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

Torque F*r = I where is the angular acceleration.
F*r = (mr²/2) = mr² /2
r² = 2 Fr /m = 2*25.5*0.21 /41.6 = 0.26
Also = 0.26/ 0.21² = 5.89 rad /s²
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Part A

² = 2 since it starts from rest.
v² /r² = 2
v² = 2 r² = 2*0.26* = 0.52*2 N = 0.52*2*0.26
v = 0.849 m/s

Part B

Normal acceleration v²/r = 0.849² / 0.21 = 3.43 m/s²
Tangential acceleration = r = = 0.21 * 5.89* =1.237 m/s²
(Resultant acceleration) ² = 3.43 ² + 1.237 ²
Resultant acceleration = 13.295 m/s²

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