Question

The speed of an object undergoing uniform circular motion is 4 m/s. The magnitude of the change in the velocity during 0.5 second is also 4 m/s. The minimum possible centripetal acceleration (in $m/s^2$) of the object
is : [Take $\pi =\frac{24}{8}$]

• A. 8
• B. 4
• C. 16
• D. 2

Answer 1

The correct option is B 4


Magnitude of change in velocity, $|\overrightarrow{\Delta V}|=2vsin\left(\frac{\theta }{2}\right)$

$4=2\times 4sin\left(\frac{\theta }{2}\right)$

$\Rightarrow \theta =\frac{\pi }{3}$ (or) $\frac{5\pi }{3}$

For $\theta =\frac{\pi }{3},\omega$ is minimum.

$\omega =\frac{\theta }{t}=\left(\frac{\frac{\pi }{3}}{0.5}\right)=\frac{2\pi }{3}rad/s$

$\therefore$ Minimum centripetal acceleration = $\omega \times v=\frac{2\pi }{3}\times 4=8m/s$.