Answer
a. $-0.6km/sec$ (moving down)
b. $\frac{18}{\pi}\approx5.73$.
Work Step by Step
a. Given $\frac{d\theta}{dt}=-0.6rad/sec$, using the figure provided by the exercise, we have $tan\theta=\frac{x}{1}$. Taking the derivative, we get $sec^2\theta\frac{d\theta}{dt}=\frac{dx}{dt}$.
With $\theta=0$ at point A, we get $\frac{dx}{dt}=(sec^20)(-0.6)=-0.6km/sec$
(negative sign means the beam is moving down)
b. With an angular speed of $0.6rad/sec$, the total angle per minute would be $T=0.6\times60=36rad$, which gives the number of revolutions per minute as $n=\frac{36}{2\pi}=\frac{18}{\pi}\approx5.73$.
