Question

Spotlight S rotates in a horizontal plane with constant
angular velocity of 0.1 rad/s. The spot of light P moves
along the wall at a distance of 3 m. The velocity of the spot P
when $\theta$ = 45$^{\circ}$ (see fig.) is.......m/s.

Answer 1

$\frac{x}{3}=tan \theta$
$\therefore \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, x==3 tan \theta$
$-\bigg(\frac{dx}{dt}\bigg)=3sec^2\theta\bigg(-\frac{d\theta}{dt}\bigg)$
$-\frac{dx}{dt}=v_p \, and$
$-\frac{d\theta}{dt}=\omega=0.1 \, rad/s.\theta=45^{\circ}$
Substituting the values, we get
$v_p=3(sec^2 45^{\circ})(0.1)=0.6m/s$