Question

A cylinder of height $H$ and diameter $\frac{H}{4}$ is kept on a frictional turntable as shown in fig. The axis of the cylinder is perpendicular to the surface of the table and the distance of axis of the cylinder is 2H from the centre of the table. The angular speed of the turntable at which the cylinder will start toppling (assume that friction is sufficient to prevent slipping) is :

• A. $\sqrt{\frac{g}{2}(\frac{1}{2}-H)}$
• B. $\sqrt{g(\frac{1}{2}-H)}$
• C. $\sqrt{\frac{g}{4H}}$
• D. $\sqrt{\frac{g}{8H}}$

Answer 1

The correct option is D $\sqrt{\frac{g}{8H}}$

Centripetal force $F=m{\omega }^2\left(2H\right)$
Net torque due to all the forces about point P is zero.
${\tau }_{centripetal}=m{\omega }^2\left(2H\right)\times \frac{H}{2}$ where $\frac{H}{2}$ is the perpendicular distance of the centripetal force from the axis passing through point P.
${\tau }_{weight}=mg\frac{H}{8}$
Equating both the torques,
$m{\omega }^2\left(2H\right)\times \frac{H}{2}=mg\frac{H}{8}$
$\Rightarrow \omega =\sqrt{\frac{g}{8H}}$