Question

A uniform disk with mass m = 8.88 kg and radius R = 1.3 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 335 N at the edge of the disk on the +x-axis, 2) a force 335 N at the edge of the disk on the –y-axis, and 3) a force 335 N acts at the edge of the disk at an angle θ = 33° above the –x-axis.

1) What is the magnitude of the torque on the disk about the z axis due to F1?
2) What is the magnitude of the torque on the disk about the z axis due to F2?
3) What is the magnitude of the torque on the disk about the z axis due to F3?
4) What is the x-component of the net torque about the z axis on the disk?
5) What is the y-component of the net torque about the z axis on the disk?
6) What is the z-component of the net torque about the z axis on the disk?
7) What is the magnitude of the angular acceleration about the z axis of the disk?
8) If the disk starts from rest, what is the rotational energy of the disk after the forces have been applied for t =1.5s?

Answer 1

Part a)

`\tau_1 = 435.5 Nm`

Part b)

`\tau_2 = 0`

Part c)

`\tau_3 = 237.2 Nm`

Part d)

`\tau_x = 0`

Part e)

`\tau_y = 0`

Part f)

`\tau_z = 198.3 Nm`

Part g)

`\alpha = 26.4 rad/s^2`

Part h)

`KE = 5892.8 J`

Step-by-step explanation:

Part a)

Torque due to F1 force is given as

`\tau = r * F`

`\tau_1 = 1.3 * 335`

`\tau_1 = 435.5 Nm`

Part b)

Torque due to F2 force is given as

`\tau = rF sin\theta`

`\tau_2 = 1.3(335)sin0`

`\tau_2 = 0`

Part c)

Torque due to F3 force is given as

`\tau = rFsin\theta`

`\tau_3 = 1.3(335)(sin33)`

`\tau_3 = 237.2 Nm`

Part d)

Net torque along x direction is given as

`\tau_x = 0`

Part e)

Net torque along y direction is given as

`\tau_y = 0`

Part f)

Net torque along x direction is given as

`\tau_z = 435.5 - 237.2`

`\tau_z = 198.3 Nm`

Part g)

angular acceleration is given as

`\tau = I \alpha`

`I = (1)/(2)mR^2`

`I = (1)/(2)(8.88)(1.3)^2`

`I = 7.5 kg m^2`

now we have

`198.3 = 7.5 \alpha`

`\alpha = 26.4 rad/s^2`

Part h)

angular speed of the disc after 1.5 s

`\omega = \alpha t`

`\omega = 26.4 * 1.5`

`\omega = 39.6 rad/s`

now rotational kinetic energy is given as

`KE = (1)/(2)I\omega^2`

`KE = (1)/(2)(7.5)(39.6)^2`

`KE = 5892.8 J`