At t = 0 a grinding wheel has an angular velocity of 26.0 . It has a constant an

Question

At t = 0 a grinding wheel has an angular velocity of 26.0 . It has a constant angular acceleration of 29.0 until a circuit breaker trips at 2.30 . From then on it turns through 504 as it coasts to a stop at constant acceleration. (There are several ways to do this question, and there is no preferred order in which to do A, B and C.)
At what time (measured from t = 0) did the grinding wheel stop?
t(top)= ? (s)
What was the angular acceleration of the grinding wheel as it slowed down?

Through what total angle did the wheel turn between t = 0 and the time it stopped?

Explanation / Answer

Given that the wheel has an angular velocity _0 = 26.0 rad /s

angular acceleration = 29.0 rad /s^2

   angular displacement = 504 rad

         time t = 2.30 s

a ) angular velocity after time t is = _0 + t

                                                            = 26.0 + ( 29.0 ) * 2.3

                                                            = 92.7 rad /s

angular deceleration is _1 = ( _f ^2 -^2 ) / 2

                                             = ( 0 - ( 92.7 rad /s) / 2( 504 )

                                              = - 8.52 rad /s^2

    time t_1 = _f - _2 / _1

                     = 0 - ( 92.7 / - 8.52 )

                    = 10.8 s

b ) The angular acceleration of the wheel to slowed it down is 8.52 rad /s^2

c ) Total angle time to stopped is T = 10.8 s + 2.30 s

                                                        = 13.1 s

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