Frictionless cylindrical flywheel, of diameter 250mm and mass 30k

[gk1989]

A frictionless cylindrical flywheel, of diameter 250mm and mass 30kg, has a string attached/wrapped on the outside diameter, with a mass of 10kg on the other end.

What is the radius of gyration of the flywheel?
If the mass starts 1.5m form the floor, what will its speed be when the mass is released & it hits the floor? and at this point how much rotational kinetic energy will the flywheel have?


Any hints or answers would be gratefully received.
many thanks
Greg

 

[minger]

The radius of gyration should be easy, it's just a simple formula. It's just a function of mass and moment of inertia if I remember correctly. OK, I looked it up, it's just:
$$R = \sqrt{\frac{I}{m}}$$
As for the problem let's try and walk through it. From the mass, you will need to convert the equivalent moment that the mass is exerting on the flywheel. From there, you can plug it into the equivalent rotational f=ma equation:
$$T = I\omega$$ With your moment of inertia and torque/moment, you can get a rotational acceleration. Your mass is 1.5m above the floor, so convert that to radians, and see how long it takes to hit the floor (from the rotational acceleration).
Given rotational acceleration and time, you can now get your final rotational velocity and the kinetic energy which is the equivalent:
$$KE = \frac{1}{2}I\omega^2$$ Good luck