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									      Physics 111
                                Tuesday,
                             November 2, 2002


• Ch 11:   Rotational Dynamics
           Torque
           Angular Momentum
           Rotational Kinetic Energy
Tues
                                          Phys
Nov
               Announcements               111
.02.




   • Wednesday, 8 - 9 pm in NSC 118/119
   • Sunday, 6:30 - 8 pm in CCLIR 468
Tues
                                                  Phys
Nov
                 Announcements                     111
.02.




       This week’s lab will be on oscillations.
       There will be a short quiz.
Tues
                                                Phys
Nov
              Ch 11: Rotational Dynamics         111
.02.

 Let’s look at a balance scale...
                 L              X?


       50 g                          250 g




 If I hang a 50 g mass at   …where should I hang
 the end of the left side   the 250 g mass to get
 of the balance...          the scale to balance?

                   Worksheet #1
Tues
                                                     Phys
Nov
             Ch 11: Rotational Dynamics               111
.02.

                  L        L/5



          50 g               250 g




         How did we arrive at such an answer?


       W1 L2     Perhaps we used one of these
         =       formulae
       W2 L1                         m1 L1 = m2 L2
Tues
                                                    Phys
Nov
              Ch 11: Rotational Dynamics             111
.02.

                     L        L/5


             50 g              250 g



       Let’s restate our balance condition so that we
       get all the subscript 1’s on the same side and
       all the subscript 2’s on the same side...

           W1 L2
             =                 L1W1 = L2W2
           W2 L1
Tues
                                                       Phys
Nov
               Ch 11: Rotational Dynamics               111
.02.

                      L       L/5


              50 g             250 g


                                       L1W1 = L2W2
  So in the case of this balance, we have...

       L(50 g)(9.8 m/s ) = ( L / 5)(250 g)(9.8 m/s )
                       2                           2


                                    Good! The balance
 490 gm/s2 L = 490 gm/s2 L            is in balance!
Tues
                                                  Phys
Nov
            Ch 11: Rotational Dynamics             111
.02.

        L        L/5
                                  What happens to
                  250 g           our balance if I
 25 g
                                  change the mass
                                  on the left side
                                  from 50 g to 25 g?
        L        L/5

 25 g
                  250 g   The scale begins
                          to rotate around
                           the pivot point.
Tues
                                                    Phys
Nov
              Ch 11: Rotational Dynamics             111
.02.

  Let’s look at one more case with our balance...
           L              L

                        X?
 100 g                    100 g     50 g   X = L/2
                                  How did we arrive at
       Worksheet #2               this answer?


             Now where should I hang the 100 g
             mass to get the scale to balance?
Tues
                                                      Phys
Nov
               Ch 11: Rotational Dynamics              111
.02.

                  L               L

                             X?
       100 g                   100 g    50 g



   First, we examined the left side of the balance:
                                               gm
       L1W1 = L(100 g)(9.8 m/s ) = 980L
                                  2
                                               s2
Tues
                                                       Phys
Nov
               Ch 11: Rotational Dynamics               111
.02.

                  L                        L

                                    X?
       100 g                          100 g     50 g



          Then, we examined the right side:

L2W2 + L3W3 = X (100 g)(9.8 m/s 2 ) + L(50 g)(9.8 m/s2 )
                          gm               gm
                  980 X   s   2   + 490L   s2
Tues
                                                               Phys
Nov
                  Ch 11: Rotational Dynamics                    111
.02.

                     L                           L

                                         X?
          100 g                              100 g   50 g



       If the scale is balanced, the two sides should
       equal one another:
                    gm                   gm               gm
           980L      s   2   =   980 X   s   2   + 490L   s2

                             X = L/2
Tues
                                                    Phys
Nov
           Ch 11: Rotational Dynamics                111
.02.


  • In our scale experiments, we’ve been playing
    around with a physical quantity that we’ve
    not yet encountered formally.

  • Our experiments show us that this quantity is
    related to a force applied at a distance.


  • The farther away from the pivot point the force
    is applied, the greater our new quantity is.
Tues
                                                   Phys
Nov
           Ch 11: Rotational Dynamics               111
.02.

  • When present and unbalanced, our new physical
    quantity causes the scale to rotate in the
    direction of the unbalanced force.

  • We’ve also seen that the quantity is additive.
    That is, if we have two forces on one side of the
    pivot at two different distances, the resulting
    physical quantity is simply the sum of the two
    force * distance products.

