Question

A uniform disk with mass m = 9.28 kg and radius R = 1.42 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 345 N at the edge of the disk on the +x-axis, 2) a force 345 N at the edge of the disk on the –y-axis, and 3) a force 345 N acts at the edge of the disk at an angle θ = 39° above the –x-axis.

a. What is the magnitude of the torque on the disk about the z axis due to F₁?

b. What is the magnitude of the torque on the disk about the z axis due to F₂?

c. What is the magnitude of the torque on the disk about the z axis due to F₃?

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

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

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

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

h. If the disk starts from rest, what is the rotational energy of the disk after the forces have been applied for t = 1.5 s?

Answer 1

a) the magnitude of the torque on the disk about the z axis due to F1,
|T1| = F1*R*sin(thea)
= 345*1.42*sin(90)
= 490 N.m <<<<<<<<<---------------Answer

b) the magnitude of the torque on the disk about the z axis due to F2,
|T2| = F2*R*sin(theta)
= 345*1.42*sin(180)
= 0 <<<<<<<<<---------------Answer

c) the magnitude of the torque on the disk about the z axis due to F3,
|T3| = F3*R*sin(theta)
= 345*1.42*sin(90 - 39)
= 381 N.m <<<<<<<<<---------------Answer

d) x-component of the net torque about the z axis on the disk, Tnetx = 0 <<<<<<<<<---------------Answer

e) y-component of the net torque about the z axis on the disk, Tnety = 0 <<<<<<<<<---------------Answer

f) z-component of the net torque about the z axis on the disk, Tnetz = 490 - 381

= 109 N.m <<<<<<<<<---------------Answer

g) moment of inertial of the disk, I = 0.5*m*r^2

= 0.5*9.28*1.42^2

= 9.26 kg.m^2

angular acceleration of the disk, alfa = Tnet/I

= 109/9.26
= 11.8 rad/s^2 <<<<<<<<<---------------Answer

h) at t = 1.5 s,

angular speed, w = wo + alfa*t

= 0 + 11.8*1.5

= 17.7 rad/s

rotational kinetic energy of the disk, KE = (1/2)*I*w^2

= (1/2)*9.26*17.7^2

= 1450 J <<<<<<<<<---------------Answer