A uniform disk with mass m = 9.1 kg and radius R = 1.37 m lies in the x-y plane

Question

A uniform disk with mass m = 9.1 kg and radius R = 1.37 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 317 N at the edge of the disk on the +x-axis, 2) a force 317 N at the edge of the disk on the –y-axis, and 3) a force 317 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?

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

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

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

Explanation / Answer

Here ,

mass of disk , m= 9.1 Kg

radius , R = 1.37 m

1) magnitude of torque due to F1 = F1 * R

magnitude of torque due to F1 = 317 * 1.37

magnitude of torque due to F1 = 434.3 N.m

the magnitude of torque due to F1 is 434.3 N.m

2)

theta = 32 degree

as the torque is perpendicular to the both R and F3

x-component of torque = 0 N.m

3)

let the angular acceleration about the z axis is a

I * a = net torque

0.5 * 9.1 * 1.37^2 * a = 317 * 1.37 - 317 * 1.37 * cos(32)

solving for a

a = 7.727 rad/s^2

the angular speed is 7.727 rad/s^2

4)

at t = 1.5 s

rotational energy of disk = 0.5 * I * (wf)^2

rotational energy of disk = 0.5 * 0.5 * 9.1 * 1.37^2 * ( 1.5 * 7.727)^2

rotational energy of disk = 573.6 J

the rotational energy of disk after 1.5 s is 573.6 J

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