Question

A rod can rotate about smooth hinge at point 'O'. A cubical block is moving in horizontal direction with constant speed 2 m/s. Find value of angular velocity of rod when $\theta=60^\circ$ (Cube has side length 2 m)

Answer 1

$\omega=\frac{3}{4} \text{ rad/s}$, (angular velocity in magnitude).

Answer 2

1. Let the point of contact be the upper left vertex of the cube at $(x,2)$ so that $\tan\theta=\dfrac{2}{x}$ or $x=2\cot\theta$.

2. Differentiate: $\frac{dx}{dt}=-2\csc^2\theta\,\frac{d\theta}{dt}$.

3. With $\frac{dx}{dt}=2$ m/s, solve to get $\frac{d\theta}{dt}=-\sin^2\theta$.

4. At $\theta=60^\circ$, $\sin^2\theta=3/4$ so that $|\omega|=0.75$ rad/s.