Question

Two masses $A$ and $B$ connected with an inextensible string of length $l$ lie on a smooth horizontal plane. $A$ is given a velocity of $vm/s$ along the ground perpendicular to line $AB$ as shown in the figure. Find the tension in string during their subsequent motion.

Answer 1

First of all we locate the center of mass of the system of masses.

It is at a distance of $\frac{\left(2m\right)l}{m+2m}=\frac{2l}{3}$ from mass $m$.

The angular momentum about the center of mass is thus $\left(2m\right)v\frac{l}{3}=\frac{2mvl}{3}$

Let the angular speed attained by $\omega$.

Thus

$\frac{2mvl}{3}=(m\times (\omega \times \frac{2l}{3}\times \frac{2l}{3}\left)\right)+(2m\times (\omega \times \frac{l}{3})\times \frac{l}{3})$

$⟹\omega =\frac{v}{l}$

The tension in the string gives rise to the centripetal acceleration = $m{\omega }^2\frac{2l}{3}=\frac{2m{v}^2}{3l}$