What fraction of the total kinetic energy is rotational for the following objects rolling without slipping on a horizontal surface? (a) A uniform solid cylinder; (b) a uniform sphere; (c) a thin-walled, hollow sphere; (d) a hollow cylinder with outer radius R and inner radius $\frac{\mathbf{R}}{\mathbf{2}}$.

We have

$$\begin{gathered} K=\frac{1}{2}m{v}^2, K_{rot}=\frac{1}{2}I{\omega }^2, \\ K_{tot}=K+K_{rot} \end{gathered}$$

For an object which is rolling without slipping, we have,

$v=R\omega$

The fraction of the total kinetic energy that is rotational is,

$$\begin{gathered} \frac{E_{rot}}{E_{tot}}=\frac{\frac{1}{2}I{\omega }^2}{\frac{1}{2}M{v}^2+\frac{1}{2}I{\omega }^2} \\ =\frac{I{\omega }^2}{M{v}^2+{\omega }^2} \\ =\frac{1}{1+\frac{M}{I}·\frac{v^2}{{\omega }^2}} \\ =\frac{1}{1+\frac{M{R}^2}{I}} \end{gathered}$$

$\therefore \frac{E_{rot}}{E_{tot}}=\frac{1}{1+\frac{M{R}^2}{I}}$.

Now, using table 9.2, we get

$I=\frac{1}{12}M{R}^2$gives the above ratio equal to $\frac{1}{3}$.

Hence, the fraction of the total kinetic energy that is rotational for a uniform solid cylinder is, $\frac{1}{3}$.

Now, using table 9.2, we get

$I=\frac{1}{25}M{R}^2$gives the above ratio equal to $\frac{2}{7}$.

Hence, the fraction of the total kinetic energy that is rotational for a uniform sphere is,$\frac{2}{7}$.

Now, using table 9.2, we get

$I=\frac{1}{23}M{R}^2$gives the above ratio equal to$\frac{2}{5}$.

Hence, the fraction of the total kinetic energy that is rotational for a thin-walled, hollow sphere is, $\frac{2}{5}$.

Now, using table 9.2, we get

$I=\frac{1}{58}M{R}^2$ gives the above ratio equal to $\frac{5}{13}$ .

Hence, the fraction of the total kinetic energy that is rotational for a hollow cylinder is, $\frac{5}{13}$.