What is the relationship between ω, v, and r when solving for KEtot ratios?

[Badre]

I have two homework problems that I've solved, but I can't reconcile the answers.

The first problem involved a sphere rolling down the hill, I must find the ratio of the translational kinetic energy to the total kinetic energy.

KEtot = KEtrans + KErot
= 1/2mv^2 + 1/2Iω^2
= 1/2mv^2 + 1/2(2/5mr^2)(v/r)^2

Working through the algebra gives me a ratio of 5/7, which was accepted as correct by the software.



The second problem is similar, this time with a string wound around a hockey puck sitting on a frictionless surface and pulling with constant force. I must determine the ratio of rotational kinetic energy to the total kinetic energy. I proceeded to do this problem the same way as the first:

KEtot = KEtrans + KErot
= 1/2mv^2 + 1/2Iω^2
= 1/2mv^2 + 1/2(1/2mr^2)(v/r)^2

This time I got a ratio of 1/3, but this is incorrect. The tutorial in the software indicated that ω does not equal (v/r), but rather is dependant on the moment of inertia of the object being rotated. Since the puck is a cylinder, ω = (2v/r). Plugging this value into the formula above gave me the accepted ratio of 2/3.

To me these look like identical problems, however in the first problem I was able to simply substitute v/r for ω, but wasn't able to do so for this problem. How do I know when to make that substitution vs. working out the relationship based on the inertia of the object?

Thanks!

 

[Doc Al]

The first problem involves rolling without slipping, which implies v = ωr. The second problem has no such condition.

 

[Badre]

Great, thank you!