Question

A hoop with mass, M, and radius, R, rolls along a level surface without slipping with a linear speed, v. What is the ratio of rotational to linear kinetic energy? (For a hoop, I = MR2.)

Answer 1

Here, linear kinetic energy of the hoop K_linear = 1/2Mv^2

where -
M = mass of the body in motion
v = velocity of the body in motion

Rotational kinetic energy K_rotational = 1/2Iω^2

where -
I = mass moment of inertia of the body in motion
ω = rotational velocity of the body in motion

For the hoop -

I = MR^2
where r = the radius of the sphere

Again we have the following relationship -

v = rω

So,  
K_linear = 1/2Mv^2
K_rotational = 1/2Iω^2
I=MR^2
v = Rω

Now substitute values for I and v into the top two equations
K_linear = 1/2M(Rω)^2
K_rotational = 1/2(MR^2)ω^2 = 1/2M(R^2)(ω^2)

So, we have -

K_rotational / K_linear = [1/2M(R^2)(ω^2)] / [1/2M(Rω)^2] = 1

Therefore, the requisite ratio = 1.