Question

A spot light $L$ rotates in horizontal plane with a constant angular velocity of $0.1rad/s$. The spot of light $S$ moves along the wall at a distance $3m$

Column I	Column II
(A) Velocity of spot $S$ in $m/s$ when it is closest to $L$ is	(p) $0.6$
(B) Acceleration of the spot in $m/s^2$ when $\varphi ={45}^o$	(q) $0.3$
(C) Value of ratio of $\varphi$ in radians to $\pi$ radians when velocity of spot is $0.6m/s$ is	(r) $0.25$
(D) Angular acceleration of the spot light $L$ when $\varphi ={30}^o$	(s) $0$
	(t) $0.12$

Answer 1

$\left(a\right)$ When $S$ is closed to $L$, it is at point $O$ which is $3$ $m$ away from $L$.

Therefore,

Velocity, $V=r\omega$

$⟹V=3\times 0.1$

$⟹V=0.3$ $m/s$

$\left(b\right)$ When $\varphi ={45}^o$:

Refer IMAGE $01$

$\cos {45}^o=\frac{3}{r}$

$⟹\frac{1}{\sqrt{2}}=\frac{3}{r}$

$⟹r=3\sqrt{2}$ $m$

$\therefore$ acceleration, $a=r{\omega }^2=3\sqrt{2}\times 0.1\times 0.1$

$⟹a=0.04$ $m/s^2$

$\left(c\right)$ When velocity of the spot is $0.6$ $m/s$:

Refer IMAGE $02$

$V=r\omega$

$⟹r=\frac{V}{\omega }$

$⟹r=\frac{0.6}{0.1}$

$⟹r=6$ $m$

$\cos \varphi =\frac{3}{6}=\frac{1}{2}$

$⟹\varphi ={60}^o=\frac{\pi }{3}$ $radians$

$\therefore \frac{\varphi }{\pi }=\frac{\pi }{3}\times \frac{1}{\pi }=\frac{1}{3}=0.33$

$\left(d\right)$ When $\varphi ={30}^o$

Refer IMAGE $03$

$\cos {30}^o=\frac{3}{r}$

$⟹\frac{\sqrt{3}}{2}=\frac{3}{r}$

$r=\frac{3\times 2}{\sqrt{3}}=2\sqrt{3}$ $m$

$\therefore$ Angular acceleration, $a=r{\omega }^2$

$⟹a=2\sqrt{3}\times 0.1\times 0.1$

$⟹a=0.034$ $rad/s^2$