11/12/2010 (Fri)
                                                                                                                     Rotational Kinetic Energy
                                                                                                        Energy associated with rotation is given by an equation 
                                                                                                        analogous to that for straight-line motion.
                                                                                                                                                           1    2
                                                                                                        For an object that is moving but not rotating:  Km=    v
                                  Rotational Kinetic Energy                                                                                                2
                                                                                                        For an object that is rotating only:    1   2
                                                                                                                                            KI=   ω
                                                                                                                                                2
                                                                                                        For an object that is rolling, i.e., translating and rotating 
                                                                                                        simultaneously, the total kinetic energy of such an object is:
                                                                                                                                11
                                                                                                                                     22
                                                                                                                            Km=+vIω
                                                                                                                                22
                                                                                    1                                                                               2
                                    Racing Shapes - Revisited                                                       Racing Shapes - Revisited
                                                                                                        Question: Which shape will have the biggest velocity after 
                           We have three objects, a solid disk, a ring, and a solid                     rolling down the slope? 
                                sphere, all with the same mass, M and radius, R. If                     Solution: Let’s use conservation of energy to analyze the race 
                                we release them from rest at the top of an incline,                     between two objects that roll without slipping down the ramp.
                                which object will win the race? Assume the objects                      Let’s analyze a generic object with a mass M, radius R, and a 
                                roll down the ramp without slipping.                                    rotational inertia of:
                                                                                                                                       2
                                                                                                                               Ic= MR
                           1.   The sphere                                                              Start with the usual five-term energy conservation equation.
                           2.   The ring                                                                                  UK+    +=W U+K
                           3.   The disk                                                                                   iincff
                           4.   It’s a three-way tie                                                    Eliminate the terms, K and U that are zero, we have U = K
                                                                                                                               i      f                         i    f
                           5.   Can't tell - it depends on mass and/or radius.                          Insert the expressions for the U and K.
                                                                                                                                        i      f
                                                                                                                                  11
                                                                                                                                       22
                                                                                    3                                     Mgh =+Mv         Iω        (1)            4
                                                                                                                                  22
                                    Racing Shapes - Revisited                                                          What does this tell us?
                                                    11
                                                         22
                                             Mgh=+Mv          Iω    (1)                                                               2        2
                                                    22                                                                    Mgh= ½Mv + ½cMv
                        Because the object rolls without slipping, we can use ω = v                                       Translational  Rotational 
                                                 2                                 R                                           KE             KE
                                         Ic= MR
                        Next, substitute                  , where c = ½ for the disc, 2/5 for 
                        the sphere and 1 for the ring.                                                                             2gh
                                               11v2                                                                          v = 1+c
                                                    22
                                        Mgh=+Mv         ()cMR
                                               22R2
                                                                                                        This result shows that the larger the value of c, the slower the 
                        Both the mass and the radius cancel out!                                        object is, because a larger fraction of the potential energy is 
                                            11                                                          directed toward the rotational kinetic energy, with less 
                                               22
                                       gh=+v        cv                                                  available for the translational kinetic energy and so the object
                                            22 2gh                                                      moves (translates) more slowly.
                        Solving for the speed at the bottom:   v =                                      Simulation
                                                                     1+c            5                                                                               6
                                                                                                                                                                                   1
                                     A Figure Skater - Revisit                                                       A Figure Skater - Revisit
                        A spinning figure skater is an excellent example of angular                        Question: When the figure skater moves her arms in 
                        momentum conservation. The skater starts spinning with her                             closer to her body while she is spinning, what 
                        arms outstretched, and has a rotational inertia of Ii and an 
                        initial angular velocity of ω. When she moves her arms close                           happens to the skater’s rotational kinetic energy?
                                                   i
                        to her body, she spins faster. Her moment of inertia 
                        decreases, so her angular velocity must increase to keep the 
                        angular momentum constant.                                                         1.   It increases
                        Conserving angular momentum:                                                       2.   It decreases
                                                                                                           3.   It must stay the same, because of conservation of 
                               vv                                                                              energy
                              LL=
                                if
                               vv
                             IIω =   ω
                              ii ff
                        Question: In this process, what happens to the skater's kinetic 
                        energy? 
                                                                                    7                                                                               8
                                     A Figure Skater - Revisit                                                     A ball rolling down a ramp
                                11
                                     2                                                                  Question: A ball with mass M and radius R rolls without 
                           KI==ω           Iωω×
                                          ()
                             iiiiii
                                22                                                                      slipping down a ramp from the top to the bottom (see 
                               11                                                                       figure). We have found that a = gsinθ/(1 + c) and fs = 
                                    2
                          KI==ω           Iωω×
                                         ()                                                             Mgsinθ/(1/c + 1), where c = 2/5. Use conservation of 
                            f     ff ff f
                               22                                                                       mechanical energy to find the non-conservative work done, 
                                                                                                        W , on the ball when it reaches the bottom.  Assume that 
                        The terms in brackets are the same, so the final kinetic                          nc
                        energy is larger than the initial kinetic energy, because                       the ball is initially at rest and at a height h above ground.
                         ω <ω .                                                                                                     α
                          if
                                                                                                                          f
                                                                                                                           s        Mgsinθ
                        Where does the extra kinetic energy come from?
                        The skater does work on her arms in bringing them closer to                                           h        a
                        her body, and that work shows up as an increase in kinetic                                              Mg         θ
                        energy.                                                     9                                                                              10
                                    A ball rolling down a ramp                                                     A ball rolling down a ramp
                       Solution:                                        α                              To find v, we use                                α
                                                                                                                f
                       Conservation of mechanical energy       f                                                                               f
                                                                s        Mgsinθ                        v2 = v2 + 2as                           s        Mgsinθ
                       gives E + W = E                                                                  f    i
                              i    nc    f                                                                = 0 + 2[-gsinθ/(1+c)][-h/sinθ] 
                       Initial mechanical energy,                 h         a                             = 2gh/(1+c)                             h         a
                       E = K + U = 0 + Mgh = Mgh                     Mg         θ                                                                   Mg         θ
                        i    i   i                                                                      E + W = Mgh-W
                       Final mechanical energy,                                                           i    nc          nc
                       E = K + U                                                                        E = (1+c)Mv2/2 = Mgh
                        f    f    f                                                                       f          f
                         = ½ mv2 + ½ Iω2 + ½ Mv2 + (c/2)Mv2
                                 f  2    f         f          f
                         = (1+c)Mv /2                                                                    E + W = E gives W = 0
                                    f                                                                     i    nc    f         nc
                                                                                                         This shows that fs actually does no work on the ball!
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