Question

A hollow cylinder with inner radius $R$ , outer radius $2R$ and mass $M$ is rolling with speed of it's axis $v$ . Its kinetic energy is

• A. $\frac{5}{6}\textit{Mv}^{2}$
• B. $\frac{5}{2}\textit{Mv}^{2}$
• C. $\frac{1}{2}\textit{Mv}^{2}$
• D. $\frac{1 3}{1 6}\textit{Mv}^{2}$

Answer 1

Answer: (D)

$\text{KE}=\frac{1}{2}\textit{Mv}^{2}+\frac{1}{2}\textit{I}\omega ^{2}$
$\text{KE}=\frac{1}{2}\left(\textit{Mv}\right)^{2}+\frac{1}{2}\textit{I \, }\left(\frac{\textit{v}}{2 \textit{R}}\right)^{2}$ ...(1)
Consider solid cylinder of radius $2R$ and mass $M'$ , from which a cylinder of radius $R$ of mass $\textit{M}\text{"}=\frac{\textit{M} \text{'}}{4}$ is taken out from a common axis.
Mass of hollow cylinder $M \, = \, M'-M''$
$\textit{M}=\textit{M}\text{'}-\frac{\textit{M} \text{'}}{4}=\frac{3 \textit{M} \text{'}}{4} \, \, \, ⇒ \, \, \, \textit{M}=\frac{3 \textit{M} \text{'}}{4} \, \, \, ⇒ \, \, \, \textit{M}\text{'}=\frac{4 \textit{M}}{3}\text{ & }\textit{M}\text{"}=\frac{\textit{M}}{3}$
$\left(\textit{I}\right)_{2 \text{R}}=\left(\textit{I}\right)_{\text{R}}+\textit{I} \, ⇒ \, \textit{I}=\left(\textit{I}\right)_{2 \text{R}}-\left(\textit{I}\right)_{\text{R}} \, ⇒ \, \textit{I}=\frac{1}{2}\textit{M}\text{'}\left(2 \textit{R}\right)^{2}-\frac{1}{2}\textit{M}\text{"}\left(\textit{R}\right)^{2}$
$\textit{I}=\left[\frac{1}{2} \times \frac{4 \textit{M}}{3} \times 4 \textit{R}^{2} - \frac{1}{2} \times \frac{\textit{M}}{3} \textit{R}^{2}\right] \, ⇒ \, \textit{I}=\frac{5}{2}\textit{MR}^{2}$ ... (2)
$\text{KE}=\frac{1 3}{1 6}\textit{Mv}^{2}$