Question

A 4.20 m diameter merry-go-round is rotating freely with angular speed of 0.800 rad/s. Its total moment of inertia is 1260 kg m squared. How much work will bring the merry-go-round to rest?

Answer 1

The rotational kinetic energy of the merry-go-round is
\[ KE_r = \frac{1}{2}I\omega^2 \] Now, you want to reduce that to zero. Hence by the work-energy theorem, you need to do exactly an equal amount of work.

Answer 2

make sense? It's analogous to asking "How much work is required to stop a 1000 kg car moving at 10 m/s?" We need to do
\[ KE = \frac{1}{2}mv^2 \] of work.