Answer
a) .25 rads/s
b) 6.5 kJ
Work Step by Step
a) Using conservation of angular momentum, we find:
$I_0\omega_0=I_f\omega_f \\ ((4,800)(8.5^2)+\frac{1}{2}(15,000)(8.5)^2)(.15)=(\frac{1}{2}(15,000)(8.5)^2)(\omega)\\ \omega=.25 \ rads/s$
b) We know that work is equal to the change in kinetic energy, so it follows:
$W=\frac{1}{2}I\omega_f^2-\frac{1}{2}I\omega_i^2$
$W=\frac{1}{2}(\frac{1}{2}(15,000)(8.5)^2)(.25)^2-\frac{1}{2}((4,800)(8.5^2)+\frac{1}{2}(15,000)(8.5)^2)(.15)^2 \approx 6.5 \ kJ$
