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

Answer:

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)

$KE = 5892.8 J$

Explanation:

Part a)

Torque due to F1 force is given as

$\tau = r \times F$

$\tau_1 = 1.3 \times 335$

$\tau_1 = 435.5 Nm$

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

$\tau = I \alpha$

$I = \frac{1}{2}mR^2$

$I = \frac{1}{2}(8.88)(1.3)^2$

$I = 7.5 kg m^2$

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

$\omega = \alpha t$

$\omega = 26.4 \times 1.5$

$\omega = 39.6 rad/s$

now rotational kinetic energy is given as

$KE = \frac{1}{2}I\omega^2$

$KE = \frac{1}{2}(7.5)(39.6)^2$

$KE = 5892.8 J$