Finals exam in 8 hours..PLEASE HELP

A ring is rolled, without slipping, down a slope and reaches the bottom at 4 m/s. A disc of the same radius will have what speed if rolled from the same place?

(for this answer, I'm assuming this is a gen-req physics course or lower. It involves simplifications that aren't acceptable in higher-level classes)

edit: Sorry, JC, I was typing up the soln when only the other two had posted. I didn't mean to step on any toes. As an aside, there's really no need to be so catty about it.

The previous answers are ignoring rotational energy. You'll want to solve problems like this using conservation of energy (Initial Energy = Final Energy). We can do this because there is no work in the system. The three kinds of energy you're working with are: Rotational Kinetic Energy (E_rot), Kinetic Energy (E_k), and Gravitational Potential Energy (U_g).

The equations for these are:

U_g = mgh
E_k = (1/2)mv^2
E_rot = (1/2)Iw^2

v = velocity, I = moment of inertia, m = mass, h = height, w = angular velocity, g = gravitational constant.

A few other equations we'll need:

w = abs(v)/r [the magnitude of v, ignoring sign]
I_disk = (1/2)mr^2
I_ring = mr^2

r = radius

You probably don't understand moment of inertia(like mass for things that spin)/rotational energy that well, because a lot of teachers skim over it. Try reading up on this to understand how we do this problem. For now, I'm just going to show you how to solve this problem.

Look at the ring first:

Initial energy is only gravitational potential energy, because it is stationary and above the ground, so-

Initial Energy = U_g = mgh

Final energy is rotational and kinetic energy, because it is travelling forward and spinning.

Final energy = (1/2)mv^2 + (1/2)m(r^2)(w^2)

But w = v/r, so

Final energy = (1/2)mv^2 + (1/2)mv^2 = mv^2