Two identical balls, A and B, are each attached to very light string, and each string is wrapped around the rim of a frictionless pulley of mass M. The only difference is that the pulley for ball A is a solid disk, while the one for ball B is a hollow disk, like part (e) in Table 9.2. If both balls are released from rest and fall the same distance, which one will have more kinetic energy, or will they have the same kinetic energy? Explain your reasoning.

In order to find in which case the kinetic energy would be greater, let’s first compare their rotational inertia.

The moment of inertial for solid cylinder is given by,

${\mathit{I}}_{\mathbf{solid}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{R}}^{\mathbf{2}}$

The moment of inertial for hollow cylinder is given by,

${\mathit{I}}_{\mathbf{solid}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}\left(R_1^2+R_2^2\right)$

From the above relation,$R<R_1+R_2$and in order to masses remain equal with the same densitiesmust be greater than. Therefore, it can be concluded that$l_{hollow}>l_{solid}$.

Draw the diagram for given situation.

By using the Law of conservation of energy, write the equation of energy as follows.

$K_{A1\text{'}}+U_{A1}+K_{solid1}=K_{A2}+U_{A2}+K_{solid2}$

Similarly, for hollow cylinder,

$K_{B1\text{'}}+U_{B1}+K_{hollow1}=K_{B2}+U_{B2}+K_{hollow2}$

Initially the blocks are at rest, then, $K_{A1},K_{solid1},K_{B1},and{K}_{hollow1}$are zero also in position 2, the object are at zero level of gravitation so $U_{A2}$, and $U_{B2}$are also zero.

localid="1663994676398" $$\begin{gathered} U_{A1}=K_{A2}+K_{solid1} \\ {U}_{B1}=K_{B2}+K_{hollow2} \end{gathered}$$

Both the objectsandhave the same mass, which means$m_A=m_B=m$and they are at the same height so$U_{A1}=U_{B1}=U$.

localid="1663995510531" $$\begin{gathered} U=\frac{m{v}_A^2}{2}+\frac{l_{solid}{\omega }_A^2}{2} \\ U=\frac{m{v}_B^2}{2}+\frac{l_{hollow}{\omega }_B^2}{2} \end{gathered}$$

It is known thatlocalid="1663995514295" $\omega =\frac{v}{R}$then,

localid="1663995518192" $$\begin{gathered} U=\frac{m{v}_A^2}{2}+\frac{l_{solid}{\left({\displaystyle \frac{v_A}{R}}\right)}^2}{2} \\ =\frac{m{v}_A^2}{2}+\frac{l_{solid}{\left(v_A\right)}^2}{2R^2} \\ U=\frac{m{v}_B^2}{2}+\frac{l_{hollow}\left(v_B\right)}{2R_2} \end{gathered}$$

Simplify the above equations forlocalid="1663995542081" $v_A$andlocalid="1663995545753" $v_B$.

localid="1663995523934" $$\begin{gathered} v_A^2=\frac{2U}{m+{\displaystyle \frac{l_{solid}}{R^2}}} \\ {v}_B^2=\frac{2U}{m+{\displaystyle \frac{l_{hollow}}{R_{{}^2}^2}}} \end{gathered}$$

Write the above equations in terms of rotational inertia.

localid="1663995529691" $$\begin{gathered} v_A^2=\frac{2U}{m+{\displaystyle \frac{{\displaystyle \frac{1}{2}m{R}^2}}{R^2}}} \\ {v}_A^2=\frac{2U}{m+{\displaystyle \frac{1}{2}m}} \\ {v}_B^2=\frac{2U}{m+{\displaystyle \frac{{\displaystyle \frac{1}{2}m\left(R_1^2+R_2^2\right)}}{R_2^2}}} \\ {v}_B^2=\frac{2U}{m+{\displaystyle \frac{1}{2}m}{\displaystyle \frac{R_1^2}{R_2^2}}+{\displaystyle \frac{1}{2}}m} \end{gathered}$$

Fro the above expressions it can be observed thatlocalid="1663995536837" $v_A^2>v_B^2$.

Therefore, the kinetic energy will be greater in case of solid cylinder than in case of hollow cylinder.