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 $a\, cm / s$ at the given instant, then find the value of $\frac{36 a}{31}$.
(Given: $\theta=30^{\circ}, \omega=1 \,rad / s , v=1\, cm / s$, $x=1 \,m )$

Answer 1

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