Question

A non-uniform cylinder of mass $m$, length $l$ and radius $r$ is having its centre of mass at a distance $l$ from the centre and lying on the axis of the cylinder. The cylinder is kept in a liquid of uniform density $\rho$. The moment of inertia of the rod about the centre of mass is $I$. The angular acceleration of point A relative to point B just after the rod is released from the position as shown in the figure is

• A. $\frac{\pi \rho g{l}^2{r}^2}{I}$
• B. $\frac{\pi \rho g{l}^2{r}^2}{4I}$
• C. $\frac{\pi \rho g{l}^2{r}^2}{2I}$
• D. $\frac{3\pi \rho g{l}^2{r}^2}{4I}$

Answer 1

Answer: (B)

$\frac{\pi \rho g{l}^2{r}^2}{4I}$

Buoyancy force will act at center and Force of gravity will act at center of mass.

Balancing torques, $\pi r^{2}l\rho g\times l/4=I\alpha$

$\Rightarrow \alpha =\pi r^2{l}^2\rho g/4I$