A uniform disk with mass m = 8.84 kg and radius R = 1.44 m lies in the x-y plane

Question

A uniform disk with mass m = 8.84 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 329 N at the edge of the disk on the +x-axis, 2) a force 329 N at the edge of the disk on the –y-axis, and 3) a force 329 N acts at the edge of the disk at an angle = 32° above the –x-axis.

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=1.8 sec

Please answer with formulas. Thank you.

Explanation / Answer

As we know:

Net Torque = Force * Distance

Net Force acting on disk is :
Fnet = F1 + F2*sin180 + F3*sin(270° + 32)
Fnet = 329 + 329*Sin180 + 329*sin(270° + 32)
Fnet = 49.99 N

Radius of disk = 1.44 m

A)

Toruque = F*r = 49.99 * 1.44
t = 71.98 N-m

B)
as
= I*
= 0.5 * M * R^2 *
= 71.98/(0.5 *8.84*1.44^2)
= 7.85 rad/s^2

C)
= o + *t
= 0 + 7.85*1.8
= 14.13 rad/s

Therefore
Rotational Energy = 0.5 * I * ^2
k = 0.5 * 0.5 * m * R^2 * ^2
k = 0.5*(0.5 *8.84*1.44^2)*14.13^2
k = 914.95 J

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