%3Cp%3EA%20grinding%20wheel%2C%20intitially%20at%20rest%20is%20rotated%20with%20
ID: 2294309 • Letter: #
Question
%3Cp%3EA%20grinding%20wheel%2C%20intitially%20at%20rest%20is%20rotated%20with%20constant%0Aangular%20acceleration%20of%202.49%20rad%2Fs%5E2%20for%206.88%20s.%20The%20wheel%20is%20then%0Abrough%20to%20rest%20with%20uniform%20deceleration%20in%209.1%20rev.%3C%2Fp%3E%0A%3Cp%3E%3Cbr%20%2F%3E%3C%2Fp%3E%0A%3Cp%3Ea)%20Find%20the%20angular%20acceleration%20required%20to%20bring%20the%20wheel%20to%0Arest.%20Note%20than%20an%20increase%20in%20angular%20velocity%20is%20consistent%20with%0Aa%20positive%20angular%20acceleration.%20%3C%2Fp%3E%0A%3Cp%3E%3Cbr%20%2F%3E%3C%2Fp%3E%0A%3Cp%3Eb)%20Determine%20the%20time%20needed%20to%20bring%20the%20wheel%20to%0Arest.%26nbsp%3B%3C%2Fp%3E%0A%3Cp%3EAnswer%20in%20units%20of%20s.%26nbsp%3B%3C%2Fp%3E%0AExplanation / Answer
heel maximum angular velocity = (2.49)(6.88) = 17.13 rad/s
angular distance traveled during wheel's deceleration = 9.1(2?) = 57.177 rad
average angular velocity during wheel's deceleration = 17.13/2 = 8.565 rad/s
time to stop wheel from maximum angular velocity = 57.177/8.565 = 6.675s
constant angular deceleration reqd to stop wheel = 17.13/6.75 = 2.5378 rad/s
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