Question

In the arrangement shown, the rod is freely pivoted at point O and is in contact with the equilateral triangular block which can move on the horizontal frictionless ground. As the block is given a speed $v$ in the forward direction, the rod rotates about point O. Find the angular velocity of the rod in rad/s at the instant when $\theta ={30}^{\circ }$. [Take $v=20\text{m/s}$, $a=1\text{m}$]

Answer 1

$\tan \theta =\frac{1}{x}$
${\sec }^2\theta \frac{d\theta }{dt}=\frac{-1}{x^2}\frac{dx}{dt}$
For $\theta ={30}^{\circ },x=\sqrt{3},\sec \theta =\frac{2}{\sqrt{3}}$
Substituting we get, $\frac{d\theta }{dt}=5rad/s$