A cockroach of mass m lies on the rim of a uniform disk of mass 4.00m that can rotate freely about its center like a merry-go-round. Initially the cockroach and disk rotate together with an angular velocity of 0.260 rad/s. Then the cockroach walks halfway to the center of the disk.(a) What then is the angular velocity of the cockroach – disk system?(b) What is the ratio K/K0 of the new kinetic energy of the system to its initial kinetic energy?(c) What accounts for the change in the kinetic energy?

Explanation:

A. And B. solutions are in the attachment.

C.

When the cockroach walks towards the center of the circle, positive work is done so kinetic energy increases.

The maximum static frictional force is 40N.

Explanation:

When an object of mass M is on a surface with a coefficient of static friction μ, there is a minimum force that you need to apply to the object in order to "break" the coefficient of static friction and be able to move the object (Called the threshold of motion, once the object is moving we have a coefficient of kinetic friction, which is smaller than the one for static friction).

This coefficient defines the maximum static friction force that we can have.

So if we apply a small force and we start to increase it, the static frictional force will be equal to our force until it reaches its maximum, and then we can move the object and now we will have frictional force.

In this case, we know that we apply a force of 40N and the object just starts to move.

Then we can assume that we are just at the point of transition between static frictional force and kinetic frictional force (the threshold of motion), thus, 40 N is the maximum of the static frictional force.