Angular Momentum and Conservation

      
Introduction

Angular Momentum, L, for a body rotating about a fixed axis              L = I
   

                                                                   /\p
With Translational Momentum, F  =   ------  or    /\t F  = /\p   = Impulse
                                                                  /\t

   
                                                             /\L
With Angular Momentum,     =  ------
                                                             /\t  
   

                                                                           /\
Angular Acceleration is defined as         =  --------
                                                                            /\t
   
   
The change in Angular Momentum is defined as   /\L =   I /\

Combining these equations we get
   

              /\L          I /\  
  =  ------  =  -----------   =  I                       Remember for Translational Motion    F  =    ma  
             /\t               /\t   
            
                                                                                           

It is possible to change the Angular Momentum by changing the shape of an object.  As we examine the equation

L = I

There is a direct relationship between Moment of Inertia and Angular Velocity.  If the Moment of Inertia decreases, the Angular Velocity increases and vice versa.

After an ice skater begins spinning, she can increase her rate of spin by pulling her arms in close to her body.  Because the mass is closer to the axis of rotation, the Moment of Inertia decreases which increases the Angular Velocity.

When the ice skater wants to slow down, she stretches out her arms.  Because the mass is now farther from the axis of rotation, the Moment of Inertia increases which decreases the Angular Velocity.

The same process is used to help divers control their dives.  After getting an initial rotational motion off the diving board, the diver goes into a tuck.  This brings the mass closer to the axis of rotation, the Moment of Inertia decreases which increases the Angular Velocity.

The diver does not want to enter the water spinning.  To slow down the spin, the diver stretches out.  Because the mass is now farther from the axis of rotation, the Moment of Inertia increases which decreases the Angular Velocity.

The diver is exposed to the force of gravity, but not a net torque.

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Problems

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Answers

1.	What is the angular momentum of a 333 g ball rotating on the end of a string in a circle with a diameter of 2.50 m at 2.89 rpm?
	2.89 rev         1 min         2 rad ------------  x  ----------  x  ------------  =  0.303 rad s -1  =    1 min             60 s           1 rev      I = mr2  =  0.333 kg x ( 1.25 m ) 2  =  0.520 kg m2     L = I   =  0.520 kg m2  x  0.303 rad s -1  =  0.158 kg m2 s -1
2.	Hurricane Bill has wind at  200. km/h.  Assuming this wind is consistent over a radius of 100. km and a height of 3.50 km with an air density of 1.3 kg/m3  in a uniform cylinder,  calculate (a) the energy, and (b) angular momentum.    Physics Charts - Moments of Inertia
	200. km          1 h             1000 m -----------  x  -----------  x  ------------  =  55.6 m/s  =  v    1 h             3600 s           1 km                 v       55.6 m s -1    =  ---- = -------------------  =  5.56 x 10 -4 rad s -1               r       1.00 x 10 5 m     mass = V x D  =  r2 h x D = ( 1.00 x 10 5 m ) x 3.50 x 10 3 m x 1.3 kg m -3  =  1.43 x 10 9 kg      I = mR2  =  x 1.43 x 10 9 kg x ( 1.00 x 10 5 m ) 2  =  7.15 x 10 18 kg m 2      (a)    KE = I 2  =  x 7.15 x 10 18 kg m2  x ( 5.56 x 10 -4 rad s -1 ) 2  =  1.11 x 10 12 J      (b)  L = I   =  7.15 x 10 18 kg m2  x  5.56 x 10 -4 rad s -1  =  3.98 x 10 15 kg m2 s -1
3.	A uniform record turntable rotates at 45.0 rpm around a frictionless spindle.  A nonrotating rod of the same mass and with a length of the turntable's diameter is dropped on to the turntable.  What is the angular velocity in rpm of the combination?    Physics Charts - Moments of Inertia
	Irtrt  =  Irt + rod'rt + rod         Conservation of Angular Momentum     mrt  =  m             mrt + rod  =  2m         Irt  =  mR2                            Irt + rod  =  mR2 + 1/12  mR2   =  7/12 mR2       Irtrt  =  Irt + rod'rt + rod        Solve for 'rt + rod                              Irtrt           mR2  x 45.0 rpm 'rt + rod  =  ------------  =  ----------------------------  = 38.6 rpm                       Irt + rod                 7/12  mR2
4.	Suppose our Sun eventually collapses into a white dwarf, in the process losing 1/2 of its mass and winding up with a radius 1.0 percent of its original radius.  What would its new rotation rate be, since its current period is 30. days?  What would its final kinetic energy be in terms of its current kinetic energy?
	I  =  I''         Conservation of Angular Momentum     m'  =  0.5 m                 R'  =  0.010 R    2 rad        1 d          1 h ----------- x --------  x  ----------  =  2.4 x 10 -6 rad s -1   30. d          24 h        3600 s    I = 2/5 mR2                 I'  =  2/5 x 0.5 m x ( 0.010 R ) 2  =  2.0 x 10 -5 mR2       I  =  I''      Solve for '                   I        2/5 mR2 x  2.4 x 10 -6 rad s -1   '  = -------  =  ------------------------------------------  =  4.8 x 10 - 2 rad s -1               I'                2.0 x 10 -5 mR2          1 s                        2 rad   ---------------------  x  -------------  =  130 seconds 4.8 x 10 - 2 rad         1 rev     KE'        0.5 m ( 4.8 x 10 - 2 rad s -1 )2  -----  =  -----------------------------------------  =  1.7 x 10 8  times greater after Nova KE          m  ( 2.4 x 10 -6 rad s -1 ) 2

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