Chapter 34, Problem 1P

To determine

The minimum angular speed of the wheel for this experiment.

Answer to Problem 1P

The minimum angular speed of the wheel is $114\text{rad}/\text{s}$.

Explanation of Solution

Given info: The distance between the light source and the mirror was $11.45\text{km}$, no of notches in the wheel is $720$ notches, experimentally determined value of $c$ was $2.998\times {10}^8\text{m}/\text{s}$.

The formula to calculate the angular speed of the wheel is,

$\omega =\frac{d\theta }{dt}$ (1)

Here,

$\omega$ is the angular speed of the wheel.

$d\theta$ is the degree movement of the wheel in consecutive notches.

$dt$ is the time for the light to travel to the stationary mirror and back.

The formula to calculate $d\theta$ is,

$d\theta =\frac{2\pi }{\text{number of notches}}$

Here,

$d\theta$ is the degree movement of the wheel in consecutive notches.

Substitute $720$ for the number of notches.

$d\theta =\frac{2\pi }{720}$ (2)

The formula to calculate $dt$ is,

$dt=\frac{2d}{c}$ (3)

Here,

$d$ is the number of notches in the wheel.

$c$ is the experimentally determined value of speed of light.

Substitute $\frac{2\pi }{720}$ for $d\theta$ and $\frac{2d}{c}$ for $dt$ in equation (1).

$\omega =\frac{\left(\frac{2\pi }{720}\right)}{\left(\frac{2d}{c}\right)}$ (4)

Substitute $11.45\text{km}$ for $d$ and $2.998\times {10}^8\text{m}/\text{sec}$ for $c$ in equation (4).

$\begin{array}{c}=\frac{2\pi /720\text{rad}}{\frac{2\times \left(11.45\text{km}\times \frac{{10}^3\text{m}}{1\text{km}}\right)}{2.998\times {10}^8\text{m}/\text{sec}}}\\ =114\text{rad}/\text{sec}\end{array}$.

Conclusion:

Therefore, the minimum angular speed of the wheel is $114\text{rad}/\text{sec}$.