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 angular acceleration at the moment when the body stops is $-x\sqrt{3}$. Find $x$.

Answer 1

$\theta =6t-2t^3$

Angular velocity, $\dot{\theta }=\frac{d\theta }{dt}=6-6t^2$

Angular acceleration, $\ddot{\theta }=\frac{d^2\theta }{d{t}^2}=-12t$

Now,

$-12t=-x\sqrt{3}$

$⟹x=\frac{12t}{\sqrt{3}}=4\sqrt{3}t$