Question

A particle is moving with constant angular acceleration $=\alpha$ in a circular path of radius $\sqrt{3}m$. At $t=0$, it was at rest and at $t=1sec$, the magnitude of its acceleration becomes $\sqrt{6}m/s^2$, then $\alpha$ is

• A. $2rad/s^2$
• B. $\sqrt{3}rad/s^2$
• C. $\sqrt{2}rad/s^2$
• D. $1rad/s^2$

Answer 1

The correct option is C $\sqrt{2}rad/s^2$

Circular path of radius $=\sqrt{3}m$

$t=0$ & (rest)

$t=1$ & $(a=\sqrt{6}m/s^2)$

Angular acceleration $\left(\alpha \right)-$ The rate of change of angular velocity with time.

$\alpha =\frac{dw}{dt}=\frac{d}{dt}\left(v/r\right)\because v=rw$

$=\frac{1}{r}\left(\frac{dv}{dt}\right)=\frac{a}{r}$

So $\alpha =\frac{\sqrt{6}}{\sqrt{3}}=\sqrt{2}rad/s^2$

Hence the angular acceleration of particle is $\sqrt{2}rad/s^2$