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

\[I_1  \omega_1  +  I_2  \omega_2  = \left( I_1 + I_2 \right)  \omega\]

\[\Rightarrow 0 . 10   \times 160 +  I_2  \times 300 = \left( 0 . 10 + I_2 \right)   \times   200\]

\[ \Rightarrow 16 + 300 I_2  = 20 + 200 I_2 \]

\[ \Rightarrow 100 I_2  = 4\]

\[ \Rightarrow  I_2  = \frac{4}{100} = 0 . 04  kg -  m^2\]