A 10.9 kg flywheel with a radius of 0.200m spins on a shaft at 10.0 rad/s....

A 10.9 kg flywheel with a radius of 0.200m spins on a shaft at 10.0 rad/s. A second 2.00 kg flywheel with a radius of 0.100 m initially at rest is on the same shaft. Assume the mass us distributed evenly throughout each fly wheel.

A) what is the moment of inertia of each flywheel ?

B) if the second flywheel is dropped onto the spinning first flywheel what is the new angular velocity of the combined system?

C) how much kinetic energy is lost when the second flywheel is dropped on the disk?

Expert Solution

A) moment of inertia of first wheel = 1/2 * 10.9 * 0.2²

= 0.218 kg.m²

moment of inertia of second wheel = 1/2 * 2.00 * 0.100²

= 0.01 kg.m²

B) Here, applying conservation of angular momentum

=> 0.218 * 10.0 = (0.218 + 0.01) * w

=> w = 9.56 rad/s

Thus, the new angular velocity of the combined system = 9.56 rad/s

C) kinetic energy is lost = 1/2 * 0.218 * 10² - 1/2 * (0.218 + 0.01) * 9.56²

= 0.481 J

genius_generous answered 2 years ago

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