This question has several parts that must be completed sequentially. If you skip

Question

This question has several parts that must be completed sequentially. If you skip.? part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. A wheel 1.8? m In diameter lies in a vertical plane and rotates about its central axis with a constant angular acceleration of 4.47 rad/s^2. The wheel starts at rest at t = 0, and the radius vector of.a certain point P on the rim makes an angle of 57.3 degree with the horizontal at this time. At t = 2.00 s, find the following. the angular speed of the wheel the tangential speed of the point P the total acceleration of the point P this angular position of the point P

Explanation / Answer

a) The radius of the wheel = 0.935 m
angular velocity = initial angular velocity + A*t where A is angular accel and t is time
ang vel = 0 + 4.47 rad/s^2 x 2s = 8.94 rad/s

b) tangential vel = r w = 0.935 m x 8.94 rad/s = 8.36 m/s

c) total accel = Sqrt[a tan^2 + a radial^2]

a tan = 4.47 rad/s^2, a radial = v^2/r = (8.36m/s)^2/0.935 m = 74.75 m/s^2

  total accel = 74.88

d) total angular displacement of the point P = w0 t + 1/2 A t^2 = 0 + 1/2 (4.47 rad/s/s)(2s)^2 = 8.94 rad
the initial position, 57.3 deg = 1 rad, so final position = 9.94 rad = 127 deg with respect to the horizontal

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