question_answer

A rod of length \[L\] rotates about an axis passing through one of its ends and perpendicular to its plane. If the linear mass density of the rod varies as\[\rho =(A{{r}^{3}}+B)kg/m\], then the moment of inertia of the rod about the given axis of rotation is

A) \[\frac{{{L}^{3}}}{3}\left[ \frac{A{{L}^{3}}}{2}+B \right]\]              

B) \[\frac{L}{3}\left[ \frac{A{{L}^{2}}}{2}+B \right]\]

C) \[\frac{{{L}^{3}}}{3}\left[ \frac{A}{2}+B \right]\]               

D)   None of these

Correct Answer: A

Solution :

Consider a small element of the rod, having length\[dr\], situated at distance \[r\] from the axis. Mass of element,\[dm=\rho \,\,dr\]                 \[=(A{{r}^{3}}+B)dr\] Moment of inertia of the rod about the axis \[I=\int\limits_{0}^{L}{dm.{{r}^{2}}}=\int\limits_{0}^{L}{(A{{r}^{3}}+B){{r}^{2}}dr}\]    \[=\int\limits_{0}^{L}{A{{r}^{5}}dr}+\int\limits_{0}^{L}{B{{r}^{2}}dr}=A\left[ \frac{{{r}^{6}}}{6} \right]_{0}^{L}+B\left[ \frac{{{r}^{3}}}{3} \right]_{0}^{L}\]    \[=\frac{A{{L}^{6}}}{6}+\frac{B{{L}^{3}}}{3}=\frac{{{L}^{3}}}{3}\left[ \frac{A{{L}^{3}}}{2}+B \right]\]