A uniform rod of length ‘ l ’ is pivoted at one of its ends on a… | JEE Main 2020 (3 Sep, Shift 2)

Q50JEE Main 2020 · 3 SEP · SHIFT 2Difficult

Rotational Kinematics & Dynamics

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 $\theta$ with it (see figure). To find $\theta$ equate the rate of change of angular momentum (direction going into the paper) $\frac{ml^{2}}{12}\omega^{2}\sin\theta\cos\theta$ about the centre of mass (CM) to the torque provided by the horizontal and vertical forces FH and FV about the CM. The value of $\theta$ is then such that:

Show answer and solution

Answer: (b)

CM moves in a circle of radius $\tfrac{l}{2}\sin\theta$ : $F_V = mg$ , $F_H = m\omega^2\tfrac{l}{2}\sin\theta$ .
Torque about CM: $F_V\tfrac{l}{2}\sin\theta - F_H\tfrac{l}{2}\cos\theta = \dfrac{ml^2}{12}\omega^2\sin\theta\cos\theta$
$\dfrac{mgl}{2} = m\omega^2 l^2\cos\theta\left(\dfrac{1}{4} + \dfrac{1}{12}\right) = \dfrac{m\omega^2 l^2\cos\theta}{3}$
$\cos\theta = \dfrac{3g}{2l\omega^2}$ , so the answer is (b).

Report a mistake

Rotational Kinematics & Dynamics question from JEE Main 2020 (3 Sep, Shift 2)

More questions on Rotational Kinematics & Dynamics