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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.
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Solution
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
\[I_1 \omega_1 + I_2 \omega_2 = \left( I_1 + I_2 \right) \omega\]
\[\Rightarrow 0 . 10 \times 160 + I_2 \times 300 = \left( 0 . 10 + I_2 \right) \times 200\]
\[ \Rightarrow 16 + 300 I_2 = 20 + 200 I_2 \]
\[ \Rightarrow 100 I_2 = 4\]
\[ \Rightarrow I_2 = \frac{4}{100} = 0 . 04 kg - m^2\]
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