Question

If angular position of a particle $\theta$ depends on time $t$ as $\theta =2t^2+3$, find the average angular velocity at the end of $3\text{sec}$ and instantaneous angular velocity at $t=3\text{sec}$ (in $rad/s$)

• A. $6,12$
• B. $6,6$
• C. $8,12$
• D. $12,4$

Answer 1

The correct option is A $6,12$

We know that,
${\omega }_{\text{avg}}=\frac{\text{Total angular displacement}}{\text{Total time}}$
or, ${\omega }_{\text{avg}}=\frac{{\theta }_{final}-{\theta }_{initial}}{t_2-t_1}$

We have from the question
$\theta =2t^2+3$
$\Rightarrow {\theta }_{final}=2(3{)}^2+3=21\text{rad}$ and
${\theta }_{initial}=2\left(0\right)+3=3\text{rad}$

Thus, ${\omega }_{\text{avg}}=\frac{21-3}{3-0}=6\text{rad/s}$

Also, ${\omega }_{\text{instantaneous}}=\frac{d\theta }{dt}=\frac{d(2t^2+3)}{dt}=4t$
Thus, at $t=3s$,${\omega }_{\text{instantaneous}}=4\times 3=12\text{rad/sec}$.