Answer
(a) $\omega = 0.120~rad/s$
(b) $K = 3.20\times 10^{-4}~J$
(c) The kinetic energy comes from the work done by the bug during the jump.
Work Step by Step
(a) We can use conservation of angular momentum to solve this question. Since the initial angular momentum of the system is zero, the magnitude of the bug's angular momentum and the magnitude of the bar's angular momentum will be equal. Let $M$ be the mass of the bar and let $m$ be the mass of the bug.
$I\omega = mvr$
$\frac{1}{3}ML^2\omega = mvL$
$\omega = \frac{3mv}{ML}$
$\omega = \frac{(3)(0.0100~kg)(0.200~m/s)}{(0.0500~kg)(1.00~m)}$
$\omega = 0.120~rad/s$
(b) We can find the total kinetic energy of the system.
$K = \frac{1}{2}I\omega^2+\frac{1}{2}mv^2$
$K = \frac{1}{2}(\frac{1}{3}ML^2)\omega^2+\frac{1}{2}mv^2$
$K = \frac{1}{6}(0.0500~kg)(1.00~m)^2(0.120~rad/s)^2+\frac{1}{2}(0.0100~kg)(0.200~m/s)^2$
$K = 3.20\times 10^{-4}~J$
(c) The kinetic energy comes from the work done by the bug during the jump.
