Question

The angular displacement of a particle is given by $\theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2$, where ${\omega }_0$ and $\alpha$ are constant and ${\omega }_0=1\text{rad/sec},\alpha =1.5{\text{rad/sec}}^2$. The angular velocity $($in $\text{rad/s})$ at time, $t=2\text{sec}$ will be

• A. $1$
• B. $5$
• C. $3$
• D. $4$

Answer 1

The correct option is D $4$

We have $\theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2$, we know that angular velocity is the rate of change of angular position of a rotating body.

$\Rightarrow \frac{d\theta }{dt}={\omega }_0+\alpha t$
This is angular velocity at time $t$. Now angular velocity at $t=2\text{sec}$ will be
$\omega ={\left(\frac{d\theta }{dt}\right)}_{t=2\text{sec}}={\omega }_0+2\alpha$

it is given that, ${\omega }_0=1\text{rad/sec},\alpha =1.5{\text{rad/sec}}^2$

$\omega =1+2\times 1.5=4\text{rad/sec}$

Hence correct answer is (D)