Answer
(a) $\omega_2 = 1.38~rad/s$
(b) $K_1 = 1080~J$
$K_2 = 495~J$
These two kinetic energies are not equal because it required work to make the disk and parachutist rotate with the same angular speed.
Work Step by Step
(a) We can use conservation of angular momentum to solve this question. Let $M$ be the mass of the disk and let $m$ be the mass of the parachutist.
$L_2 = L_1$
$I_2\omega_2 = I_1\omega_1$
$(\frac{1}{2}MR^2+mR^2)\omega_2 = \frac{1}{2}MR^2\omega_1$
$(M+2m)\omega_2 = M\omega_1$
$\omega_2 = \frac{M\omega_1}{M+2m}$
$\omega_2 = \frac{(120~kg)(3.00~rad/s)}{(120~kg)+(2)(70.0~kg)}$
$\omega_2 = 1.38~rad/s$
(b) We can find the initial kinetic energy.
$K_1 = \frac{1}{2}I_1\omega_1^2$
$K_1 = \frac{1}{2}(\frac{1}{2}MR^2)\omega_1^2$
$K_1 = \frac{1}{4}(120~kg)(2.00~m)^2(3.00~rad/s)^2$
$K_1 = 1080~J$
We can find the final kinetic energy.
$K_2 = \frac{1}{2}I_2\omega_2^2$
$K_2 = \frac{1}{2}(\frac{1}{2}MR^2+mR^2)\omega_2^2$
$K_2 = \frac{1}{2}[(\frac{1}{2})(120~kg)(2.00~m)^2+(70.0~kg)(2.00~m)^2](1.38~rad/s)^2$
$K_2 = 495~J$
These two kinetic energies are not equal because it required work to make the disk and parachutist rotate with the same angular speed.
