Calculate the angular velocity (a) of a clock’s second hand, (b) its minute hand, and (c) its hour hand. State in \({{{\bf{rad}}} \mathord{\left/{\vphantom {{{\bf{rad}}} {\bf{s}}}} \right.} {\bf{s}}}\). (d) What is the angular acceleration in each case?

The angular velocity may describe as angular displacement divided by a corresponding time interval. The expression for the angular velocity is as follows:

\(\omega = \frac{{\Delta \theta }}{{\Delta t}}\) … (i)

Here,\(\Delta \theta \)is the angular displacement, and\(\Delta t\)is the time interval.

(a)

The second hand makes one complete revolution in one minute. One revolution is equal to (\(2\pi \;{\rm{rad}}\), and one minute is equal to (\(60\;{\rm{s}}\). Thus, substitute (\(2\pi \;{\rm{rad}}\)for (\(\Delta \theta \)and (\(60\;{\rm{s}}\)for (\(\Delta t\)in equation (i).

\(\begin{aligned}{l}{\omega _{{\rm{second}}}} = \frac{{\left( {2\pi \;{\rm{rad}}} \right)}}{{\left( {1\;\min }\right)\left( {\frac{{60\;{\rm{s}}}}{{1\;\min }}} \right)}}\\{\omega _{{\rm{second}}}} = 0.105\;{{{\rm{rad}}}\mathord{\left/{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}}\end{aligned}\)

Thus, the angular velocity of the second hand is about \(0.105\;{{{\rm{rad}}} \mathord{\left/ {\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}}\).

(b)

The minute hand makes one complete revolution in an hour. One revolution is equal to (\(2\pi \;{\rm{rad}}\), and one hour is equal to (\(3600\;{\rm{s}}\). Thus, substitute (\(2\pi \;{\rm{rad}}\) for (\(\Delta \theta \) and (\(3600\;{\rm{s}}\) for (\(\Delta t\) in equation (i).

\(\begin{aligned}{l}{\omega _{{\rm{minute}}}} = \frac{{\left( {2\pi \;{\rm{rad}}} \right)}}{{\left( {1\;{\rm{h}}}\right)\left( {\frac{{3600\;{\rm{s}}}}{{1\;{\rm{h}}}}} \right)}}\\{\omega _{{\rm{minute}}}} = 1.75 \times {10^{ - 3}}\;{{{\rm{rad}}} \mathord{\left/{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}}\end{aligned}\)

Thus, the angular velocity of the minute hand is about \(1.75 \times {10^{ - 3}}\;{{{\rm{rad}}} \mathord{\left/{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}}\).

(c)

The hour hand makes one complete revolution in 12 hours. One revolution is equal to\(2\pi \;{\rm{rad}}\), and 12 hours is equal to\(12 \times 3600\;{\rm{s}}\). Thus, substitute\(2\pi \;{\rm{rad}}\)for\(\Delta \theta \)and\(12 \times 3600\;{\rm{s}}\)for\(\Delta t\)in equation (i).

\(\begin{aligned}{l}{\omega _{{\rm{hour}}}} = \frac{{\left( {2\pi \;{\rm{rad}}} \right)}}{{\left( {12\;{\rm{h}}}\right)\left( {\frac{{3600\;{\rm{s}}}}{{1\;{\rm{h}}}}} \right)}}\\{\omega _{{\rm{hour}}}} = 1.45 \times {10^{ - 4}}\;{{{\rm{rad}}} \mathord{\left/ {\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}}\end{aligned}\)

Thus, the angular velocity of the hour hand is about \(1.45 \times {10^{ - 4}}\;{{{\rm{rad}}} \mathord{\left/{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}}\).

(d)

The expression for the angular acceleration is as follows:

\(\alpha = \frac{{\Delta \omega }}{{\Delta t}}\) … (ii)

Here,\(\Delta \omega \)is the change in the angular velocity, and\(\Delta t\)is the time interval.

Since the second, minute, and hour hands of the clock move with constant angular velocity, the variation in the angular velocity will be \(\Delta {\omega _{{\rm{second}}}} = \Delta {\omega_{{\rm{minute}}}} = \Delta {\omega _{{\rm{hour}}}} = 0\). Use these values in equation (ii) to find the angular acceleration in each case.

\(\begin{aligned}{c}{\alpha _{{\rm{second}}}} = {\alpha _{{\rm{minute}}}} = {\alpha _{{\rm{hour}}}} =\frac{{\left( 0 \right)}}{{\Delta t}}\\{\alpha _{{\rm{second}}}} = {\alpha _{{\rm{minute}}}} = {\alpha_{{\rm{hour}}}} = 0\end{aligned}\)

Thus, the angular acceleration in each case is zero.