Question

+ A uniform disk with mass m = 8.76 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 317 N at the edge of the disk on the +X-axis, 2) a force 317 N at the edge of the disk on the -y- axis, and 3) a force 317 N acts at the edge of the disk at an angle 0 = 39° above the -X-axis. +X >

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 = 1.6 s?

J

Answer 1

mass = m = 8.76 Kg. Radius = r = 1.42 m. F1 = 317 N at (r,0,0)

F2 = 317 N at ( 0, - r , 0). F3 = 317 N at location ( r cos 39 , r sin 39 , 0)

1)

Torque due to F1 = 317 N x 1.42 m = 450 N-m along + z axis

2)

Torque due to F2 = 317 N x 0 m = 0 N-m

3)

Torque due to F3  = 317 N x 1.42 x cos 39 = 350 N-m along - z axis

4) component of net torque along x-axis = 0

5) component of net torque along y- axis = 0

6) component of net torque along x- axis = (450-350) N-m = 100 N-m in + z- direction

7) angular acceleration along z axis = 100 N-m / moment of inertia about z-axis

moment of inertia = 0.5 x M x r^2 = 0.5 x 8.76 Kg x (1.42 m)^2 = 8.83 Kg-m^2

angular acceleration = 100 N-m/ (0.5 x 8.76 Kg x (1.42 m)^2 )

= 11.3 R/s^2 about + Z axis

8)

t = 1.6 s angular velocity of disk after 1.6 sec = 0+ 11.3 R/s^2 x 1.6 s = 18.1 R/s

rotational KE = 0.5 x 0.5 x 8.76 Kg x (1.42 m)^2 x (18.1 R/s)^2 = 1147 J