Answer
25%
Work Step by Step
a) We find the final angular velocity:
$\omega_f = \frac{(.044)(.035)^2}{(.044)(.035)^2+(.27)(.023)^2}\times180 = 140\ rpm$
b) We find:
$percent \ lost = 100\times\frac{k_i-k_f}{k_i}$
Use the fact that $k=\frac{1}{2}I\omega^2$, it follows:
$percent \ lost \approx \fbox{25%}$
