91Ó°ÊÓ

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:

$\mathrm{Ó\neg }=\frac{∆\mathrm{θ}}{∆t}...\left(\mathrm{i}\right)$

Here,$∆\mathrm{θ}$ 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, $∆\mathrm{θ}=2\mathrm{Ï€°ù²¹»å}$

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

$$\begin{gathered} \mathrm{Ó\neg }=\frac{2\mathrm{Ï€°ù²¹»å}}{60\mathrm{s}} \\ =0.1074\mathrm{rad}/\mathrm{s} \\ ≈0.105\mathrm{rad}/\mathrm{s} \end{gathered}$$

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, $∆\mathrm{θ}=2\mathrm{Ï€°ù²¹»å}$

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

$$\begin{gathered} \mathrm{Ó\neg }=\frac{2\mathrm{Ï€°ù²¹»å}}{60\min ×\left({\displaystyle \frac{60\mathrm{s}}{1\min }}\right)} \\ ≈1.74×{10}^{-3}\mathrm{rad}/\mathrm{s} \end{gathered}$$

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

$$\begin{gathered} \mathrm{The}\mathrm{hour}\mathrm{hand}\mathrm{completes}\mathrm{one}\mathrm{revolution}\mathrm{in}∆\mathrm{t}=12\mathrm{h}\mathrm{and}\mathrm{the}\mathrm{angular}\mathrm{distance}\mathrm{covered} \\ \mathrm{is},∆\mathrm{θ}=2\mathrm{Ï€°ù²¹»å} \\ \mathrm{Using}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{the}\mathrm{angular}\mathrm{speed}\mathrm{of}\mathrm{hour}\mathrm{hand}\mathrm{is}\mathrm{calculated}\mathrm{as}: \\ \mathrm{Ó\neg }=\frac{2\mathrm{Ï€°ù²¹»å}}{12\mathrm{h}×\left({\displaystyle \frac{3600\mathrm{s}}{1\mathrm{h}}}\right)} \\ =1.45×{10}^{-4}\mathrm{rad}/\mathrm{s} \\ \mathrm{Thus},\mathrm{the}\mathrm{angular}\mathrm{speed}\mathrm{of}\mathrm{the}\mathrm{hour}\mathrm{hand}\mathrm{is}1.45×{10}^{-4}\mathrm{rad}/s. \end{gathered}$$