A flywheel with a radius of 0.429 m starts from rest and accelerates with a cons

Question

A flywheel with a radius of 0.429 m starts from rest and accelerates with a constant angular acceleration of 0.528 rad/ s^2 . Compute the magnitude of the tangential acceleration, the radial acceleration, and the resultant acceleration of a point on its rim at the start.Compute the magnitude of the tangential acceleration, the radial acceleration, and the resultant acceleration of a point on its rim after it has turned through 45.0 degrees. Compute the magnitude of the tangential acceleration, the radial acceleration, and the resultant acceleration of a point on its rim after it has turned through 140 degrees.

Explanation / Answer

The radius of the flywheel is r = 0.429 m with initial velocity u = 0 The angular acceleration of the flywheel is z = 0.528 rad/s^2 The flywheel has turned through A = 45 degrees = 45 * (2pi/180) radians = 1.57 radians The tangential acceleration of the flywheel is at = r * z = 0.429 * 0.528 = 0.226 m/s^2 We know that S = r * A = 0.429 * 1.57 = 0.673 m We know from the relation v^2 - u^2 = 2at * S or v^2 = 2at * S or v = (2at * S)^1/2 The radial acceleration is ar = (v^2/r) The resultant acceleration is a = (at^2 + ar^2)^1/2 When A = 140 degrees = 140 * (2pi/180) radians = 4.88 radians The tangential acceleration is same,that is,at = 0.226 m/s^2 We know that S1 = r * A = 0.429 * 4.88 = 2.1 m We know from the relation v1^2 - u1^2 = 2at * S1 where u1 = 0 or v1^2 = 2at * S1 or v1 = (2at * S1)^1/2 The radial acceleration is ar = (v1^2/r) The resultant acceleration is a = (at^2 + ar^2)^1/2

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