91Ó°ÊÓ

We have

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

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

$v=R\mathrm{Ó\neg }$

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

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

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

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, $\frac{5}{13}$.