Sum

A diver having a moment of inertia of 6⋅0 kg-m^{2} 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-m^{2}, what will be the new angular speed?

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#### Solution

Initial moment of inertia of diver, I_{1} = 6 kg-m^{2}

Initial angular velocity of diver, ω_{1} = 2 rad/s

Final moment of inertia of diver, I_{2} = 5 kg-m^{2}

Let ω_{2} be the final angular velocity of the diver.

We have

External torque = 0

∴ I_{1}ω_{1} = I_{2}ω_{2}

\[\Rightarrow \omega_2 = \frac{6 \times 2}{5} = 2.4 \text{ rad/s}\]

Concept: Moment of Inertia

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