A uniform disk with mass 37.6 kg and radius 0.200 m is pivoted at its center abo

Question

A uniform disk with mass 37.6 kg and radius 0.200 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 30.0 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 0.290 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 0.290 revolution?

Explanation / Answer

A) torque = I x alpha

I = moment of inertia

0.200 x 30 = (37.6 x 0.2^2 / 2) alpha

alpha = 7.98 rad/s^2

using wf^2 - wi^2 = 2 x alpha x theta

theta = 0.290 x 2pi rad = 1.82 rad

wf^2 - 0 = 2 x 7.98 x 1.82

wf = 5.39 rad/s

v =wf *r = 5.39 x 0.20 = 1.08 m/s

B) tangential acc. a_t = alpha*r = 7.98*0.20 = 1.60 m/s^2

centripetal acc. a_c = v^2 / r = 1.08^2 / 0.2 =5.83 m/s^2

resultant = sqrt(a_t^2 + a_c^2) = 6.05 m/s^2

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