Question

A non-uniform cylinder of mass $m$, length $l$ and radius $r$ is having its center of mass at distance $l/4$ from the center and lying on the axis of the cylinder. The cylinder is kept in a liquid of uniform density $\rho$. The movement of inertia of the rod about the center of mass is $l$. 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. $\frac{\pi \rho g{l}^2{r}^2}{l}$
• B. $\frac{\pi \rho g{l}^2{r}^2}{4l}$
• C. $\frac{\pi \rho g{l}^2{r}^2}{2l}$
• D. $\frac{3\pi \rho g{l}^2{r}^2}{4l}$

Answer 1

The correct option is B $\frac{\pi \rho g{l}^2{r}^2}{4l}$

$F_{buoyancy}=V_{in}\rho g$ where $V_{in}$ is volume

$F_{buoyancy}=\pi r^{2}l\rho g$ act upward at Center

$F_{gravitational}=mg$ act downward at COM .

Torque equation at COM

$\tau =F_{buoyancy}\times \frac{l}{4}$

$\tau =\pi r^{2}lg\times \frac{l}{4}$ and $\tau =I\alpha$

substituting $\tau$ in above equation

$\alpha =\pi r^{2}lg\times \frac{l}{4I}$

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

angular acceleration $\alpha =\frac{\pi \rho g{l}^2{r}^2}{4I}$.

Option B is correct.