A uniform disk with mass m = 9.23 kg and radius R = 1.41 m lies in the x-y plane

Question

A uniform disk with mass m = 9.23 kg and radius R = 1.41 m lies in the x-y plane and centered at the origin. Three forces act in the +y-direction on the disk: 1) a force 337 N at the edge of the disk on the +x-axis, 2) a force 337 N at the edge of the disk on the –y-axis, and 3) a force 337 N acts at the edge of the disk at an angle ? = 37° above the –x-axis.

6)What is the z-component of the net torque about the z axis on the disk?

7)What is the magnitude of the angular acceleration about the z axis of the disk?

8)

If the disk starts from rest, what is the rotational energy of the disk after the forces have been applied for t = 1.8 s?

Explanation / Answer

1) Torque due to F1 about the z-axis, 1 = r1 x F1 = (1.41 i^) x (337 j^) = 475.2 k^

2) Torque due to F2 about the z-axis, 2 = r2 x F2 = (-1.41 j^) x (337 j^) = 0

3) Torque due to F3 about the z-axis, 3 = r3 x F3 = [1.41(-cos 37o i^ + sin 37o j^)] x (337 j^) = -379.5 k^

4) Net torque, net = 1 + 2 + 3 = [475.2 + 0 + (-379.5)] k^ = 95.7 k^

So, x-component of net torque = 0

5) y-component of net torque = 0

6) z-component of net torque = 95.7 k^

7) Moment of inertia of the disk about the z-axis, Iz = MR2/2 = 9.23*1.412/2 = 9.175 kg-m2

So, Magnitude of angular acceleration around z-axis, = net/Iz = 95.7/9.175 = 10.43 rad/s2

8) Angular speed of the disk after 1.8 seconds is, = o + t = 0 + 10.43*1.8 = 18.774 rad/s

So, rotational energy of disk after 1.8 seconds is, Er = Iz2/2 = 9.175*(18.774)2/2 = 1616.9 J

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