Answer
Since the final kinetic energy is less than the initial kinetic energy, the mechanical energy is not conserved as the cockroach stops.
Work Step by Step
We can find the rotational inertia of the cockroach:
$I_c = m~r^2$
$I_c = (0.17~kg)(0.15~m)^2$
$I_c = 0.003825~kg~m^2$
We can find the initial angular velocity of the cockroach:
$\omega_c = \frac{v}{r}$
$\omega_c = \frac{2.0~m/s}{0.15~m}$
$\omega_c = 13.33~rad/s$
Let $I_0$ be the rotational inertia of the lazy Susan.
We can find the initial kinetic energy of the system:
$K_i = \frac{1}{2}I_c~\omega_c^2+\frac{1}{2}I_0~\omega_0^2$
$K_i = \frac{1}{2}(0.003825~kg~m^2)(13.33~rad/s)^2+\frac{1}{2}(5.0\times 10^{-3}~kg~m^2)(-2.8~rad/s)^2$
$K_i = 0.36~J$
In part (a), we found that the angular speed after the cockroach stops is $~~\omega_f = 4.2~rad/s$
We can find the final kinetic energy of the system:
$K_f = \frac{1}{2}I_f~\omega_f^2$
$K_f = \frac{1}{2}(0.003825~kg~m^2+5.0\times 10^{-3}~kg~m^2)(4.2~rad/s)^2$
$K_f = 0.16~J$
Since the final kinetic energy is less than the initial kinetic energy, the mechanical energy is not conserved as the cockroach stops.
