A uniform disk with mass m = 8.85 kg and radius R = 1.44 m lies in the x-y plane

Question

A uniform disk with mass m = 8.85 kg and radius R = 1.44 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 330 N at the edge of the disk on the +x-axis, 2) a force 330 N at the edge of the disk on the –y-axis, and 3) a force 330 N acts at the edge of the disk at an angle = 32° above the –x-axis.

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

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

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

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

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

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

7) What is the magnitude of the angular acceleration about the z axis of the disk? rad/s2

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? J

Explanation / Answer

1) Torque on the disk about the z axis due to F1 = F1*R

= 330*1.44

= 475.2 Nm in +z direction

2) Torque on the disk about the z axis due to F2 = F2*R*cos 90 degree = 0

3) Torque on the disk about the z axis due to F3 = F1*R cos 32 degree

=300*1.44*cos 32 degree

= 366.36 Nm in -z direction

4) Net torque = 475.2 - 366.36 Nm in +z direction

= 108.84 Nm  in +z direction

Therefore x component of torque is zero

5) y component of torque is zero

6) z component of torque is 108.84 Nm

7) Angular acceleration = 108.84/(0.5*8.85*1.44*1.44) = 11.86 rad/s

8) w = 0+at = 11.86*1.8 = 21.35 rad/s

Rotational Energy = 0.5 I w^2

= 0.5* (0.5*8.85*1.44*1.44) * 21.35^2

= 2091 J

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