91Ó°ÊÓ

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" $$\begin{gathered} U=\frac{m{v}_A^2}{2}+\frac{l_{solid}{\mathrm{Ó\neg }}_A^2}{2} \\ U=\frac{m{v}_B^2}{2}+\frac{l_{hollow}{\mathrm{Ó\neg }}_B^2}{2} \end{gathered}$$

It is known thatlocalid="1663995514295" $\mathrm{Ó\neg }=\frac{v}{R}$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.