A uniform disk with mass m = 8.73 kg and radius R = 1.41 m lies in the x-y plane

Question

A uniform disk with mass m = 8.73 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 311 N at the edge of the disk on the +x-axis, 2) a force 311 N at the edge of the disk on the –y-axis, and 3) a force 311 N acts at the edge of the disk at an angle ? = 37° above the –x-axis.What is the x-component of the net torque about the z axis on the disk?

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

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

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?

Explanation / Answer

4. There is no x-component.

5. There is no y-component.

6. = sum of [R x Fi]
= (R)(F1) + (R)(F2)(sin180°) + (R)(F3)(sin(270° + 37°))
= (R)[(F1) + (F2)(sin180°) + (F3)(sin(270° + 37°))]
= (1.41 m)[(311 N) + 0 + (311 N)(sin(270° + 37°))]
= (1.41 m)(311 N)[1 + sin307°]
= 88.3 N-m

7)
= I
= (1/2)(mR²)
88.3 N-m = (1/2)(8.73 kg)(1.41 m)²
=10.18 rad/s²

8)
= (o) + t
= (10.18 rad/s²)(1.5 s)
=15.27 rad/s

K = (1/2)I²
K = (1/2)(1/2)(mR²)²
K = (1/4)(8.53 kg)(1.43 m)²(12.8 rad/s)²
K = 714 J

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