Question

In a situation shown in figure, if the length of the minute's hand is $\frac{10}{\pi }\text{cm}$, find the magnitude of change in velocity of the tip of the minute's hand.

• A. $\frac{1}{\sqrt{3}}\text{cm/min}$
• B. $\sqrt{3}\text{cm/min}$
• C. $\sqrt{2}\text{cm/min}$
• D. $\frac{1}{\sqrt{2}}\text{cm/min}$

Answer 1

The correct option is A $\frac{1}{\sqrt{3}}\text{cm/min}$

For one complete revolution of minute hand, $\theta =2\pi \text{radians}$
$\therefore$ For $20$ minutes, $\theta =\frac{2\pi }{60}\times 20=\frac{2\pi }{3}\text{radians}$
Magnitude of change in velocity
$|\overrightarrow{v_2}-\overrightarrow{v_1}|=\sqrt{v_2^2+v_1^2-2v_2{v}_1\cos \theta }$
For uniform circular motion, $v_1=v_2=v$
$|\overrightarrow{v_2}-\overrightarrow{v_1}|=\sqrt{v^2+v^2-2v^2\cos \frac{2\pi }{3}}=\sqrt{3}v=\sqrt{3}r\omega$
($\because v=r\omega$)

To find $\omega$:
$\omega =\frac{\Delta \theta }{\Delta t}=\frac{\frac{2\pi }{3}}{20}=\frac{\pi }{30}\text{rad/min}$.
$\therefore |\overrightarrow{v_2}-\overrightarrow{v_1}|=\sqrt{3}\times \frac{10}{\pi }\times \frac{\pi }{30}=\frac{1}{\sqrt{3}}\text{cm/min}$.