Q 11-06-177JEE MainJEE Main 2020 (6 Sep, Shift 1)Medium
Four point masses, each of mass $m$, are fixed at the corners of a square of side $l$. The square is rotating with angular frequency $\omega$ about an axis passing through one of the corners of the square and parallel to its diagonal, as shown in the figure. The angular momentum of the square about the axis is:
Answer: (C) $3ml^2\omega$
The axis passes through one corner and is parallel to the diagonal that does not pass through that corner. Perpendicular distances of the masses from the axis:
Corner on the axis: $0$.
Two neighbouring corners: $\dfrac{l}{\sqrt2}$ each.
Opposite corner: the full diagonal, $\sqrt2\,l$.
$$I = m\left(\frac{l^2}{2} + \frac{l^2}{2} + 2l^2\right) = 3ml^2$$
$$L = I\omega = 3ml^2\omega$$
Solution by Sreeraj P, M.Sc Physics