Question

A solid body rotates about a stationary axis according to the law $\theta =6t-2t^3$. Here $\theta$ is in radian
and $t$ in seconds.
Hint: If $Y=y\left(t\right)$, then mean/average value of $y$ between $t_1$, and $t_2$ is $<Y>\frac{{\int }_{t_1}^{t_2}Y\left(t\right)dt}{t_2-t_1}$. The mean values of angular acceleration averaged over the time interval between $t=0$ and the complete stop is $-x$. Find $x$.

Answer 1

$\omega =\frac{d\theta }{dt}=6-6t^2$
$\alpha =\frac{d\omega }{dt}=-12t$
$\omega =0$
at $t=1s$
a) $(\omega {)}_{0-1}=\frac{{\int }_0^1\omega dt}{1}={\int }_0^1(6-6t^2)dt$
$=4rad/s$
$(\alpha {)}_{0-1}=\frac{{\int }_0^1(-12t)dt}{1}$
$=-6rad/s^2$