Question

A wheel of moment of inertia 0⋅10 kg-m² is rotating about a shaft at an angular speed of 160 rev/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 200 rev/minute. Find the moment of inertia of the second wheel.

Answer 1

Given:
For the first wheel,
I₁ = 10 kg-m² and ω₁ = 160 rev/min
For the second wheel, ω₂ = 300 rev/min
Let I₂ be the moment of inertia of the second wheel.
After they are coupled, we have:
ω = 200 rev/min
If we take the two wheels to be an isolated system, we get:
Total external torque = 0
Therefore, we have:
$I_1{\omega }_1+I_2{\omega }_2=\left(I_1+I_2\right)\omega$
$$\begin{gathered} \Rightarrow 0.10\times 160+I_2\times 300=\left(0.10+I_2\right)\times 200 \\ \Rightarrow 16+300I_2=20+200I_2 \\ \Rightarrow 100I_2=4 \\ \Rightarrow I_2=\frac{4}{100}=0.04\mathrm{kg}-{\mathrm{m}}^2 \end{gathered}$$