## anonymous 4 years ago 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?

1. JamesJ

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.

2. JamesJ

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.