Question

In the above figure, $x$ is any instantaneous position. The mirror is rotating about $z$-axis with angular velocity $\omega$ in anticlockwise direction and the object is moving with velocity $v$ along $x$-axis. If the magnitude of velocity of image is $acm/s$ at the given instant, then find the value of $\frac{36a}{31}$.
(Given: $\theta =30^{\circ },\omega =1rad/s,v=1cm/s$, $x=1m)$

Answer 1

Answer: 4

Let the angle of incidence be $\theta$
Position vector of image
$=r_I=(-x\cos 2\theta \hat{i})+(-x\sin 2\theta )\hat{j}$
$\frac{d}{dt}\left(r_I\right)=v_I,\frac{dx}{dt}=v,\frac{d\theta }{dt}=\omega$
$\frac{d}{dt}\left(r_I\right)=\left[\frac{-dx}{dt}\cdot \cos 2\theta -x(-\sin 2\theta )\cdot 2\frac{d\theta }{dt}\right]\hat{i}$
$+\left[\left(\frac{-dx}{dt}\right)\sin 2\theta -x\cdot \cos 2\theta \times 2\frac{d\theta }{dt}\right]\hat{j}$
$V_I=[-V\cos 2\theta +2\omega x\sin 2\theta ]\hat{i}$.
$+[-V\sin 2\theta -2\omega x\cos 2\theta ]\hat{j}$
Putting the given values of $\theta ,\omega$ and $v$,
$\left|V_I\right|=a=\frac{31}{9}cm/s$