A uniform disk with mass 44.0 kg and radius 0.260 m is pivoted at its center abo

Question

A uniform disk with mass 44.0 kg and radius 0.260 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 27.0 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 0.300 revolution?

Express your answer with the appropriate units.

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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 0.300 revolution?

Express your answer with the appropriate units.

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A uniform disk with mass 44.0 kg and radius 0.260 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 27.0 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 0.300 revolution?

Express your answer with the appropriate units.

v =

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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 0.300 revolution?

Express your answer with the appropriate units.

a =

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Explanation / Answer

torque = r x F

t = 0.260 * 27 = 7.02 N-m

we know t = I*alpha

I = moment of inertia

for disk I = mr^2/2

7.02 = mr^2 *alpha

alpha = 2.36 rad/s^2

wf^2 = wi^2 + 2*alpha*theta

wi = 0

theta = 0.300 rev = 0.300 * 2pi rad

alpha = 2.36 rad/s^2

wf = 2.11 rad/s

v = wf*r = 0.548 m/s

part b )

tangential acceleration = ar = alpha*r = 0.6136 m/s^2

centripetal acceleration = ac = w^2*r = 1.156 m/s^2

|a| = sqrt(ac^2 + ar^2)

|a| = 1.3088 m/s^2

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