Answer
(a) $K = 1.33\times 10^6~J$
(b) The new angular speed is 2770 rpm.
Work Step by Step
(a) We can find the angular speed $\omega$
$\omega = (2400~rpm)(2~\pi~rad)(\frac{1~min}{60~s})$
$\omega = 80~\pi~rad/s$
We can find the rotational kinetic energy.
$K = \frac{1}{2}I\omega^2$
$K = \frac{1}{2}(\frac{1}{12}mL^2)\omega^2$
$K = \frac{1}{2}[\frac{1}{12}(117~kg)(2.08~m)^2](80~\pi~rad/s)^2$
$K = 1.33\times 10^6~J$
(b) We can find the new angular speed $\omega_2$.
$\frac{1}{2}(\frac{1}{12}(0.75~m)L^2)\omega_2^2 = \frac{1}{2}(\frac{1}{12}mL^2)\omega_1^2$
$0.75~\omega_2^2 = \omega_1^2$
$\omega_2 = \frac{\omega_1}{\sqrt{0.75}}$
$\omega_2 = \frac{2400~rpm}{\sqrt{0.75}}$
$\omega_2 = 2770~rpm$
The new angular speed is 2770 rpm.
