Question

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.

Short answer

$$\begin{gathered} \left(\mathrm{a}\right)\mathrm{The}\mathrm{angular}\mathrm{speed}\mathrm{of}\mathrm{the}\mathrm{second}\mathrm{hand}\mathrm{is}0.105\mathrm{rad}/\mathrm{s}. \\ \left(\mathrm{b}\right)\mathrm{The}\mathrm{angular}\mathrm{speed}\mathrm{of}\mathrm{the}\mathrm{minute}\mathrm{hand}\mathrm{is}1.74\times {10}^{-3}\mathrm{rad}/\mathrm{s}. \\ \left(\mathrm{c}\right)\mathrm{The}\mathrm{angular}\mathrm{speed}\mathrm{of}\mathrm{hour}\mathrm{hand}\mathrm{is}1.45\times {10}^{-4}\mathrm{rad}/\mathrm{s}. \end{gathered}$$

Step-by-step solution

Understanding the angular speed

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.

(a) Determination of angular speed of the second hand.

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$0.105\mathrm{rad}/\mathrm{s}.$

(b) Determination of angular speed of the minute hand

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$1.74\times {10}^{-3}\mathrm{rad}/\mathrm{s}.$

(c) Determination of angular speed of the hour hand

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