What is the angular speed of (a) the second hand, (b) the minute hand, and (c) the hour hand of a smoothly running analog watch? Answer in radians per second.

Angular speed may be defined as the rate of change of angular distance with time. The angular distance is taken in radians.

The expression for angular speed is given as:

$\omega =\frac{∆\theta }{∆t}...\left(\mathrm{i}\right)$

Here,$∆\theta$ is the angular distance and is the time interval.

The second hand completes one revolution in$∆\mathrm{t}=60\mathrm{s}$and the angular distance covered is, $∆\mathrm{\theta }=2\mathrm{\pi rad}$

Using equation (i), the angular speed of second hand is calculated as:

$$\begin{gathered} \mathrm{\omega }=\frac{2\mathrm{\pi rad}}{60\mathrm{s}} \\ =0.1074\mathrm{rad}/\mathrm{s} \\ \approx 0.105\mathrm{rad}/\mathrm{s} \end{gathered}$$

Thus, the angular speed of second hand is$0.105\mathrm{rad}/\mathrm{s}.$

The minute hand completes one revolution in $∆\mathrm{t}=60\min$and the angular distance covered is, $∆\mathrm{\theta }=2\mathrm{\pi rad}$

Using equation (i), the angular speed of minute hand is calculated as:

$$\begin{gathered} \mathrm{\omega }=\frac{2\mathrm{\pi rad}}{60\min \times \left({\displaystyle \frac{60\mathrm{s}}{1\min }}\right)} \\ \approx 1.74\times {10}^{-3}\mathrm{rad}/\mathrm{s} \end{gathered}$$

Thus, the angular speed of minute hand is$1.74\times {10}^{-3}\mathrm{rad}/\mathrm{s}.$

$$\begin{gathered} \mathrm{The}\mathrm{hour}\mathrm{hand}\mathrm{completes}\mathrm{one}\mathrm{revolution}\mathrm{in}∆\mathrm{t}=12\mathrm{h}\mathrm{and}\mathrm{the}\mathrm{angular}\mathrm{distance}\mathrm{covered} \\ \mathrm{is},∆\mathrm{\theta }=2\mathrm{\pi rad} \\ \mathrm{Using}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{the}\mathrm{angular}\mathrm{speed}\mathrm{of}\mathrm{hour}\mathrm{hand}\mathrm{is}\mathrm{calculated}\mathrm{as}: \\ \mathrm{\omega }=\frac{2\mathrm{\pi rad}}{12\mathrm{h}\times \left({\displaystyle \frac{3600\mathrm{s}}{1\mathrm{h}}}\right)} \\ =1.45\times {10}^{-4}\mathrm{rad}/\mathrm{s} \\ \mathrm{Thus},\mathrm{the}\mathrm{angular}\mathrm{speed}\mathrm{of}\mathrm{the}\mathrm{hour}\mathrm{hand}\mathrm{is}1.45\times {10}^{-4}\mathrm{rad}/s. \end{gathered}$$