A uniform disk with mass 39.0 kg and radius 0.240 m is pivoted at its center abo

Question

A uniform disk with mass 39.0 kg and radius 0.240 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 33.0 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 0.400 revolution?

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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 0.400 revolution?

Explanation / Answer

angular displacement theta = 0.4*2*3.142 = 2.5136 rad

initial angular velocity wi = 0 rad/s

Ftan = m*atan

33 = 39*atan

tangential accelaration atan = 33/39 = 0.846 m/s^2

but atan =r*alpha = 0.24*alpha = 0.846

alpha = 0.846/0.24 = 3.525 rad/s^2

using wf^2-wi^2 = 2*alpha*theta = 2*3.525*2.5136 = 17.72 rad/s

then v = r*wf = 0.24*17.72 = 4.25 m/sec

B) radial accealration a_rad = v^2/r = 4.25^2/0.24 = 75.3 m/s^2

a_tan = 0.846 m/s^2

then a = sqrt(a_tan^2+a_rad^2) = sqrt(0.846^2+75.3^2) = 75.3 m/s^2

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