91Ó°ÊÓ

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 $c\text{is}3.0×{10}^5\raisebox{1ex}{${\displaystyle \text{Km}}$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\text{is}2.998×{10}^8 \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

By calculating the angular displacement and time, we can find the angular velocity$\mathrm{Ó\neg }$. From, we can find the linear speed v.

1. Whereis the velocity of light.
2. The angular displacement$\mathrm{θ}$is$\mathrm{θ}=\frac{2\mathrm{π}}{500}$
3. The angular velocity $\mathrm{Ó\neg }$is$\mathrm{Ó\neg }=\frac{\mathrm{θ}}{t}$
4. The linear speed vis$v=\mathrm{Ó\neg }r$

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

$$\begin{gathered} \mathrm{θ}=\frac{2\mathrm{π}}{500} \\ =1.26×{10}^{-2} \text{rad} \end{gathered}$$

That time is

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

Substitute all the value in the above equation.

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

Thus, the angular velocity of wheel isgiven by

$\mathrm{Ó\neg }=\frac{\mathrm{θ}}{t}$

Substitute all the value in the above equation.

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

Hence the angular speed is, $=3.8×{10}^3\raisebox{1ex}{${\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

$v=\mathrm{Ó\neg }r$

Substitute all the value in the above equation.

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

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