Who is good at Physics...?

Question

I have a physics question I do not understand and neither do my friends. A ball rolls along the street with a constant linear speed v. What is the fraction of its total kinetic energy that is in the form of rotational kinetic energy about the center of the ball. Also, if the linear speed of the ball is doubled to 2v, what happens to the answer in part a?
I think the answer to b is that it increases, but I'm not sure.
Can you show me how to do it all?

Answer 1

"In the rotating system, the moment of inertia, I, takes the role of the mass, m, and the angular velocity, ω, takes the role of the linear velocity, v. The rotational energy of a rolling cylinder varies from one half of the translational energy (if it is massive) to the same as the translational energy (if it is hollow)."
And yes, you are correct that it increases

Answer 2

veghead566-

I found a really easy to understand article on moment of inertia, cited below. (it discusses a solid, not a hollow, ball)

Also from the same source, a good article on rolling motion, also cited.

(they have used the letter w interchangeably with the Greek letter omega, angular velocity)

since, as the rolling motion article states, the total kinetic energy is the sum of two terms, both including v^2, the fraction of total kinetic energy in the form of rotational kinetic energy is given by some kind of ratio of M, I, and R. You can probably solve that yourself.

If the speed of the ball is doubled, that very same fraction would apply, so there would be no difference in the fraction of kinetic energy in the form of rotational kinetic energy. (both the translational and rotational kinetic energies would quadruple if the the speed were doubled because they each go like v^2)

hope that helps some!

:-)

Answer 3

The answer to your question is: NOT I.

Answer 4

m/v
m/2v