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)
$6 - 6t^{2} = 0$
$t^{2}=\pm1$
As, $t =-1\, s$ is not possible, hence $t = 1 \,s$.
Substituting $t = 1\, s$ in Eq. (i), we get
$\theta_{1}=6-2=4$ rad
and $\theta_{0}=6\times0-2\times0^{2}\left(t=0\right)=0$
$\therefore $ Mean value of $\omega=\frac{\theta_{1}+\theta_{0}}{t_{1}+t_{0}}=\frac{4+0}{1+0}=4 rads^{-1}$