Question

A wheel of moment of inertia $0.10{kg-m}^2$ is rotating about a shaft at an angular speed of $160rev/minute$. A second wheel is set into rotation at 300 rev/minute and is coupled to the same shaft so that both the wheels finally rotate with a common angular speed of $200rev/minute$. Find the moment of inertia of the second wheel.

Answer 1

Wheel $\left(1\right)$ has
$I_1=0.10kg-m^2,{\omega }_1=160rev/min$
Wheel $\left(2\right)$ has
$I_2?,{\omega }_2=300rev/min$
Given that after they are coupled $\omega =200rev/min$
Therefore if we take the two wheels to bean isolated system
Total external torque $=0$
Therefore $I-1{\omega }_1+I_1{\omega }_2=(I_1+I_1)\omega$
$\Rightarrow 0.10\times 160+I_2\times 300=(0.10+I_2)\times 200$
$\Rightarrow 5I_2=1-0.8\Rightarrow I_2=0.40kg-m^2$