Difference between a spinning sphere and rolling one

[Gabriel Maia]

Consider this: We have a sphere rolling down a slant, released from some height h with null velocity. At the end of the slant its potential energy will have been fully converted to kinetic energy, part translational and part rotational.

Now consider this: at the end of the slant the ball enters a loop. As it travels around the loop reaching its top it will have a potential energy, alright, but more importantly for our present problem, it will have again a translational and a rotational kinetic energy. The rotational part is due to the ball spinning around its center (as it rolls along its path) or is it due to the rotation around the loop's center? Shouldn't be 2 rotationa kinetic energies?

Thank you

 

[A.T.]

The rotational part is due to the ball spinning around its center

This is how it's usually defined, but you can decompose the KE any way you like. Note that the moment of inertia will depend on the center you chose.

 

[Gabriel Maia]

I see...

But the moment of inertia of a particle on a circular trajectory is $I = m R^2$ (where R is the trajectory radius) and the moment of inertia of a sphere spinning around its own center is $I = \frac{2}{5} m r^2$(where r is the sphere radius). So the total kinetic energy will be

$K = \frac{1}{2} mv^2 + \frac{1}{2} mR^2 \omega^2 + \frac{1}{2}\left[\frac{2}{5}mr^2\right]\omega'^2$?

(where $\omega$ is the angular speed of the ball around the loop and $\omega'^2$ around itself)

 

[A.T.]

I see...

But the moment of inertia of a particle on a circular trajectory is $I = m R^2$ (where R is the trajectory radius) and the moment of inertia of a sphere spinning around its own center is $I = \frac{2}{5} m r^2$(where r is the sphere radius). So the total kinetic energy will be

$K = \frac{1}{2} mv^2 + \frac{1}{2} mR^2 \omega^2 + \frac{1}{2}\left[\frac{2}{5}mr^2\right]\omega'^2$?

No, you are accounting twice for the movement along the circular path. The first two terms are identical, and represent the same thing.

 

[Gabriel Maia]

Splendid, how foolish of me. Thank you very much.