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?

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$}\right.$

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

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$v=\omega r$

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$}\right.$ .