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Question
A boy is seated in a revolving chair revolving at an angular speed of 120 revolutions per minute. Two heavy balls form part of the revolving system and the boy can pull the balls closer to himself or may push them apart. If by pulling the balls closer, the boy decreases the moment of inertia of the system from 6 kg-m2 to 2 kg-m2, what will be the new angular speed?
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Solution
Given
Initial angular speed of the system,
\[\omega_1 = 120\text{ rpm }= 120 \times \left( \frac{2\pi}{60} \right) = 4\pi\text{ rad/s}\]
Initial moment of inertia of the system,
\[I_1 = 6 kg - m^2\]
Final moment of inertia of the system,
\[I_2 = 2 kg - m^2\]
Two balls are inside the system; therefore, we get
Total external torque = 0
\[I_2 = 2 kg - m^2\]
\[\therefore I_1 \omega_1 = I_2 \omega_2\]
\[\Rightarrow 6 \times 4\pi = 2 \omega_2 \]
\[ \Rightarrow \omega_2 = 12\pi rad/s\]
\[\text{Or, }\omega_2 = 6\text{ rev/s }= 360\text{ rev/minute}\]
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