General Physics?

Question

A 2.2 kg cylinder (radius = 0.12 m, length = 0.50 m) is released from rest at the top of a ramp and allowed to roll without slipping. The ramp is 0.81 m high and 5.0 m long.
(a) When the cylinder reaches the bottom of the ramp, what is its total kinetic energy? 17.4J
(b) What is its rotational kinetic energy? ?J
(c) What is its translational kinetic energy? ?J
Please help me on paarts b&c

Answer 1

If it rolls without slipping, then there is a linear relationship between its angular velocity and its linear velocity: in rolling 2pi radians, a cylinder rolls one circumference, or a distance of 2piR.

Thus linear distance = (angular distance in radians) x radius, so
linear velocity v = (angular velocity) x R

The cylinder's rotational kinetic energy is Er = (1/2)I(w^2) where:

I is the moment of inertia
w is the angular velocity

http://en.wikipedia.org/wiki/Rotational_energy

Its translational kinetic energy is Et = (1/2)M(v^2) where:

M is the mass
v is the velocity

Conservation of energy says total energy = Et + Er

You have one equation connecting v (velocity) and w (angular velocity).

There is another connecting I and M if you assume that the cylinder is homogeneous:

http://en.wikipedia.org/wiki/List_of_moments_of_inertia

So you can get an equation for v^2 in terms of the total energy and the quantities you know.

Solve for v^2, then v, then w, and w^2. Finally the answers you want.