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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.

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प्रश्न

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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उत्तर

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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पाठ 10: Rotational Mechanics - Exercise [पृष्ठ १९८]

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एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
पाठ 10 Rotational Mechanics
Exercise | Q 52 | पृष्ठ १९८

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