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22. Problem 12.43 E (HRW) The angular momentum of a flywheel having a rotational inertia of 0.140 kg m2 about its axis decreases from 3.00 to 0.800 kg m2/s in 1.50 s. What is the average torque acting on the flywheel about its central axis during this period? Assuming a uniform acceleration, through what angle the flywheel will have turned? How much work was done on the wheel? What is the average power of the flywheel? |
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Solution: Click For PDF Version (a) Equation of motion for rotation gives that change in angular momentum per
second is the average torque acting on the flywheel. The change in angular
momentum of the flywheel in 1.50 s is (0.800-3.00) kg m2/s. Therefore, the
average torque acting on the flywheel during this period will be
(b) Let
For rotational motion with constant acceleration the solution of the equation of motion is
Therefore,
(c) Work done on the flywheel, W, is equal to the change in its kinetic energy during the period. Therefore,
(d) The average power delivered by the flywheel is change in kinetic energy per second; Power
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be the
average angular acceleration. The equation of motion is
,
where I is the moment of inertia and
is the
average acceleration. Therefore,
respectively.
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