Question

In a situation shown below, if the length of seconds hand is $\frac{30}{\pi }\text{cm}$, find the magnitude of change in velocity of the tip of the seconds hand.

• A. $1.732\text{cm/s}$
• B. $2\text{cm/s}$
• C. $1.414\text{cm/s}$
• D. $0\text{cm/s}$

Answer 1

The correct option is B $2\text{cm/s}$


For one complete revolution of seconds hand, $\theta =2\pi \text{radians}$.
$\therefore$ It is clear the $\theta =\pi$ radians between $t=15\text{s}$ and $t=45\text{s}$
To find the magnitude of change in velocity:
$|v_2-v_1|=\sqrt{v_2^2+v_1^2-2v_2{v}_1\cos \theta }$
For uniform circular motion, $|v_2|=|v_1|=v$
$\Rightarrow |v_2-v_1|=\sqrt{v^2+v^2-2v^2\cos \pi }=2v$
$\omega =\frac{\Delta \theta }{\Delta t}=\frac{\pi }{30}\text{rad/s}$
$\Rightarrow v=rw=\frac{30}{\pi }\times \frac{\pi }{30}\text{cm/s}=1\text{cm/s}$
$\Rightarrow |v_2-v_1|=2v=2\text{cm/s}$