Answer
(a) The mass of the rod is 36.0 grams.
(b) The mass of the bug is 3.00 grams.
Work Step by Step
(a) We can use the moment of inertia of the rod to find the mass of the rod.
$I = \frac{1}{3}ML^2 = 3.00\times 10^{-3}~kg~m^2$
$M = \frac{9.00\times 10^{-3}~kg~m^2}{L^2}$
$M = \frac{9.00\times 10^{-3}~kg~m^2}{(0.500~m)^2}$
$M = 0.0360~kg = 36.0~grams$
The mass of the rod is 36.0 grams.
(b) We can find the final angular speed.
$\omega_2 = \frac{v}{L}$
$\omega_2 = \frac{0.160~m/s}{0.500~m}$
$\omega_2 = 0.320~rad/s$
We can use conservation of angular momentum to solve this question. Let $M$ be the mass of the rod and let $m$ be the mass of the bug.
$L_2 = L_1$
$I_2\omega_2 = I_1\omega_1$
$(\frac{1}{3}ML^2+mL^2)\omega_2 = \frac{1}{3}ML^2\omega_1$
$(M+3m)\omega_2 = M\omega_1$
$m = \frac{M(\omega_1-\omega_2)}{3\omega_2}$
$m = \frac{(0.0360~kg)(0.400~rad/s-0.320~rad/s)}{(3)(0.320~rad/s)}$
$m = 0.00300~kg = 3.00~grams$
The mass of the bug is 3.00 grams.
