Question

A particle moves from point $A$ to $B$ in uniform circular motion as shown in figure. The ratio of angular velocity of particle w.r.t. observed $O^{\prime }$ and $O^{\prime \prime }$ is

• A. 2:1
• B. 3:1
• C. 4:1
• D. 1:1

Answer 1

The correct option is A 2:1

In the given figure

$\begin{array}{c}\angle A{O}^{\prime }B=\alpha \\ \angle A{O}^{{}^{\prime \prime }}B=\beta \end{array}$

Tangential velocity at $B=W{o}^{{}^{\prime }}r$

$\begin{array}{c}BC=r\sin \alpha \\ O^{{}^{\prime }}C=r\cos \alpha \end{array}$

Tangential velocity at $B$ is same

$\begin{array}{c}{O}^{\prime \prime }B\sin \beta =r\sin \alpha \\ so,O^{\prime \prime }B=\frac{r\sin \alpha }{\sin \frac{\alpha }{2}}\end{array}$

$\begin{array}{c}{W}_{O^{{}^{\prime }}}r=W_{O^{\prime \prime }}{O}^{\prime \prime }B=\frac{W_{O^{{}^{\prime }}}}{W_{O^{\prime \prime }}}\\ =\frac{2r\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}}{r\sin \frac{\alpha }{2}}=2\frac{\cos \alpha }{2}\end{array}$

For particle of $A$ $\alpha =0$

we get.

$\frac{2\cos \alpha }{2}$

so option $A$ is correct.