   • Before we name our new physical quantity,
     let’s examine one more thought experiment.
Tues
                                                      Phys
Nov
             Ch 11: Rotational Dynamics                111
.02.
                                   L


                                          50 g
  Let’s put a block on
    one side of our
  balance and let go.




       What will eventually happen to this system?
       (Assume there is SOME friction in the pivot,
       but that the pivot is free to rotate 360o.)
Tues
                                             Phys
Nov
             Ch 11: Rotational Dynamics       111
.02.


       The block eventually
          comes to rest
         with the balance
        aligned vertically.


 We still have a mass of 50 g
 at a distance L from the pivot   L
 point, so why has the motion
 stopped? What has
 happened to the value of our
 new physical quantity?               50 g
Tues
                                                Phys
Nov
            Ch 11: Rotational Dynamics           111
.02.

  For this case, we notice that
  the gravitational force on the
  mass is acting along the
  length of the balance.

  In other words, the force is
  parallel to the radius of length
  L around the pivot point.
                                     L
   In this case, the
   magnitude of our
   new physical
   quantity must be                      50 g
Tues
                                                    Phys
Nov
            Ch 11: Rotational Dynamics               111
.02.

  Let’s finally give a name to
  this new physical quantity.      How about...
         
       | τ |= F⊥ d
  Where d is the distance from the pivot point (the
  point of rotation) to the point at which the force
  is applied and…
                  is the component of the applied
          F       force that is perpendicular to the
             ⊥ line joining the pivot point to the
                  point at which the force is applied.
Tues
                                                   Phys
Nov
             Ch 11: Rotational Dynamics             111
.02.


          
        | τ |= F⊥ d
          
       [| τ |] = [F⊥ ][d]      
                            [| τ |] = N m
   Although torque has the same units as energy
   (J), we generally denote torques as forces
   acting at a distance, and therefore leave the
   units in the form N m.
Tues
                                                      Phys
Nov
            Ch 11: Rotational Dynamics                 111
.02.
                                  0.5 m
                        pivot                    Fg
 What is the                              50 g
 magnitude of the
 torque experienced               Here, the force of
 by this unbalanced                   gravity acts
 balance?                          perpendicularly
                                to the radius joining
     
   | τ |= F⊥ d                      the two yellow
                                     points above.
     
   | τ |= mgd = (0.05 kg)(9.8 m/s )(0.5 m) = 0.245 Nm
                                 2
Tues
                                          Phys
Nov
       Ch 11: Rotational Dynamics          111
.02.

                           Worksheet #3
   ≥
Tues
                                                           Phys
Nov
                       Ch 11: Rotational Dynamics           111
.02.

                                            Worksheet #3

  You are using a wrench and
  trying to loosen a rusty nut.
  Which of the arrangements
  shown is most effective in
  loosening the nut? List in order
  of descending efficiency the
  following arrangements:

       1) 1, 2, 3, 4     5) 1, 3, 4, 2
       2) 4, 3, 2, 1     6) 4, 2, 3, 1
       3) 2, 1, 4, 3     7) 2, 3, 4, 1
       4) 2, 4, 1, 3     8) 2, 3, 1, 4
PI, Mazur (1997)
Tues
                                              Phys
Nov
            Ch 11: Rotational Dynamics         111
.02.




                                   m
                                    Ft
                       r


                 pinned to
                   table


   The tangential force results
   in a tangential acceleration.   Ft = mat
Tues
                                                      Phys
Nov
            Ch 11: Rotational Dynamics                 111
.02.




                                      m
                                       Ft
                       r


                 pinned to
                   table

   It also creates a torque about the pinned point.
                                 
            | τ |=| Ft || r |= m | at || r |
Tues
                                                       Phys
Nov
                Ch 11: Rotational Dynamics              111
.02.




                                         m
                                          Ft
                           r


                     pinned to
                       table

       Recalling that tangential acceleration is related
       to angular acceleration, we get:
                                        2 
 | τ |= m | at || r |= m(| r || α |) | r |= mr | α |
Tues
                                            Phys
Nov
            Ch 11: Rotational Dynamics       111
.02.




                                  m
                                   Ft
                       r


                 pinned to
                   table

   This expression is valid so        2
   long as our connecting        τ = mr α
   rod/string is massless.
Tues
                                                   Phys
Nov
            Ch 11: Rotational Dynamics              111
.02.

