A uniform disk with mass m = 8.65 kg and radius R = 1.38 m lies in the x-y plane
ID: 1409125 • Letter: A
Question
A uniform disk with mass m = 8.65 kg and radius R = 1.38 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 300 N at the edge of the disk on the +x-axis, 2) a force 300 N at the edge of the disk on the –y-axis, and 3) a force 300 N acts at the edge of the disk at an angle = 34° above the –x-axis. What is the magnitude of the torque on the disk about the z axis due to F1? What is the magnitude of the torque on the disk about the z axis due to F2? What is the magnitude of the torque on the disk about the z axis due to F3?
What is the x-component of the net torque about the z axis on the disk?
What is the y-component of the net torque about the z axis on the disk?
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). T = F*r = 300*1.38 = 414 N-m (anticlockwise)
b). since F2 is passing from the origin, torque due to F2 is 0.
Torque = 300 N x 0 m (already on the y axis) = 0 N-m
c).torque due to F3 = 300*1.38*cos34 = 343.22 N-m (clockwise)
d). all the torques are in z-direction only.
=> 0 torque in x-direction
e). 0
since the disk is in a 2-D plane and the rotational components can come only in z-direction..
so, the torque is 0 in x & y-direction
f). net torque in z-direction = 414 - 343.22 = 70.78 N-m (anticlockwise)
g).T = 70.78 N-m
I = m*r^2/2 = 8.65*1.38^2/2 = 8.236 kg-m^2
angular acceleration = T/I = 70.78 / 8.236 = 8.6 rad/s^2
h).
final angular velocity after 1.7s = 0 + 8.6*1.8 = 15.47 rad/s
rotational energy = 0.5*I*w^2 = 0.5*8.236*15.47^2 = 985.3 J
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