Two fly wheels A and B are mounted side by side with frictionless bearings on a common shaft. Their moments of inertia about the shaft are 5.0 kg m2 and 20.0 kg m2 respectively. Wheel A is made to rotate at 10 revolution per second. Wheel B, initially stationary, is now coupled to A with the help of a clutch. The rotation speed of the wheels will become

By conservation of angular momentum, 5×10=5ω+20ω⇒ω=5025=2rps

Two fly wheels A and B are mounted side by side with frictionless bearings on a common shaft. Their moments of inertia about the shaft are 5.0 kg m2 and 20.0 kg m2 respectively. Wheel A is made to rotate at 10 revolution per second. Wheel B, initially stationary, is now coupled to A with the help of a clutch. The rotation speed of the wheels will become

By conservation of angular momentum, 5×10=5ω+20ω⇒ω=5025=2rps

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Two fly wheels A and B are mounted side by side with frictionless bearings on a common shaft. Their moments of inertia about the shaft are 5.0 kg m² and 20.0 kg m² respectively. Wheel A is made to rotate at 10 revolution per second. Wheel B, initially stationary, is now coupled to A with the help of a clutch. The rotation speed of the wheels will become

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$2 \sqrt{5} rps$

0.5 rps

2 rps

None of these

answer is C.

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Detailed Solution

By conservation of angular momentum, $5 \times 10 = 5 ω + 20 ω \Rightarrow ω = \frac{50}{25} = 2 rps$

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