A uniform disk with mass m = 8.88 kg and radius R = 1.3 m lies in the x-y plane

Question

A uniform disk with mass m = 8.88 kg and radius R = 1.3 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 335 N at the edge of the disk on the +x-axis, 2) a force 335 N at the edge of the disk on the –y-axis, and 3) a force 335 N acts at the edge of the disk at an angle = 33° above the –x-axis.

1) What is the magnitude of the torque on the disk about the z axis due to F1?

2)  What is the magnitude of the torque on the disk about the z axis due to F2?

3) What is the magnitude of the torque on the disk about the z axis due to F3?

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?

8) If the disk starts from rest, what is the rotational energy of the disk after the forces have been applied for t =1.5s?

Explanation / Answer

1) Torque, T1 = R x F1 = 1.3 * 335 = 435.5 N-m

2) Torque, T2 = R x F2 = 0 (since F2 passes through the z-axis)

3) Torque, T3 = R x F3 = 1.3 * 335 * sin(90o - 33o) = 365.24 N-m

4) x-component of net torque is zero

5) y-component of net torque is zero

6) z-component of net torque, Tz = (435.5 + 0 + 365.24) N-m = 800.74 N-m

7) Moment of inertia of the disk about the z-axis, Iz = MR2/2 = 8.88 * 1.32 / 2 = 3.752 kg-m2

Angular acceleration, = Tz / Iz = 800.74 / 3.752 = 213.42 rad/s2

8) Angular speed after 1.5 seconds, = t = 213.42 * 1.5 = 320.13 rad/s

Rotational energy of the disk after 1.5 seconds, E = I2/2 = 3.752 * 320.132 / 2 = 192.26 kJ

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