A uniform disk with mass m = 8.95 kg and radius R = 1.32 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 346 N at the edge of the disk on the +x-axis, 2) a force 346 N at the edge of the disk on the –y-axis, and 3) a force 346 N acts at the edge of the disk at an angle θ = 36° 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 = 1.6 s?
To answer these questions, we need to use the concept of torque and the equations of rotational motion. We'll begin by calculating the torque due to each force.
a. The magnitude of the torque (τ) on the disk due to force F1 can be calculated using the formula:
τ = r * F * sin(θ)
Here, r is the perpendicular distance from the axis of rotation to the line of action of the force, F is the magnitude of the force, and θ is the angle between the force vector and the line connecting the axis of rotation to the point of application of the force.
In this case, since the force F1 is acting at the edge of the disk on the +x-axis, the perpendicular distance from the axis of rotation (z-axis) to the line of action of force F1 is equal to the radius of the disk, R.
Therefore, the torque due to F1 can be calculated as:
τ1 = R * F1 * sin(θ1)
= R * 346 N * sin(0°)
Since sin(0°) = 0, we find that the torque due to F1 is zero.
b. The torque (τ2) due to force F2 can be calculated similarly as:
τ2 = R * F2 * sin(θ2)
= R * 346 N * sin(270°)
Since sin(270°) = -1, we find that the torque due to F2 is:
τ2 = R * 346 N * (-1)
c. The torque (τ3) due to force F3 can be calculated as:
τ3 = R * F3 * sin(θ3)
= R * 346 N * sin(36°)
Substituting the given values, we find the torque due to F3.
Next, we need to find the net torque on the disk. In order to calculate the net torque, we need to consider the individual torques in terms of their components along the x, y, and z-axes.
d. The x-component of the net torque (τx) can be obtained by summing the x-components of the individual torques. In this case, since τ1 = 0, the total x-component is simply the x-component of τ3.
e. The y-component of the net torque (τy) is the sum of the y-components of the individual torques.
f. The z-component of the net torque (τz) can be obtained by summing the z-components of the individual torques. In this case, τ1 = 0 and τ2 = 0, so the total z-component is the z-component of τ3.
Now that we have calculated the components of the net torque, let's move on to the next part.
g. The magnitude of the angular acceleration (α) of the disk can be calculated using the equation:
τ = I * α
Here, I is the moment of inertia of the disk, and τ is the net torque on the disk about the z-axis.
By rearranging the equation, we can solve for α:
α = τ / I
To find the magnitude of α, we need to know the moment of inertia of the disk. The moment of inertia for a uniform disk rotating about its central axis can be calculated using the formula:
I = (1/2) * m * R^2
Using the given values of m and R, we can calculate the moment of inertia.
Finally, for part h, we can calculate the rotational energy of the disk using the equation:
Rotational energy (E) = (1/2) * I * ω^2
Here, ω is the angular velocity of the disk, which is related to the angular acceleration (α) by the equation:
ω = α * t
Substituting the values of α and t, we can find the rotational energy of the disk after t = 1.6 s.
By following these steps, you should be able to answer all the questions related to the torque and rotational energy of the disk.