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in Rigid Body Dynamics by (49.6k points)
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M and N two wheels are on the same axis. The moment of inertia of M is 6 kg m2 and is rotating with 600 cycles/min and N is at rest. Joining both of them together by a clutch they do motion by 400 cycles/min. Calculate the moment of inertia of N.

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Moment of inertia of wheel M,
I1 = 6kg m2
Frequency of rotation,
n1 = 600 cycles/min.
\(\frac{600}{60}\) c.s-1 = 10 c.s-1
∴ Initial angular velocity,
ω1 = 2πn1 = 2π × 10 = 20n rad/s
Angular velocity of joint wheels,
ω2 = 2πn2
where, n2 = 400 c/min = \(\frac{400}{60}\) c/s = \(\frac{20}{3}\) c/ s
∴ ω2 = 2π × \(\frac{20}{3}\) = \(\frac{40}{3}\)π rad/s 

If the joint moment of inertia be I2, then by the law of conservation of angular momentum,
I2ω2 = I1ω1
or I2 × \(\frac{40}{3} \pi\) = 6020π
or I2 = 9 kg m2
but I2 = IM + IN
∴ 9 = 6 + IN
or IN = 9 – 6 = 3kg m2.

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