   If we examine a more general system and plot
   the relationship, we find

                               The slope of this
             τ                 line is known as
                               the moment of
                               inertia, I.
                          α
                               For our previous
                               example,
  Note: r is the distance to
  the axis of rotation.            I = mr    2
Tues
                                                            Phys
Nov
              Ch 11: Rotational Dynamics                     111
.02.


  2Imr=             [I ] = [m][r ]   2



                    [I ] = kg m      2


  If we have an extended object, we can compute
  its moment of inertia about some axis from the
  sum of the moments of each of the individual
  pieces of that object:
                                                 N
       I = m1r12 + m2r22 + m3r32 + ... + mN rN = ∑ mi ri2
                                             2

                                                 i=1
Tues
                                            Phys
Nov
               Ch 11: Rotational Dynamics    111
.02.




       So we can rewrite our
       expression for torque          
       in terms of our new
       quantity, the moment of
                                  τ = Iα
       inertia...

       Notice the similarity          
       to our linear motion
       expression of              F = ma
Tues
                                          Phys
Nov
       Ch 11: Rotational Dynamics          111
.02.

                           Worksheet #4
   ≥
Tues
                                                                      Phys
Nov
                     Ch 11: Rotational Dynamics                        111
.02.

                                                Worksheet #4

       Two wheels with fixed hubs, each having a mass of 1 kg,
       start from rest, and forces are applied as shown. Assume the
       hubs and spokes are massless, so that the rotational inertia
       is I = mR2. In order to impart identical angular
       accelerations, how large must F2 be?


         1. 0.25 N
         2. 0.5 N
         3. 1 N
         4. 2 N
         5. 4 N

PI, Mazur (1997)
Tues
                                                    Phys
Nov
               Ch 11: Rotational Dynamics            111
.02.




       For the ball to fall into the
       basket, the linear acceleration
       of the end of the stick must
       exceed that of gravity.

        Is that possible?       1) Yes      2) No

       The moment of inertia of a stick        1
       rotated about its end is given by    I = ML2

                                               3
Tues
                                                         Phys
Nov
            Ch 11: Rotational Dynamics                    111
.02.




                                
                 τ net = F⊥ R = Iα
   The force creating the torque in this case is
   that of gravity, acting at the center of the stick.

                       L 1  2
       τ net = Mg(cosθ ) = ML α
                        2 3
Tues
                                                       Phys
Nov
              Ch 11: Rotational Dynamics                111
.02.




                       To get the linear
          3g           acceleration of the end of
       α=    (cosθ )   the stick, we simply multiply
          2L
                       by the length of the stick.

             atan = α L = 1.5g(cosθ )
 So, if the cos θ > 2/3, the end of the stick
                                                θ < 480
 will move with a linear acceleration > g.
Tues
                                            Phys
Nov
               Ch 11: Rotational Dynamics    111
.02.




                                         
  Recall in our                      Δv Δp
  discussion of force,     F = ma = m    =
  we discovered                       Δt Δt

       where p was the                 
       momentum                    p = mv
       in the system
Tues
                                                 Phys
Nov
             Ch 11: Rotational Dynamics           111
.02.




       We can construct a completely analogous
       argument to define angular momentum
                              
                       Δω ΔL
              τ = Iα = I    =
                         Δt   Δt

     where L is the
                                      
   angular momentum               L = Iω
Tues
                                          Phys
Nov
             Ch 11: Rotational Dynamics    111
.02.


             
         L = Iω
                  
       [ L] = [I ][ω ]
                              kg m 2
       [ L] = (kg m )(rad/s) =
                   2

                                 s
Tues
                                                    Phys
Nov
              Ch 11: Rotational Dynamics             111
.02.




               When there is no NET external torque
               on a system, the angular momentum
               of the system will be conserved.
                                
                Li = Iω i = Iω f = L f

       So when examining isolated systems,
       total energy, linear momentum, and angular
       momentum are all conserved!

             Conservation of Angular
Tues
                                                          Phys
Nov
            Ch 11: Rotational Dynamics                     111
.02.




  We’ve studied this quantity
  already in looking at masses
  at some fixed distance, r,             K = mv
                                              1
                                              2
                                                      2

  from an axis of rotation.


   When looking at an object rotating at a constant
   angular velocity ω at a distance r from the axis,

        K = m(rω ) = mr ω = Iω
              1
              2
                       2   1
                           2
                                 2   2    1
                                          2
                                                  2
Tues
                                             Phys
Nov
                Ch 11: Rotational Dynamics    111
.02.


