A child's top is spun with angular acceleration
`alpha=5t^(3)-4t`,
with t in seconds and `alpha` in radians per second-squared. At t = 0, the top has angular velocity 5 rad/s, and a reference line on it is at angular position `theta=2` rad.
(a) Obtain an expression for the angular velocity `omega(t)` of the top. That is, find an expression that explicitly indicates how the angular velocity depends on time.
(b) Obtain an expression for the angular position `theta(t)` of the top.

Explanation

(a) By definition, `alpha(t)` is the derivative of `omega(t)` with respect to time. Thus, we can find `omega(t)` by integrating `alpha(t)` with respect to time.
Calculations: `domega=alphadt`
so `intdomega=intalphadt`.
From this we find
`omega=int(5t^(3)-4t)dt=5/4t^(4)-4/2t^(2)+C`.
To evaluate the constant of integration C, we note that `omega=5rad//s` at t = 0. Substituting these values in our expression for `omega` yields
`5rad//s=0-0+C`,
so C = 5 rad/s. Then
`omega=5/4t^(4)-2t^(2)+5`.
(b) By definition, `omega(t)` is the derivative of `theta(t)` with respect to time. Therefore, we can find `theta(t)` by integrating `omega(t)` with respect to time.
Calculations: `d theta=omegadt`,
`theta=intomegadt=int(5/4t^(4)-2t^(2)+5)dt`
= `1/4t^(5)-2/3t^(3)+5t+C.`
= `1/4t^(5)-2/3t^(3)+5t+2`,
where C" has been evaluated by noting that `theta=2` rad at t = 0.