Question

Angular displacement of any particle as given $\theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2$ where ${\omega }_0$ & $\alpha$ are constant if ${\omega }_0=1rad/sec,\alpha =1.5rad/se{c}^2$ then in $t=2sec$, angular velocity will be (in$rad/sec$)

• A. $1$
• B. $5$
• C. $3$
• D. $4$

Answer 1

The correct option is D $4$

Given that,

$\alpha =1.5rad/{\sec }^2$

${\omega }_0=1rad/\sec$

$t=2\sec$

Now, the angular displacement is

$\theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2.....\left(I\right)$

Now, the angular velocity is

$\frac{d\theta }{dt}={\omega }_0+\frac{2}{2}\alpha t$

$\frac{d\theta }{dt}=1+1.5\times 2$

$\frac{d\theta }{dt}=4rad/\sec$

Hence, the angular velocity is $4rad/sec$