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A non uniform cylinder of mass $m$ , length $l$ and radius $r$ is having its cetnre of mass at a distance $l/4$ 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 shown in figure is
$\frac{{\pi \rho g{l^2}{r^2}}}{I}$
$\frac{{\pi \rho g{l^2}{r^2}}}{4I}$
$\frac{{\pi \rho g{l^2}{r^2}}}{2I}$
$\frac{{3\pi \rho g{l^2}{r^2}}}{4I}$
Solution
Torque about $\mathrm{CM}:$
$F_{b} \cdot \frac{\ell}{4}=\mathrm{I} \alpha$
$\Rightarrow \quad \alpha=\frac{1}{I}\left(\pi r^{2}\right)(\ell)(\rho)(g) \cdot \frac{\ell}{4}$
$\alpha=\frac{\pi r^{2} \ell^{2} g \rho}{4 I}$
$'\alpha'$ will be same for all points.
