Chapter 22, Problem 14P

To determine

The minimum angular speed.

Answer to Problem 14P

The answer is $23585.65\text{rps}$

Explanation of Solution

Given data:

  

Distance, $d=2d=70km$

Speed of light, $c=2.99792458\times {10}^{8}m/s$

The Mirror has 8 sides, $\Delta Q=\frac{1}{8}\text{rev}=\frac{\pi }{4}\text{rad}$

Formula used:

It is known that:

  $t=\frac{2d}{c}$ and $\omega =\frac{\Delta Q}{t}$

Where,

  $\omega$ is the angular speed.

  $d$ is the total distance travelled.

  $c$ is speed of the light.

  $\Delta t$ is change in time.

Calculation:

Get the time that is taken by the light to reach from one mirror to the observer, $t=\frac{d}{c}$

  $t=\frac{2d}{c}=\frac{70\times {10}^{3}m}{2.99792458\times {10}^{8}m/s}=2.335\times {10}^{-4}\mathrm{s}$

Since, each side of the mirror takes up $\frac{\pi }{4}\text{rad}$ . Hence, the for $8$ sided-rotation will be $\frac{7\pi }{4}\text{rad}$ .

Therefore, the minimum angular speed at which the mirror rotate:

  $\omega =\frac{\Delta Q}{t}=\frac{\left(\frac{7}{4}\right)\pi }{2.335\times {10}^{-4}\mathrm{s}}=23585.65\text{rps}$

Conclusion: The required angular speed is $23585.65\text{rps}$ .