A solid disk and a solid sphere, each of the same mass, are rolling along a surface.Which one is faster?

Question

NOTE: They have equal kinetic energies!!!

Answer 1

If both have the same mass, same kinetic energies to start the acceleration and are both rolling on the same surface with the same coefficient of friction, therefore, no one is faster than the other. Both have the same velocity.

Answer 2

There are two motions going on:
- Translation, at velocity v: the kinetic energy of this is:
KE_t = m*v^2/2

- Rotation, at angular velocity w = v/R: the kinetic energy of this is:
KE_r = I*w^2/2 = I*(v/R)^2 /2 = I/(2*R^2)*v^2

For a solid disk:
I_d = m*R^2/2
so KE_r_d = m*R^2*(1/2)*(1/2)*(1/R^2)*v^2
= m*v^2/4

For a solid sphere:
I_s = m*R^2*(2/5)
so KE_r_s = m*R^2*(2/5)*(1/(2*R^2)) * v^2
= m*(2/5)*(1/2)*v^2
= mv^2/5

So, in total, the KE:
- solid disk: KE = (1/4 + 1/2)*m*v^2 = (3/4)*m*v^2 = (15/20)*mv^2
- solid sphere: KE = (1/5 + 1/2)*m*v^2 = (7/10)*m*v^2 = (14/20)*mv^2

Since the KE for the disk is a little larger than for the sphere for the same speed, the equality of these two KEs implies that the sphere must be going a little faster. In fact,
v_sphere/v_disk = sqrt(15/14)

Answer 3

Assuming your shapes are not in a vacuum, the sphere has a greater sail area and would be more restricted.
If this is for a test, find a better source than me. :)