Question

Two particles of mass m each are attached to a light rod of length d, one at its centre and the other at a free end. The rod is fixed at the other end and is rotated in a plane at an angular speed $\omega$. Calculate the angular momentum of the particle at the end with respect to the particle at the centre (i.e. angular momentum of B with respect to A).

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

The situation is shown in figure. The velocity of the particle A with respect to the fixed end O is $vA=\omega \left(\frac{d}{2}\right)$ and that of B with respect to O is $vB=\omega d$. Hence the velocity of B with respect to A is $vB-vA=\omega \left(\frac{d}{2}\right)$. The angular momentum of B with respect to A is, therefore,

$L=m\left(\frac{\omega d}{2}\right)\frac{d}{2}$
$=m\frac{\omega d^2}{4}$