A solid body rotates about a stationary axis accordig to the law `theta=6t-2t^(3)`. Here `theta`, is in radian and `t` in seconds. Find
(a). The mean values of thhe angular velocity and angular acceleration averaged over the time interval between `t=0` and the complete stop.
(b). The angular acceleration at the moment when the body stops.
Hint: if `y=y(t)`. then mean/average value of `y` between `t_(1)` and `t_(2)` is `ltygt=(int_(t_(1))^(t_(2))y(t)dt))/(t_(2)-t_(1))`

Explanation

Let us take the rotation axis as z-axis whose positive direction is associated with the positive direction of the coordinate `varphi`, the rotation angle, in accordance with the right-hand screw rule (figure).
(a) Defferentiating `varphi(t)` with respect to time.
`(dvarphi)/(dt)=a-3bt^2=omega_z` (1) and
`(d^2varphi)/(dt^2)=(domega_z)/(dt)=beta_z=-6bt` (2)
From (1) the solid comes to stop at `Deltat=sqrt((a)/(3b))`
The angular velocity `omega=a-3bt^2`, for `0letlesqrt(a//3b)`
So, `lt omega ge(int omegadt)/(int dt)=(underset0overset(sqrt(a//3b))int (a-3bt^2)dt)/(underset(0)overset(sqrt(a//3b))intdt)=[at-bt^3]_0^(sqrt(a//3b))//sqrt(a//3b)=2a//3`
Similarly `beta=|beta_z|=6bt` for all values of t.
So `lt beta gt =(int betadt)/(int dt)=(underset(0)overset(sqrt(a//3b))int 6btdt)/(underset(0)overset(sqrt(a//3b))intdt)=sqrt(3ab)`
(b) From Eq. (2) `beta_x=-6bt`
So, `(beta_x)_t=sqrt(a//3b)=-6bsqrt((a)/(3b))=-2sqrt(ab)`
Hence `beta=|(beta_x)_(t-sqrt(a//3b))|=2sqrt(3ab)`