Question:
A wheel of moment of inertia $0.10 \mathrm{~kg}^{-m^{2}}$ is rotating about a shaft at an angular speed of 160 rev/minute. A second wheel is set into rotation at $300 \mathrm{rev} / \mathrm{minute}$ and is coupled to the same shaft so that both the wheels finally rotate with a common angular speed of $200 \mathrm{rev} / \mathrm{minute}$. Find the moment of inertia of the second wheel.
Solution:
Taking wheels as a system
$\tau_{\text {ext }}=0$
$L_{i}=L_{f}$
$I_{1} \omega_{1}=I_{2} \omega_{2}=\left(I_{1}+I_{2}\right) \omega$
$(0.1)(160)+I_{2}(300)=\left(0.1+I_{2}\right)(200)$
$I_{2}=0.04 \mathrm{~kg}^{-} m^{2}$