Answer
(a) t = 4.23 s
(b) $\alpha = -78.1~rad/s^2$
(c) The motor shaft turns through 93.3 revolutions.
(d) $\omega = 250~rad/s$
(e) $\omega_{ave} = 139~rad/s$
Work Step by Step
(a) $\theta (t) = (250~rad/s)~t-(20.0~rad/s^2)~t^2-(1.50~rad/s^3)~t^3$
$\omega (t) = \frac{d\theta}{dt}$
$\omega (t) = (250~rad/s)-(40.0~rad/s^2)~t-(4.50~rad/s^3)~t^2$
We can find $t$ when $\omega (t) = 0$
$\omega (t) = 0$
$(4.50~rad/s^3)~t^2+(40.0~rad/s^2)~t -(250~rad/s) = 0$
We can use the quadratic formula to find $t$.
$t = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$t = \frac{-(40.0)\pm \sqrt{(40.0)^2-(4)(4.50)(-250)}}{(2)(4.50)}$
$t = -13.1~s, 4.23~s$
Since the negative value is unphysical, the solution is $t = 4.23~s$.
(b) $\alpha (t) = \frac{d\omega}{dt}$
$\alpha (t) = -(40.0~rad/s^2)-(9.00~rad/s^3)~t$
We can find $\alpha$ when t = 4.23 s:
$\alpha = -(40.0~rad/s^2)-(9.00~rad/s^3)(4.23~s)$
$\alpha = -78.1~rad/s^2$
(c) We can find $\theta$ when t = 4.23 s:
$\theta = (250~rad/s)(4.23~s)-(20.0~rad/s^2)(4.23~s)^2-(1.50~rad/s^3)(4.23~s)^3$
$\theta = 586.11~rad$
We can find the number of revolutions.
$rev = \frac{\theta}{2 \pi}$
$rev = \frac{586.11~rad}{2 \pi}$
$rev = 93.3$
The motor shaft turns through 93.3 revolutions.
(d) We can find $\omega$ when $t = 0$:
$\omega = (250~rad/s)-(40.0~rad/s^2)(0)-(4.50~rad/s^3)(0)^2$
$\omega = 250~rad/s$
(e) We can find the average angular velocity between $t = 0$ and $t = 4.23~s$:
$\omega_{ave} = \frac{\theta_2-\theta_1}{t}$
$\omega_{ave} = \frac{586.11~rad-0}{4.23~s}$
$\omega_{ave} = 139~rad/s$
