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.

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

Please explain your answers with Forumlas. Thank you.

Explanation / Answer

the formula for torque is

T = Fr

the magnitude of the torque on the disk about the z axis due to F1 is

a). T = F*R

= 329*1.44

= 473.76 Nm (anticlockwise)

(b)
the magnitude of the torque on the disk about the z axis due to F2 is
since F2 is passing from the origin,

. T = F*R = 329*0 =0

torque due to F2 is 0.

c).torque due to F3 is

T = FR cos theta

= 329*1.44*cos32

= 401.77 Nm (clockwise)

d)all the torques are in z-direction only.
0 torque in x-direction

e)0

according guidelines answered first four question please post next time remaining questions

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