A uniform disk with mass m = 9.35 kg and radius R = 1.44 m lies in the x-y plane

Question

A uniform disk with mass m = 9.35 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 305 N at the edge of the disk on the +x-axis, 2) a force 305 N at the edge of the disk on the –y-axis, and 3) a force 305 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? = ____ N-m

8,If the disk starts from rest, what is the rotational energy of the disk after the forces have been applied for t = 1.6 s? = ____ J.

I neeed for help Thank you

Explanation / Answer

1) |torque| = |r x F| = 1.44 i x 305 j = 439.2 Nm.

2)  |torque| = 0

3) |torque| = r F cos e = 1.44*305*cos 32o = 372.46 Nm

4) Net torque = (439.2 - 372.46)k = 66.74 k

As the net torque is along z-axis only, x-component about z-axis is ZERO, 0 Nm.

5) Net torque = (439.2 - 372.46)k = 66.74 k

As the net torque is along z-axis only, y-component about z-axis is ZERO, 0 Nm.

6) Net torque = (439.2 - 372.46)k = 66.74 k

As the net torque is along z-axis only, z-component about z-axis is 66.74 Nm.

7) torque = moment of inertia * angular acceleration

moment of inertia, I = MR2/2 = 9.35*1.442/2 = 9.69 kg.m2

torque = 66.74 Nm

angular acceleration = 66.74 Nm / 9.69 kg.m2

the magnitude of the angular acceleration about the z axis of the disk = 6.88 rad/s2

8) w = wo + (angular acceleration*t)

wo = 0;

w = 6.88 rad/s2*1.6s

w = 11.0154 rad/s

rotational energy of the disk = 1/2*I*w2 = 0.5*9.69*11.01542

rotational energy of the disk = 587.88 J

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