JEE Main 2020 (3 Sep, Shift 2) Physics Q4: Angular Momentum

Question

A uniform rod of length 'l' is pivoted at one of its ends on a vertical shaft of negligible radius. When the shaft rotates at angular speed $\omega$ the rod makes an angle $	heta$ with it (see figure). To find $	heta$ equate the rate of change of angular momentum (direction going into the paper) $\frac{ml^2}{12}\omega^2\sin	heta\cos	heta$ about the centre of mass (CM) to the torque provided by the horizontal and vertical forces $F_H$ and $F_V$ about the CM. The value of $	heta$ is then such that :

Accepted Answer

Option 3: $\cos	heta = \frac{3g}{2l\omega^2}$

Answer

$\cos	heta = \frac{g}{2l\omega^2}$

Answer

$\cos	heta = \frac{g}{l\omega^2}$

Answer

$\cos	heta = \frac{3g}{2l\omega^2}$

Answer

$\cos	heta = \frac{2g}{3l\omega^2}$

JEE Main 3 September 2020, Shift 2, Physics Q4

Same idea in other shifts