Answer
(a) $\tau = 0.704~N~m$
(b) $\theta = 157~rad$
(c) $Work = 111~J$
(d) $K_{rot}= 111~J$
Work Step by Step
(a) $\tau = I\alpha$
$\tau = (\frac{1}{2}mR^2)(\frac{\omega-\omega_0}{t})$
$\tau = [\frac{1}{2}(2.80~kg)(0.100~m)^2][\frac{(20~rev/s)(2\pi~rad/rev)}{2.5~s}]$
$\tau = 0.704~N~m$
(b) $\theta = \omega_{ave}~t$
$\theta = (\frac{\omega_f+\omega_0}{2})~t$
$\theta = (\frac{20~rev/s+0}{2})(2\pi~rad/rev)~t$
$\theta = (10~rev/s)(2\pi~rad/rev)(2.5~s)$
$\theta = 157~rad$
(c) $Work = \tau~\theta$
$Work = (0.704~N~m)(157~rad)$
$Work = 111~J$
(d) $K_{rot} = \frac{1}{2}I\omega^2$
$K_{rot} = \frac{1}{2}(\frac{1}{2}mR^2)\omega^2$
$K_{rot} = \frac{1}{4}(2.80~kg)(0.100~m)^2[(20~rev/s)(2\pi~rad/rev)]^2$
$K_{rot}= 111~J$
The work done by the torque is equal to the rotational kinetic energy of the wheel.
