Question

A solid body rotates an angle $\theta$ about a stationary axis according to the law $\theta =6t-2t^3$ .What is the mean value of angular velocity over the time interval between $t=0$ and the time when the body comes to rest ?

• A. 1 rad/s
• B. 2 rad/s
• C. 3 rad/s
• D. 4 rad/s

Answer 1

Answer: (D)

Given, $\theta =6t-2t^3$...(i)
$\therefore \frac{d\theta }{dt}=\omega =6-6t^2$
As, $\omega =0$ (given)
$6-6t^2=0$
$t^2=\pm 1$
As, $t=-1s$ is not possible, hence $t=1s$.
Substituting $t=1s$ in Eq. (i), we get
${\theta }_1=6-2=4$ rad
and ${\theta }_0=6\times 0-2\times 0^2\left(t=0\right)=0$
$\therefore$ Mean value of $\omega =\frac{{\theta }_1+{\theta }_0}{t_1+t_0}=\frac{4+0}{1+0}=4rad{s}^{-1}$