Question

If angular velocity of a disc depends an angle rotated $\theta$ as $\omega ={\theta }^2+2\theta$, then its angular acceleration $\alpha$ at $\theta =1\text{rad}$ is

• A. $8{\text{rad/sec}}^2$
• B. $10{\text{rad/sec}}^2$
• C. $12{\text{rad/sec}}^2$
• D. None

Answer 1

The correct option is C $12{\text{rad/sec}}^2$

It is given that,
$\omega ={\theta }^2+2\theta$
differentiating both sides w.r.t $t$
$\frac{d\omega }{dt}=2\theta \frac{d\theta }{dt}+2\frac{d\theta }{dt}$
We know that, angular acceleration is the differention of angular velocity with respect to time and angular velocity is the differential of angular displacement.

$\alpha =2\theta \omega +2\omega$

At $\theta =1\text{rad}$, $\omega =1^2+2\times 1=3\text{rad/s}$

So,
$\alpha =2\theta \omega +2\omega$
$\alpha =2\times 1\times 3+2\times 3$
$\alpha =12{\text{rad/s}}^2$

Hence option C is the correct answer.


(OR)


Angular acceleration $\alpha =\omega \frac{d\omega }{d\theta }$
Given $\omega ={\theta }^2+2\theta$
$\frac{d\omega }{d\theta }=\frac{d({\theta }^2+2\theta )}{d\theta }=2\theta +2$

Now at $\theta =1$,
$\omega {|}_{\theta =1}=1^2+2\left(1\right)=3$
$\frac{d\omega }{d\theta }{|}_{\theta =1}=2\left(1\right)+2=4$

$\alpha =3\left(4\right)=12rad/s^2$