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.

• A. $\frac{1}{2}m\omega d^2\hat{j}$
• B. $\frac{1}{4}m\omega d^2\hat{j}$
• C. $\frac{1}{3}m\omega d^2\hat{j}$
• D. $\frac{2}{3}m\omega d^2\hat{j}$

Answer 1

The correct option is B

$\frac{1}{4}m\omega d^2\hat{j}$

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

$L=mvr=m\omega \left(\frac{d}{2}\right)\frac{d}{2}=\frac{1}{4}m\omega d^2$

Along the direction perpendicular to the plane of rotation i.e. j