Question

The length of second's hand in a watch is 1 cm. The change in velocity of its tip in 15 seconds is?
A- zero
B- \[\dfrac{\pi }{30\sqrt{2}}\]cm/s
C- \[\dfrac{\pi }{30}\] cm/s
D- \[\dfrac{\pi \sqrt{2}}{30}\]cm/s

Answer 1

Hint: we know that the minute hand, the hour hand and the second’s hand of the clock moves around a complete a \[2\pi \] rotation. We do not know anything about translation because the hands are in the rotational motion. So, we can use the relationship between rotation and translation for velocity to solve this problem.

Complete step by step answer:
The length given is 1 cm so it will be the radius. The hand completes \[2\pi \]radians in 60 seconds.
\[\therefore \omega =2\pi \times \dfrac{1}{60}=\dfrac{\pi }{30}rad/s\]
Let us assume that initially the hand was at 12 in the watch. So, it points in a positive x direction, after 12 s the hand rotates and covers some distance. The angle covered by it is given as:
\[\Rightarrow \dfrac{2\pi }{60}\times 15=\dfrac{\pi }{2}rad\]
While covering this much angle, the direction of the velocity changes towards the negative direction of Y axis.
We know the relationship between linear velocity and angular velocity is given by: \[v=r\omega \]
$ \Rightarrow v=1\times \dfrac{\pi }{30} \\ $
$ \therefore v=\dfrac{\pi }{30}cm/s \\ $
Initial velocity= \[\overrightarrow{v}\widehat{i}\]
Final velocity= \[-\overrightarrow{v}\widehat{j}\]
Change in velocity= \[-\overrightarrow{v}\widehat{j}-\overrightarrow{v}\widehat{i}\]
Magnitude= \[\left| -\overrightarrow{v}\widehat{j}-\overrightarrow{v}\widehat{i} \right|\]
$ \Rightarrow \sqrt{{{v}^{2}}+{{v}^{2}}} \\
 \Rightarrow \sqrt{2{{v}^{2}}} \\
 \Rightarrow v\sqrt{2} \\
 \therefore \dfrac{\sqrt{2}\pi }{30}cm/s \\ $

So, the correct answer is “Option D”.

Note:
Here we have used the relationship between the linear velocity and the angular velocity. The length of the hand was used in cm because the options are given in cm/s otherwise we would have converted it into metres. The angle is always measured in radians and not in degrees. To find the magnitude of any vector we take its dot product with itself.