Question

Calculate the angular velocity (a) of a clock’s second hand, (b) its minute hand, and (c) its hour hand. State in $\mathbf{r}\mathbf{a}\mathbf{d}/\phantom{\mathbf{r}\mathbf{a}\mathbf{d}\mathbf{s}}\mathbf{s}$. (d) What is the angular acceleration in each case?

Short answer

(a) The angular velocity of the second hand is about $0.105\mathrm{r}\mathrm{a}\mathrm{d}/\phantom{\mathrm{r}\mathrm{a}\mathrm{d}\mathrm{s}}\mathrm{s}$.

(b) The angular velocity of the minute hand is about $1.75\times {10}^{-3}\mathrm{r}\mathrm{a}\mathrm{d}/\phantom{\mathrm{r}\mathrm{a}\mathrm{d}\mathrm{s}}\mathrm{s}$.

(c) The angular velocity of the hour hand is about $1.45\times {10}^{-4}\mathrm{r}\mathrm{a}\mathrm{d}/\phantom{\mathrm{r}\mathrm{a}\mathrm{d}\mathrm{s}}\mathrm{s}$.

(d) The angular acceleration in each case is zero.

Step-by-step solution

Determination of angular velocity

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

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

Here,$\mathrm{\Delta }\theta$is the angular displacement, and$\mathrm{\Delta }t$is the time interval.

Calculate the angular velocity of the second hand

(a)

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

$\begin{array}{r}l{\omega }_{\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{d}}=\frac{\left(2\pi \mathrm{r}\mathrm{a}\mathrm{d}\right)}{\left(1min\right)\left(\frac{60\mathrm{s}}{1min}\right)}\\ {\omega }_{\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{d}}=0.105\mathrm{r}\mathrm{a}\mathrm{d}/\phantom{\mathrm{r}\mathrm{a}\mathrm{d}\mathrm{s}}\mathrm{s}\end{array}$

Thus, the angular velocity of the second hand is about $0.105\mathrm{r}\mathrm{a}\mathrm{d}/\phantom{\mathrm{r}\mathrm{a}\mathrm{d}\mathrm{s}}\mathrm{s}$.

Calculate the angular velocity of the minute hand

(b)

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

$\begin{array}{r}l{\omega }_{\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{u}\mathrm{t}\mathrm{e}}=\frac{\left(2\pi \mathrm{r}\mathrm{a}\mathrm{d}\right)}{\left(1\mathrm{h}\right)\left(\frac{3600\mathrm{s}}{1\mathrm{h}}\right)}\\ {\omega }_{\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{u}\mathrm{t}\mathrm{e}}=1.75\times {10}^{-3}\mathrm{r}\mathrm{a}\mathrm{d}/\phantom{\mathrm{r}\mathrm{a}\mathrm{d}\mathrm{s}}\mathrm{s}\end{array}$

Thus, the angular velocity of the minute hand is about $1.75\times {10}^{-3}\mathrm{r}\mathrm{a}\mathrm{d}/\phantom{\mathrm{r}\mathrm{a}\mathrm{d}\mathrm{s}}\mathrm{s}$.

Calculate the angular velocity of the hour hand

(c)

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

$\begin{array}{r}l{\omega }_{\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}}=\frac{\left(2\pi \mathrm{r}\mathrm{a}\mathrm{d}\right)}{\left(12\mathrm{h}\right)\left(\frac{3600\mathrm{s}}{1\mathrm{h}}\right)}\\ {\omega }_{\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}}=1.45\times {10}^{-4}\mathrm{r}\mathrm{a}\mathrm{d}/\phantom{\mathrm{r}\mathrm{a}\mathrm{d}\mathrm{s}}\mathrm{s}\end{array}$

Thus, the angular velocity of the hour hand is about $1.45\times {10}^{-4}\mathrm{r}\mathrm{a}\mathrm{d}/\phantom{\mathrm{r}\mathrm{a}\mathrm{d}\mathrm{s}}\mathrm{s}$.

Calculate the angular acceleration in each case

(d)

The expression for the angular acceleration is as follows:

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

Here,$\mathrm{\Delta }\omega$is the change in the angular velocity, and$\mathrm{\Delta }t$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 $\mathrm{\Delta }{\omega }_{\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{d}}=\mathrm{\Delta }{\omega }_{\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{u}\mathrm{t}\mathrm{e}}=\mathrm{\Delta }{\omega }_{\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}}=0$. Use these values in equation (ii) to find the angular acceleration in each case.

$\begin{array}{r}c{\alpha }_{\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{d}}={\alpha }_{\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{u}\mathrm{t}\mathrm{e}}={\alpha }_{\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}}=\frac{\left(0\right)}{\mathrm{\Delta }t}\\ {\alpha }_{\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{d}}={\alpha }_{\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{u}\mathrm{t}\mathrm{e}}={\alpha }_{\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}}=0\end{array}$

Thus, the angular acceleration in each case is zero.