The angular position of a point on the rim of a rotating wheel is given by `theta=4t^(3)-2t^(2)+5t+3` rad. Find
(a) the angular velocity at `t=1 s`,
(b) the angular acceleration at `t=2 s`.
(c ) the average angular velocity in time interval `t=0` to `t=2 s` and
(d) the average angular acceleration in time interval `t=1` to `t=3 s`.

Explanation

`theta=4t^(3)-2t^(2)+5t+3`
`omega=(d theta)/(dt)=12t^(2)-4t+5`
`alpha=(domega)/(dt)=24t-4`
(a) At `t=1 s`,
`omega=12(1)^(2)-4(1)+5=13 rad//s`
(b) At `t=2 s`,
`alpha=24(2)-4=44 rad//s^(2)`
(c ) `t_(1)=0, theta_(1)=3 rad`
`t_(2)=2 s, theta_(2)=4(2)^(3)-2(2)^(2)+5(2)+3=37 rad`
`undersetomega(-)=(theta_(2)-theta_(1))/(t_(2)-t_(1))=(37-3)/(2-0)=17 rad//s`
(d) At `t_(1)=1 s`,
` omega_(1)=12(1)^(2)-4(1)+5=13 rad//s`
At `t_(2)=3 s`,
`omega_(2)=12(3)^(2)-4(3)+5=101 rad//s`
`undersetalpha(-)=(omega_(2)-omega_(1))/(t_(2)-t_(1))=(101-13)/(3-1)=44 rad//s^(2)`