A uniform disk with mass 40.0 kg and radius 0.200 m is pivoted at its center. Th

Question

A uniform disk with mass 40.0 kg and radius 0.200 m is pivoted at its center. The disk is initially at rest and then a constant force F = 25.0 N is applied tangent to the rim of the disk. (For a uniform disk I = 1/2 MR^2.) 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 3.00 revolutions? b) What is the magnitude of the tangential acceleration a_tan of a point on the rim of the disk after the disk has turned through 3.00 revolutions?

Explanation / Answer

(a)

First we find the angular acceleration of the disk.

We have I = MR2/2 = (40 kg)(0.2 m)2/2

or, I = 0.8 kg-m2

Torque applied on the disk is,

T = FR = (25 N)(0.2 m)

or, T = 5 N-m

So we have,

T = I

or, = T/I = (5 N/m)/(0.8 kg-m2)

or, = 6.25 rad/s2

So angular velocity after 3 revolutions is .

We use the formula,

f2 - i2 = 2, here i = 0, f = , = 3X2 = 18.84 rad

So we have,

2 = 2(6.25 rad/s2)(18.84 rad)

or, = 15.35 rad/s

So tangential speed at this instant is v = R

or, v = (15.35 rad/s)(0.2 m)

or, v = 3.07 m/s

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(b)

angular acceleration is = 6.25 rad/s2

so, tangential acceleration is,

a = R = (6.25 rad/s2)(0.2 m)

a = 1.25 m/s2

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This concludes the answers. Check the answer and let me know if it's correct. If you need any more clarification or correction, feel free to ask.....

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