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Question
A diver having a moment of inertia of 6⋅0 kg-m2 about an axis thorough its centre of mass rotates at an angular speed of 2 rad/s about this axis. If he folds his hands and feet to decrease the moment of inertia to 5⋅0 kg-m2, what will be the new angular speed?
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Solution
Initial moment of inertia of diver, I1 = 6 kg-m2
Initial angular velocity of diver, ω1 = 2 rad/s
Final moment of inertia of diver, I2 = 5 kg-m2
Let ω2 be the final angular velocity of the diver.
We have
External torque = 0
∴ I1ω1 = I2ω2
\[\Rightarrow \omega_2 = \frac{6 \times 2}{5} = 2.4 \text{ rad/s}\]
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