A uniform disk with mass m = 8.67 kg and radius R = 1.39 m lies in the x-y plane

Question

A uniform disk with mass m = 8.67 kg and radius R = 1.39 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 303 N at the edge of the disk on the +x-axis, 2) a force 303 N at the edge of the disk on the –y-axis, and 3) a force 303 N acts at the edge of the disk at an angle = 35° above the –x-axis.

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

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

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?

Explanation / Answer

a)

1)the magnitude of the torque on the disk due to F1
torq=R x F -- (1.39)i x (303)j = (421.17)k

2)the magnitude of the torque on the disk due to F2
(-1.39)j x (303)j = 0k

3)magnitude of the torque on the disk due to F3:

(-1.39cos35 i + 1.39sin35 j) x (303)j = (-345.002 )k

z-component of the net torque on the disk = 421.17 -345.002
= 76.168 N-m

b)
I=1/2*mR^2
I = 1/2*8.67*(1.39)^2 = 8.3756
torq total = I*alpha
so alpha=76.168 /8.3756 = 9.09 rad/s^2

c)
= (o) + t
= (9.09 rad/s²)(1.8 s)
= 16.362 rad/s

K = (1/2)I²
K = (1/2)(1/2)(mR²)²
K = (1/4)(8.67 kg)(1.39 m)²(16.362 rad/s)²
K = 1121.14 J

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