Chapter 21, Problem 105A

(a)

To determine

The time taken by light to travel the distance to the mountain and back.

Answer to Problem 105A

  $t=2.3\times {10}^{-4}\text{ s}$

Explanation of Solution

Given:

The distance between the rotating mirror and the mountain is $l=35\text{ km}$

Formula used:

The time taken by light to travel the distance to the mountain and back can be calculated by using the formula,

  $t\text{ =}\frac{L}{v}$ $\left(1\right)$

Where, $v$ is the velocity of light, $v=c=3\times {10}^8\text{ m/s}$

  $L$ is the distance travelled by light

Calculation:

Since light has to travel from rotating mirror to mountain and reflects from the mirror in the mountain back to the rotating mirror. Therefore the distance travelled by the mirror is $L=2l=70\text{ km}$

Substituting the numerical values in equation $\left(1\right)$

  $t=\frac{\left(\text{70 km}\right)\left(\text{1000m/1 km}\right)}{\left(3\times {10}^8\text{ m/s}\right)}=2.3\times {10}^{-4}\text{ s}$

Conclusion:

The time taken by light to travel the distance to the mountain and back is $2.3\times {10}^{-4}\text{ s}$ .

(b)

To determine

The rotating speed of the mirror.

Answer to Problem 105A

  $f=5.6\times {10}^2\text{ revolution/s}$

Explanation of Solution

Given:

Rotating octagon has mirror on each face. Octagon has $\text{8}$ faces. Hence there are totally $\text{8}$ rotating mirror.

Formula used:

The period of revolution can be calculated by,

  $T=t\times 8\text{ mirrors/revolution}$ $\left(2\right)$

Where, the time taken by light to travel the distance to the mountain and back.

The speed of revolution (frequency of revolution) can be calculated using the equation,

  $f=\frac{1}{T}$ $\left(3\right)$

Calculation:

From part (a),

  $t=2.3\times {10}^{-4}\text{ s}$

Substituting the numerical values in equation $\left(2\right)$ ,

  $T=\left(\frac{2.3\times {10}^{-4}\text{ s}}{1\text{ mirror}}\right)\left(8\text{ mirrors/revolution}\right)=1.8\times {10}^{-3}\text{ s/revolution}$

Substituting the value of $T$ in equation $\left(3\right)$ ,

  $f=\frac{1}{1.8\times {10}^{-3}\text{ s/revolution}}=5.6\times {10}^2\text{ revolution/s}$

Conclusion:

The rotating speed of the mirror is $5.6\times {10}^2\text{ revolution/s}$ .

(c)

To determine

The approximate centripetal force needed to hold the mirrors when it is rotating.

Answer to Problem 105A

  $F_c=1.2\times {10}^4\text{ N}$

Explanation of Solution

Given:

Mass of each mirror is $m=1.0\times {10}^1\text{ g = }1.0\times {10}^1\times {10}^{-3}\text{ kg = }1.0\times {10}^{-2}\text{ kg}$

The radius of the circle through which the mirrors rotate is

  $r\text{ = }1.0\times {10}^1\text{ cm = }1.0\times {10}^1\times {10}^{-2}\text{ m = 0}\text{.1 m}$

Formula used:

The centripetal force can be calculated using the equation,

  $F_c=4{\pi }^{2}m{f}^{2}r$ $\left(4\right)$

Where, $m$ is mass of the mirror

  $f$ is revolution speed

  $r$ is radius of the circle

Calculation:

Substituting the numerical values in equation $\left(4\right)$ ,

  $F_c=4{\pi }^2\left(1.0\times {10}^{-2}\text{ kg}\right){\left(5.6\times {10}^2\text{ revolution/s}\right)}^2\left(\text{0}\text{.1 m}\right)$

  $=1.2\times {10}^4\text{ N}$

Conclusion:

The approximate centripetal force needed to hold the mirrors when it is rotating is $1.2\times {10}^4\text{ N}$ .