Initial angular momentum
= (1/2) m R^2 omega
= (1/2)85 * 1.7^2 * 4
final angular momentum with kid on rim
= [(1/2) 85*1.7^2 + 45*1.7^2]*new omega
there is no torque, so no change in angular momentum
so
solve for new omega (part a) (turns slower)
then
kid walks to center
still no change in angular momentum but the moment of inertia reverts to the original so the omega goes back up to the original
energy kid on rim = (1/2) [(1/2) 85*1.7^2 + 45*1.7^2]* (new omega)^2
energy kid at center = (1/2)[(1/2) 85*1.7^2] *(original omega)^2
A child of mass 45 kg stands beside a circular platform of mass 85 kg and radius 1.7m spinning at 4 rad/s. Treat the platform as a disk. The child steps on the rim.
a) What is the new angular speed; (rad/s)
b) She then walks to the center and stay there. What is the angular velocity of the platform then? (rad/s)
c) What is the change in kinetic energy when she walks from the rim to the center of the platform? (J)
1 answer