Question

Figure $\mathbf{10}\mathbf{-}\mathbf{32}$ shows an early method of measuring the speed of light that makes use of a rotating slotted wheel. A beam of light passes through one of the slots at the outside edge of the wheel, travels to a distant mirror, and returns to the wheel just in time to pass through the next slot in the wheel. One such slotted wheel has a radius of $\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{\text{cm}}\mathbf{}$ and $500$ slots around its edge. Measurements taken when the mirror is from the wheel indicate a speed of light of$\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\raisebox{1ex}{${\displaystyle \mathbf{\text{km}}}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{s}$}\right.$ . (a) What is the (constant) angular speed of the wheel? (b) What is the linear speed of a point on the edge of the wheel?

Short answer

1. The constant angular speed of the wheel is $3.8\times {10}^3\raisebox{1ex}{$rad$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.
2. The linear speed of a point on the edge of the wheel is $1.9\times {10}^2\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.

Step-by-step solution

Understanding the given information

1. The radius$r\text{is 0}.050 \text{m}$ .
2. The slot around its edge is$500$.
3. The distance between wheel and mirror,$L\text{is}500 \text{m}$ .
4. The speed of light $c\text{is}3.0\times {10}^5\raisebox{1ex}{${\displaystyle \text{Km}}$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\text{is}2.998\times {10}^8 \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Concept and Formula used for the given question

By calculating the angular displacement and time, we can find the angular velocity$\omega$. From, we can find the linear speed v.

1. The time tis$t=\frac{2L}{c}$

1. Whereis the velocity of light.
2. The angular displacement$\theta$is$\theta =\frac{2\pi }{500}$
3. The angular velocity $\omega$is$\omega =\frac{\theta }{t}$
4. The linear speed vis$v=\omega r$

(a) Calculation for the (constant) angular speed of the wheel

Inthetime that light takes to go from the wheel to the mirror and back again, the wheel turns through an angle, which is

$$\begin{gathered} \theta =\frac{2\pi }{500} \\ =1.26\times {10}^{-2} \text{rad} \end{gathered}$$

That time is

$t=\frac{2L}{c}$

Substitute all the value in the above equation.

$$\begin{gathered} t=\frac{{\displaystyle 2\times 500 \text{m}}}{{\displaystyle 2.998\times {10}^8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.}} \\ =3.34\times {10}^{-6} \text{s} \end{gathered}$$

Thus, the angular velocity of wheel isgiven by

$\omega =\frac{\theta }{t}$

Substitute all the value in the above equation.

role="math" localid="1660971022005" $$\begin{gathered} \omega =\frac{1.26\times {10}^{-2} \text{rad}}{3.34\times {10}^{-6} \text{s}} \\ =3.8\times {10}^3\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \end{gathered}$$

Hence the angular speed is, $=3.8\times {10}^3\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.

Step 3: (b) Calculation for the linear speed of a point the edge of the wheel

If r is the radius of the wheel, the linear speed of a point on its rim is given by

$v=\omega r$

Substitute all the value in the above equation.

$$\begin{gathered} v=3.8\times {10}^3\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{$s{\displaystyle \times }{\displaystyle 0}{\displaystyle .}{\displaystyle 050}{\displaystyle }{\displaystyle \text{m}}$}\right. \\ =1.9\times {10}^2\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \end{gathered}$$

Hence the linear speed is,$1.9\times {10}^2 \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$ .