Answer
(a) $0.52\frac{rad}{s}$
(b ) $370J, 2.0\times 10^2J$
Work Step by Step
(a)According to law of conservation of angular momentum
$L_i=L_f$
$\implies I{\omega}+I{\omega}=I{\omega}+mr^2{\omega}$
We plug in the known values to obtain:
$820(0.95)+0=820{\omega}+75(3)^2\omega$
$779=1495\omega$
$\omega=\frac{779}{1495}=0.52\frac{rad}{s}$
(b) Initial Rotational Kinetic energy of the system of platform plus person:
$\frac{1}{2}I\omega^2=\frac{1}{2}(820)(0.95)^2=370J$
Final Rotational Kinetic energy of the system of platform plus person:
$\frac{1}{2}(820)(0.52)^2+\frac{1}{2}(75)(3)^2(0.52)^2=2.0\times 10^2J$