Question

A particle is moving in a circular path with velocity varying with time as $v=1.5t^2+2t$. If 2cm is the radius of circular path, the angular acceleration at $t=2s$ will be

• A. $0.4rad/s^2$
• B. $40rad/s^2$
• C. $400rad/s^2$
• D. $4rad/s^2$

Answer 1

The correct option is C $400rad/s^2$

Given:

$r=2cm=0.02m,t=2s$

and $v=1.5t^2+2t$

Since, the tangential acceleration is given by,

$a_t=\frac{dv}{dt}=\alpha r$

From the given equation,

$a_t=\frac{dv}{dt}=3t+2$

$\alpha =\frac{a_t}{r}=\frac{3t+2}{0.02}$

$\alpha =\frac{3t+2}{0.02}=\frac{3\left(2\right)+2}{0.02}=400rad/s$

Final answer (c)