In my physics class, we have seen experimentally that objects with lower $I$ values (like spheres) will reach the bottom of a ramp sooner and with a higher final velocity than objects with higher $I$ values (like a hoop) assuming total mass and radius are the same. However when looking at energy, I see that gravitational potential = translational KE + rotational KE. Since the sphere has a higher translational velocity, it has more translational kinetic energy. But since it’s moving faster, it must be rotating faster, and therefore have a higher $\omega$. Looking at the equation for rotational KE $= \frac{1}{2} I \omega^2$, I would assume that since $\omega$ is squared it would have a much bigger impact than that of decreasing the $I$ value.
Why does rotational kinetic energy not increase as quickly with objects that have a lower I value?
Best Answer
Generalizing ratios between translational and rotational kinetic energy makes sense only when object is rotating without slipping. In that case translational and rotational velocities are proportional $v = \omega r$, and you should compare moment of inertia $I$ against mass $m$.
Total kinetic energy of a rotating body is combination of translational and rotational kinetic energy
$$K = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2$$
Moment of inertia for most (simple) rigid objects can be generalized as $I = c m r^2$:
Total kinetic energy can then be written as
$$K = \frac{1}{2} m v^2 + \frac{1}{2} c m (\omega r)^2 = \frac{1+c}{2} m v^2$$
From the work-energy theorem $\Delta K = W$ and if we assume that initial kinetic energy is zero, the final translational velocity is then
$$v = \sqrt{\frac{1}{1 + c} \frac{2 W}{m}} = \sqrt{\frac{2 W}{m + I/r^2}}$$
From this we can conclude that translational velocity $v$ will be larger for smaller values of $c$, which also means smaller moment of inertia $I$.