Question

One end of a rod of length $L=1m$ is fixed to a point on the circumference of a wheel of radius $R=1/\sqrt{3}m$. The other end is sliding freely along a straight channel passing through the center $O$ of the wheel as shown in the figure below. The wheel is rotating with a constant angular velocity $\omega$ about $O$.
The speed of the sliding end $P$ when $\theta ={90}^{\circ }$ is

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

Answer 1

The correct option is D $\frac{\omega }{\sqrt{3}}$

As velocity of both the ends of rod along the length should be same so,

$\omega \times R\times cos{45}^o=v\times cos{45}^o$ $for\theta ={90}^o$

Hence v=$\omega \times 1/\sqrt{2}$ $=\frac{\omega }{\sqrt{2}}$