Problem 1:

Prof. Armstrong sits on a (frictionless) chair that is spinning on its axis with an angular speed of 2.0 rev/s. His arms are outstretched and he holds a heavy weight in each hand. The moment of inertia of the Prof, the weights, and the chair is (in total) 10 kg·m². Armstrong pulls the weights closer towards his body so that the moment of inertia decreases to 3 kg·m².

a) What is the resulting angular speed of the chair?
b) What is the change in kinetic energy of the system?
c) Where did the increase in kinetic energy come from?

Solution:

We did this demonstration in class, and did these calculations...

a) Angular momentum is conserved (there are no external torques), so we have

     L = I = constant

         I_{i i} = I_{f f}

Thus    _f = (I_{i i})/I_f

       _f  =  (10 kg m²)(2 rev/s)/(3 kg m²) =  6.67 rev/s
             = (6.67 rev/s)(2 rad/rev)  =  41.9 rad/s

b)    K_i = ½ I_ii² = ½ (10 kg m²)(2 x 2 rad/s)² = 790 J

        K_f = ½ I_ff² = ½ (3 kg m²)(41.9 rad/s)² = 2633 J

Change in kinetic energy is    K_f - K_i = 1843 J

c) The energy came from chemical energy in Armstrong's not-so-strong arms; he does work in moving the weights inward. This was discussed in class...

Problem 2
Test 3
Physics 101 Home page
Physics Department Home Page