A uniform disk with mass 41.6 kg and radius 0.210 m is pivoted at its center abo

Question

A uniform disk with mass 41.6 kg and radius 0.210 m is pivoted at its center about a horizontal, frictionless axle that is stationary. The disk is initially at rest, and then a constant force 25.5 N is applied tangent to the rim of the disk.

Part A

What is the magnitude v of the tangential velocity of a point on the rim of the disk after the disk has turned through 0.260 revolution?

Part B

What is the magnitude a of the resultant acceleration of a point on the rim of the disk after the disk has turned through 0.260 revolution?

Explanation / Answer

Part A) Torque, T = RF = 0.21 * 25.5 = 5.355 N-m

Also, T = I = MR2/2

=> angular acceleration, = 2T/MR2 = 2 * 5.355 / (41.6 * 0.2102) = 5.838 m/s2

Angular speed of the disk, = (2)1/2 = [2 * 5.838 * (0.260 * 2)]1/2 = 4.367 rad/s2

Tangential velocity, v = R = 4.367 * 0.210 = 0.917 m/s

Part B) Tangential acceleration, a = R = 5.838 * 0.210 = 1.23 m/s2

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