Question

The second’s hand of a watch has $6cm$ length. The speed of its tip and magnitude of difference in velocity of it at any two perpendicular positions will be
$A)\,2\pi \text{ and }0\text{ }mm/s$
$B)\,2\sqrt{2}\pi \text{ and 4}\text{.44 }mm/s$
$C)\,2\sqrt{2}\pi \text{ and 2}\pi \text{ }mm/s$
$D)\,2\pi \text{ and 2}\sqrt{2}\pi \text{ }mm/s$

Answer 1

Hint: This problem can be solved by using the formula for the speed of a body in circular motion in terms of its time period and radius of circle. The difference in the velocity can be calculated by using the formula for the magnitude of the resultant of two vectors.

Formula used:
$v=\dfrac{2\pi R}{T}$
$\overrightarrow{\left| R \right|}=\sqrt{{{\left| \overrightarrow{A} \right|}^{2}}+{{\left| \overrightarrow{B} \right|}^{2}}+2\left| \overrightarrow{A} \right|\left| \overrightarrow{B} \right|\cos \theta }$

Complete step by step answer:
We will use the direct formula for the speed of a body in uniform circular motion and for the second part of the problem, we will use the direct formula for the magnitude of two resultant vectors.
The speed $v$ of a body going around a circle of radius $R$ in uniform circular motion with time period $T$ is given by
$v=\dfrac{2\pi R}{T}$ --(1)
Now, let us analyze the question.
The length of the second’s watch will essentially be the radius of the circle for its tip. Therefore, radius of the circle $R=6cm=6\times {{10}^{-2}}m$ $\left( \because 1cm={{10}^{-2}}m \right)$
The time period for the motion will be $T=60s$, since the second’s hand completes one rotation in one minute, that is sixty seconds.
Let the required speed of its tip be $v$.
Therefore, using (1), we get
$v=\dfrac{2\pi R}{T}$
$\therefore v=\dfrac{2\pi \times 6\times {{10}^{-2}}}{60}=2\pi \times {{10}^{-3}}m/s=2\pi \text{ }mm/s$ $\left( \because {{10}^{-3}}m/s=1mm/s \right)$ --(2)
Now, we will find the magnitude of the difference in velocities of the tip at two perpendicular positions. When the tip is at two perpendicular positions, the velocities of the tip will also be at two perpendicular directions.
Let these two mutually perpendicular velocity vectors be represented as $\overrightarrow{{{v}_{A}}}$ and $\overrightarrow{{{v}_{B}}}$.
Now, the magnitudes of these two velocities will be the same and equal to the speed of the tip which remains constant throughout the full circular motion.
$\therefore \left| \overrightarrow{{{v}_{A}}} \right|=\left| \overrightarrow{{{v}_{B}}} \right|=v=2\pi \text{ }mm/s$ --(3) [Using (2)]
Let the resultant of $\overrightarrow{{{v}_{A}}}$ and $\overrightarrow{{{v}_{B}}}$ be $\overrightarrow{{{v}_{R}}}$
Now, the magnitude $\left| \overrightarrow{R} \right|$ of the resultant $\overrightarrow{R}$ of the difference of two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ is given by
$\overrightarrow{\left| R \right|}=\sqrt{{{\left| \overrightarrow{A} \right|}^{2}}+{{\left| \overrightarrow{B} \right|}^{2}}-2\left| \overrightarrow{A} \right|\left| \overrightarrow{B} \right|\cos \theta }$ --(4)
Where \[\left| \overrightarrow{A} \right|,\left| \overrightarrow{B} \right|\] are the magnitudes of $\overrightarrow{A}$ and $\overrightarrow{B}$ respectively and $\theta $ is the angle between $\overrightarrow{A}$ and $\overrightarrow{B}$.
Now, using (4), we get
$\overrightarrow{\left| {{v}_{R}} \right|}=\sqrt{{{\left| \overrightarrow{{{v}_{A}}} \right|}^{2}}+{{\left| \overrightarrow{{{v}_{B}}} \right|}^{2}}-2\left| \overrightarrow{{{v}_{A}}} \right|\left| \overrightarrow{{{v}_{B}}} \right|\cos {{90}^{0}}}$ ($\theta ={{90}^{0}}$, since $\overrightarrow{{{v}_{A}}}$ and $\overrightarrow{{{v}_{B}}}$ are perpendicular to each other)
Using (3) in the above equation, we get
\[\overrightarrow{\left| {{v}_{R}} \right|}=\sqrt{{{v}^{2}}+{{v}^{2}}-2vv\left( 0 \right)}\] $\left( \because \cos {{90}^{0}}=0 \right)$
$\therefore \overrightarrow{\left| {{v}_{R}} \right|}=\sqrt{{{v}^{2}}+{{v}^{2}}-0}=\sqrt{2{{v}^{2}}}=\sqrt{2}v$
Using (2) in the above equation, we get
$\therefore \overrightarrow{\left| {{v}_{R}} \right|}=\sqrt{2}\times 2\pi =2\sqrt{2}\pi \text{ }mm/s$
Hence, the required magnitude of the difference in velocity is $2\sqrt{2}\pi \text{ }mm/s$.
Therefore, the correct answer is $D)\,2\pi \text{ and 2}\sqrt{2}\pi \text{ }mm/s$.

Note: A common mistake that students may make in this problem is that they write the difference in the velocities of the tip at two perpendicular directions as zero because they see that the magnitude of the velocity, that is, the speed is constant regardless of the position of the tip. However, they must realize that velocity is a vector quantity and the magnitude along with the direction has to be taken into account and the formula for the vector resultant has to be used.