       K = Iω
            1
            2
                     2



       [K ] = [ ][I][ω ]
                 1           2
                 2

       [K ] = (kg m )(rad/s)
                         2       2


                     2
              kg m
       [K ] =     2
                    = Nm = J
                s
Tues
                                                       Phys
Nov
               Ch 11: Rotational Dynamics               111
.02.




  So we’ve defined another form of kinetic energy.

  Before, we studied translational kinetic energy.
  Now we have rotational kinetic energy, too.

       So the total kinetic energy of a system now
       includes two terms (if the system is rotating
       and translating):
                              Ktot = Kr + Kt
Tues
                                          Phys
Nov
       Ch 11: Rotational Dynamics          111
.02.

                           Worksheet #5
   ≥
Tues
                                                                           Phys
Nov
                     Ch 11: Rotational Dynamics                             111
.02.

                                                        Worksheet #5

       A figure skater stands on one spot on the ice (assumed
       frictionless) and spins around with her arms extended.
       When she pulls in her arms, she reduces her rotational
       inertia and her angular speed increases so that her
       angular momentum is conserved. Compared to her initial
       rotational kinetic energy, her rotational kinetic energy
       after she has pulled in her arms must be

                   1. the same.
                   2. larger because she’s rotating faster.
                   3. smaller because her rotational inertia is smaller.

PI, Mazur (1997)
Tues
                                                    Phys
Nov
             Ch 11: Rotational Dynamics              111
.02.




       So, when we examine the total energy of
       a system, we must take care to include the
       rotational kinetic energy term as well.

  The total energy (including rotational kinetic
  energy) will be conserved so long as NO non-
  conservative forces are acting upon the system.

           Etot = Kr + Kt + U g ( +U spring )
Tues
                                                        Phys
Nov
               Ch 11: Rotational Dynamics                111
.02.

   We have studied objects in “equilibrium” before.

           What is the defining characteristic
           of an object in equilibrium?




       And from our earlier studies, the fact that it
       did not accelerate indicated that:
                       
                       Fnet = 0
Tues
                                                      Phys
Nov
               Ch 11: Rotational Dynamics              111
.02.


       But is this a sufficient condition to ensure
       that in fact an object does not accelerate
       in some way?


       This “zero net force” condition
       only guarantees that the center of mass
       of the object on which the forces are
       being exerted does not accelerate.

        Let’s look at the meter stick again...
Tues
                                                           Phys
Nov
            Ch 11: Rotational Dynamics                      111
.02.

                                      Worksheet #6
       Exert a constant force
       tangential to the circle
       of motion.                        Top view



                      Holding pencil through the
                      pivot point.

  Draw a free-body
                              Assume the system is
  diagram for this        Oriented in a horizontal plane
  system.
Tues
                                              Phys
Nov
            Ch 11: Rotational Dynamics         111
.02.


   Is the net force in this system zero?


   So why does the ruler rotate with
   increasing speed?


       Because the net torque is NOT zero!

       We have a new equilibrium condition:
                    
                    τ net = 0
Tues
                                                        Phys
Nov
              Ch 11: Rotational Dynamics                 111
.02.




       For an object to be in STATIC equilibrium
       (i.e., at rest, not moving, not rotating), the
       following TWO conditions must be met.
                                   
         Fnet = 0                   τ net = 0
        (We’re free to pick ANY origin about which to
             computer our torques in this case.)
Tues
                                                       Phys
Nov
        Ch 11: Rotational Dynamics                      111
.02.

       A student sits on a rotating stool holding two
       weights, each of mass 3.00 kg. When his arms
       are extended horizontally, the weights are each
       1.00 m from the axis of rotation, and he rotates
       at an angular speed of 0.750 rad/s. The
       moment of inertia of the student + stool is 3.00
       kg m2 and is assumed to be constant. The
       student now pulls the weights horizontally in to
       a distance of 0.300 m from the rotation axis. (a)
       What’s the new angular speed? (b) What are the
       initial and final kinetic energies of this system?


                    Problem #1
Tues
                                                           Phys
Nov
              Ch 11: Rotational Dynamics                    111
.02.

         (a) What’s the new angular speed? (b) What are the
         initial and final kinetic energies of this system?

