Question

A dumbell consists of two masses A and B each of mass $m$ joined by a light rod of length $L$ with B attached to a suspension point O by a light rod of length $L$ as shown. The minimum speed with which mass A to be projected horizontally so that system completes the vertical circular motion is $V=4\sqrt{\frac{3gL}{n}}$. Find $n$.

Answer 1

CM represents the centre of mass of the dumbell which is performing the vertical cirecular motion.

Velocity of centre of mass, $V_{CM}=\frac{m\times 0+m\times V}{m+m}=\frac{V}{2}$

Distance of CM from the suspension point, $x=L+\frac{L}{2}=\frac{3L}{2}$

Minimum velocity required by a body at the lowermost position of the rod to complete the vertical circular moton is $\sqrt{4gR}$.

$\therefore$ $\frac{V}{2}=\sqrt{4gR}$ where $R=\frac{3L}{2}$

OR $V=2\sqrt{4g\times \frac{3L}{2}}$

$⟹$ $V=4\sqrt{\frac{3gL}{2}}$

Hence, $n=2$