# Energy in rotational motion

#### nuebie1

A frictionless pulley has the shape of a uniform solid disk of mass 2.50 kg and radius 20.0 cm. A 1.50-kg stone is attached to a very light wire that is wrapped around the rim of the pulley, and the system is released from rest. (a) How far must the stone fall so that the pulley has 4.50 J of kinetic energy? (b) What percent of the total kinetic energy does the pulley have?

Help, I really don't know how to start on this one. I only know that the pulley's kinetic energy is given by
K = 1/2 w^2I where w is the angular velocity and I is the moment of inertia. From this I can find the pulley's angular velocity when it's kinetic energy is 4.50, but I don't know how/if that can help me and what to do from there....

#### studiot

You have to assume there is no slip between the wire and the pulley.
Then the wire is taught from stone to pulley circumference so all parts of that portion of the wire are travelling at the same velocity.

You have just said you know how to calculate the angular velocity so you can calculate the linear speed of the pulley circuference and hence the linear speed of the wire and stone.

Thus can you now calculate the distance fallen from rest?