Finding the kinetic energy of a hollow sphere Consider the setup where a solid sphere of mass $m$ and radius $R$ rolls down a plane that is at an angle $\alpha$ to the horizontal. I'm given that the sphere rolls without slipping and that the kinetic energy is given by $$T=\frac{7}{10}m\dot{s}^2$$ where $s$ is the distance the sphere has travelled down the plane. I'm now asked to find the kinetic energy for a hollow sphere of thickness $a \ll R$ and the same mass. I'm given that the moment of inertia for such an object is given by $\frac{2}{3}mR^2$. Using this I have found the kinetic energy to be given by $$T=\frac{5}{6}m\dot{s}^2$$ which is greater than the previous kinetic energy. Is this right though? It was my understanding that the spherical shell would have less kinetic energy since it is harder to roll, but my answer doesn't match that idea. What's going on here?

General formula for KE of rotating body is

$$K = \frac{1}{2} \ I \omega ^2+\frac{1}{2}mv_{\text{com}}^2$$

where

* $\omega=\dfrac{v}{r}$ for rolling without slipping, and

* $v_{\text{com}} $ is velocity of centre of mass (here $v_{\text{com}}=v$)




Thus for **spherical shell** ,

$$K = \frac{1}{2} \cdot \dfrac{2mr^2}{3}\cdot \dfrac{v^2}{r^2}+\frac{1}{2}mv^2$$

$$\therefore K = \dfrac{5}{6}mv^2$$

* * *

Regarding your intuition, since both the rotating bodies have same velocity and mass, then moment of inertia determines which body will have greater $K$.

Moment of inertia is _rotational analogue_ of translational inertia, i.e. mass. When two bodies have same velocity, then the body with higher mass will have greater $K$.