Question

The second-hand of a tower clock is $1.0m$ long. Which of the following is the CORRECT statement.
A. Linear velocity of the tip of the hand is $\dfrac{{10\pi }}{3}\,cm\,{\sec ^{ - 1}}$.
B. Angular velocity of the hand is $\dfrac{\pi }{{30}}\,rad\,{\sec ^{ - 1}}$.
C. Average acceleration of the tip of the hand in $15$ minute is zero.
D. Instantaneous acceleration of the tip of the hand is $\dfrac{{{\pi ^2}}}{9}\,cm\,{\sec ^{ - 2}}$.

Answer 1

In a clock, there is one second-hand which measures each second and then a minute-hand which measures minute and later a smallest one hour-hand which measures hour. We will use the concept of second-hand which completes a full angle of $2\pi$ radians in one minute or we can say it covers $2\pi$ radians in $60$ seconds.

Formula used:
Angular velocity is given by $\omega = \dfrac{{Angl{e_{total}}}}{{Tim{e_{taken}}}}$
Linear velocity is given by $v = r\omega$ where $r$ is radius of circular path.
Average acceleration is given by ${a_{average}} = \dfrac{{\Delta v}}{t}$
Instantaneous acceleration is given by $a = {\omega ^2}r$

Complete step by step answer:
Let us examine each option one by one. Since, the total angle covered by second-hand is $2\pi$ radians and time taken is a total $60$ seconds.So, using formula of angular velocity we have,
$\omega = \dfrac{{Angl{e_{total}}}}{{Tim{e_{taken}}}}$
$\Rightarrow \omega = \dfrac{{2\pi }}{{60}}$
$\Rightarrow \omega = \dfrac{\pi }{{30}}\,rad\,{\sec ^{ - 1}}$
So, option (B) is correct.

Now, using formula of linear velocity we have, $v = r\omega$ where $r$ is length of second-hand and it’s given as $r = 1m = 100cm$ and we find $\omega = \dfrac{\pi }{{30}}rad{\sec ^{ - 1}}$ so,
$v = r\omega$
$\Rightarrow v = 100 \times \dfrac{\pi }{{30}}$
$\Rightarrow v = \dfrac{{10\pi }}{3}\,cm\,{\sec ^{ - 1}}$
So, option (A) is correct.

Now, using formula of average acceleration ${a_{average}} = \dfrac{{\Delta v}}{t}$ we have, at $t = 15\min$ since linear velocity is independent of time and has a constant magnitude so, $\Delta v = 0$
$\Rightarrow {a_{average}} = 0$
So, option (C) is correct.

Now, using the formula of instantaneous acceleration we have, $a = {\omega ^2}r$ where,
$\omega = \dfrac{\pi }{{30}}rad{\sec ^{ - 1}}$ And $r = 1m = 100cm$ we get,
$\Rightarrow a = \dfrac{\pi }{{30}} \times \dfrac{\pi }{{30}} \times 100$
$\therefore a = \dfrac{{{\pi ^2}}}{9}\,cm\,{\sec ^{ - 2}}$
So, option (D) is correct

Hence, the correct options are A, B,C and D.

Note: It should be remembered that, all the second-hand minute-hand and hour-hands of the clock rotate in circular direction and the basic unit of conversions are $1m = 100cm$ a complete angle in two dimensional Cartesian plane is of ${360^0}$ which thus in units of radian is written as $2\pi radians$ Also, minute hand covers an angle of $2\pi radians$ in $3600$ seconds.