W
                   
                          I

In this case,     I = 1 kg - m2
                  W = mgh = (2 kg)(9.81 m/s2)(1 m) = 19.6 J


                   = 6.26 rad/s ~ 1 rev/s

                                                   Physics 211: Lecture 18, Pg 33
                  Recap of today’s lecture

   More about rolling

   Direction and the right hand rule

   Rotational dynamics and torque

   Work and energy with example




                                        Physics 211: Lecture 18, Pg 34

								
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