At a time t = 3.40 s , a point on the rim of a wheel with a Radius of 0.250 m ha

Question

At a time t = 3.40 s , a point on the rim of a wheel with a Radius of 0.250 m has a tangential speed of 55.0 m/s as the wheel slows down with a tangential acceleration of constant magnitude 10.9 m/s2 . Part A Calculate the wheel's constant angular acceleration. =
rad/s2 SubmitGive Up Part B Calculate the angular velocity at t = 3.40 s . =
rad/s SubmitGive Up Part C Calculate the angular velocity at t=0. 0 =
rad/s SubmitGive Up Part D Through what angle did the wheel turn between t=0 and t = 3.40 s ? =
rad SubmitGive Up Part E Prior to the wheel coming to rest, at what time will the radial acceleration at a point on the rim equal g = 9.81 m/s2 ? t =
At a time t = 3.40 s , a point on the rim of a wheel with a Radius of 0.250 m has a tangential speed of 55.0 m/s as the wheel slows down with a tangential acceleration of constant magnitude 10.9 m/s2 . Part A Calculate the wheel's constant angular acceleration. =
rad/s2 SubmitGive Up Part B Calculate the angular velocity at t = 3.40 s . =
rad/s SubmitGive Up Part C Calculate the angular velocity at t=0. 0 =
rad/s SubmitGive Up Part D Through what angle did the wheel turn between t=0 and t = 3.40 s ? =
rad SubmitGive Up Part E Prior to the wheel coming to rest, at what time will the radial acceleration at a point on the rim equal g = 9.81 m/s2 ? t =
At a time t = 3.40 s , a point on the rim of a wheel with a Radius of 0.250 m has a tangential speed of 55.0 m/s as the wheel slows down with a tangential acceleration of constant magnitude 10.9 m/s2 . Part A Calculate the wheel's constant angular acceleration. =
rad/s2 SubmitGive Up Part B Calculate the angular velocity at t = 3.40 s . =
rad/s SubmitGive Up Part C Calculate the angular velocity at t=0. 0 =
rad/s SubmitGive Up Part D Through what angle did the wheel turn between t=0 and t = 3.40 s ? =
rad SubmitGive Up Part E Prior to the wheel coming to rest, at what time will the radial acceleration at a point on the rim equal g = 9.81 m/s2 ? t =

Explanation / Answer

A] we know that = a/r = -10.9/0.25 = -43.6 rad/s^2

B] angular velocity w = v/r = 55/0.250 = 220 rad/s

C] w0 = w - t = 220 + 43.6*3.4 = 368.24 rad/s

D] = [w^2 - wo^2]/2 = [220^2 - 368.24^2]/(2*-43.6) = 1000 rad

E] for this acceleration, w =sqrt(g/r) = sqrt(9.81/0.25) = 6.2642 rad/s

time t = [w-wo]/ = [6.2642-368.24]/-43.6 = 8.302 s answer

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