A uniform disk with mass m = 8.57 kg and radius R = 1.36 m lies in the x-y plane
ID: 1376278 • Letter: A
Question
A uniform disk with mass m = 8.57 kg and radius R = 1.36 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 338 N at the edge of the disk on the +x-axis, 2) a force 338 N at the edge of the disk on the y-axis, and 3) a force 338 N acts at the edge of the disk at an angle ? = 31 above the x-axis.
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 = 1.6 s?
J
Explanation / Answer
Torque = Force * Perpendicular Disdtance
(1)
Magnitude of Torque due to F1 =
= F1 * R
= 338 * 1.36
= 459.68 Nm
Diretion - Anticlockwise
(2)
Magnitude of Torque due to F2 = 0
(3)
Magnitude of Torque due to F3 =
= F3 * R * cos(31)
= 338 * 1.36 * cos(31)
= 394.02 Nm
Diretion - Clockwise
(4)
x-component of the net torque about the z axis on the disk = 0
(5)
Y-component of the net torque about the z axis on the disk = 0
(6)
z-component of the net torque about the z axis on the disk = 459.68 - 394.02 Nm
z-component of the net torque about the z axis on the disk = 65.66 Nm
(7)
Torque = Moment of Inertia * Angular Acceleration
= I*
Moment of Inertia of Disk = 1/2 * M*R^2
I = 1/2 * 8.57 * 1.36^2
I = 7.93 kG m^2
= /I
= 65.66 / 7.93
= 8.28 rad/s2
Angular acceleration about the z axis of the disk, = 8.28 rad/s2
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