Question

One end of rod of length L=1 m is fixed to a point on the circumference pf a wheel of radius $R=1/\sqrt{3}$ m. The othrt end is sliding freely along a straight channel passing through the centre O of the wheel as shown in the figure below. The wheel is rotating with a constant velocity $\omega$ about O.
The speed of the sliding end P when $\theta ={60}^o$ is

• A. $\frac{2\omega }{3}$
• B. $\frac{\omega }{3}$
• C. $\frac{2\omega }{\sqrt{3}}$
• D. $\frac{\omega }{\sqrt{3}}$

Answer 1

Answer: (C)

$\frac{2\omega }{\sqrt{3}}$

$\begin{array}{c}\text{Let the speed of sliding end of the boint be}v\\ \text{along op -}\\ \Rightarrow \text{velocity of Rod at point}p\text{will be}\\ \text{equal to the speed of point s along its length}\\ \text{as the Rod is rigid}\\ \Rightarrow R\omega =v\cos {30}^{\circ }\\ \Rightarrow v=\frac{R\omega }{\cos {30}^{\circ }}=\frac{2\omega }{\sqrt{3}}\end{array}$