  Pictorial Representation:        Knowns:
                ωi                 Istudent = 3.00 kg m2
 Initially:
                                   m = 3.00 kg
        ri                m        ri = 1.00 m
                                   rf = 0.30 m
                                   ωi = 0.75 s-1
  Finally:       ωf
         rf           m           Unknowns:
                                  ωf
Tues
                                                                 Phys
Nov
                  Ch 11: Rotational Dynamics                      111
.02.

         (a) What’s the new angular speed? (b) What are the
         initial and final kinetic energies of this system?

         Itotal = Iweights + Istudent = 2 (mr2) + (3 kg m2)
 Initially:
              Itotal = 2 (3 kg)(1 m)2 + (3 kg m2) = 9.00 kg m2

 Finally:
              Itotal = 2 (3 kg)(0.3 m)2 + (3 kg m2) = 3.54 kg m2

  Ii ωi = If ωf
                        Ii      (9kg m 2 )
                   ω f = ωi =             2
                                            (0.75rad/s) = 1.91rad/s
                        If    (3.54kg m )
Tues
                                                               Phys
Nov
              Ch 11: Rotational Dynamics                        111
.02.

         (a) What’s the new angular speed? (b) What are the
         initial and final kinetic energies of this system?


Initially:
       Ki = 0.5 Ii ωi2 = 0.5 (9 kg m2)(0.75 rad/s)2 = 2.53 J

Finally:

       Kf = 0.5 If ωf2 = 0.5 (3.54 kg m2)(1.91 rad/s)2 = 6.45 J
Tues
                                                        Phys
Nov
            Ch 11: Rotational Dynamics                   111
.02.

       A uniform horizontal beam weighs 300 N, is 5.00 m
       long, and is attached to a wall by a pin connection
       that allows the beam to rotate. Its far end is
       supported by a cable that makes an angle of 53o
       with the horizontal. If a 600-N person stands
       1.50 m from the wall, find the tension in the cable
       and the force exerted by the wall on the beam.



        1.5 m
                                   Problem #2


                        53o
                5.0 m
Tues
                                                        Phys
Nov
           Ch 11: Rotational Dynamics                    111
.02.

       A uniform horizontal beam weighs 300 N, is 5.00 m
       long, and is attached to a wall by a pin connection
       that allows the beam to rotate. Its far end is
       supported by a cable that makes an angle of 53o
       with the horizontal. If a 600-N person stands
       1.50 m from the wall, find the tension in the cable
       and the force exerted by the wall on the beam.

  Free-body                                         T
   diagram        f                   5.0 m   53o
    for the
                      N       Fc          W
     plank
                            1.5 m

                              2.5 m
Tues
                                                                       Phys
Nov
             Ch 11: Rotational Dynamics                                 111
.02.
                                         y
 From the free-body                                                    T
                          f                    x
 diagram we can write                                    5.0 m   53o
 down the torque equation
                                     N         Fc            W
 and Newton’s 2nd Law in
                                             1.5 m
 the x and y directions.
                                               2.5 m

 Compute the torques about the left end of the
 plank, where the plank meets the wall:
 τ net = f (0m) − Fc (1.5m) − W (2.5m) + (T sin53 )(5.0m) = 0
                                                     0


 0 = 0 − (600N)(1.5m) − (300N)(2.5m) + (4.0m)T

                       T = 412.5 N
Tues
                                                                   Phys
Nov
           Ch 11: Rotational Dynamics                               111
.02.
                                   y
 From the free-body                                                T
                          f                  x
 diagram we can write                                5.0 m   53o
 down the torque equation
                               N             Fc          W
 and Newton’s 2nd Law in
                                           1.5 m
 the x and y directions.
                                             2.5 m

 Newton’s 2nd Law in the x-direction:

           Fnet,x = N − T cos53 = 0
                               0


           0 = N − (412.5N)cos53       0



                  N = 248 N
Tues
                                                                  Phys
Nov
           Ch 11: Rotational Dynamics                              111
.02.
                                      y
 From the free-body                                               T
                          f                 x
 diagram we can write                               5.0 m   53o
 down the torque equation
                                  N         Fc          W
 and Newton’s 2nd Law in
                                          1.5 m
 the x and y directions.
                                            2.5 m

 Newton’s 2nd Law in the y-direction:

       Fnet, y = f + T sin53 − Fc − W = 0
                          0


       0 = f + (412.5N)sin53 − 600N − 300N
                              0



                    f = 571 N

								
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