A angular positio of a point on the rim of a rotating wheel is given by `theta=4t-3t^(2)+t^(3)` where `theta` is in radiuans and `t` is in seconds. What are the angualr velocities at
(a).`t=2.0` and
(b). `t=4.0s`
(c). What is the average angular acceleration for the time interval that begins at `t=2.0s` and ends at `t=4.0s`?
(d). What are the instantaneous angular acceleration at the biginning and the end of this time interval?

Explanation

Angular velocity `omega=(dtheta)/(dt)=(d)/(dt)(4t-3t^(2)+t^(3))` or `omega=4-6t+3t^(2)`
(a). At `t=2.0s,omega=4-6xx2+3(2)^(2)` or `omega=4rad//s`
(b). At `t=4.0s,omega=4-6xx5+3(4)^(2)` or `omega=28rad//s`
(c). Average angular acceleration `alpha_(av)=(omega_(f)-omega_(i))/(t_(f)-t_(i))=(28-4)/(4-2)` or `alpha_(av)=12rad//s^(2)`
(d). Instantaneous angular acceleartion is ,
`alpha=(domega)/(dt)=(d)/(dt)(4-6t+3t^(2))` or `alpha=-6+6t`
at `t=2.0s` `alpha=-6+6xx2=6rad//s^(2)`
at `t=4.0s` `alpha=-6+6xx4=18rad//s`