A uniform disk with mass m = 8.54 kg and radius R = 1.35 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 333 N at the edge of the disk on the +x-axis, 2) a force 333 N at the edge of the disk on the –y-axis, and 3) a force 333 N acts at the edge of the disk at an angle θ = 39° above the –x-axis.
1)What is the magnitude of the torque on the disk about the z axis due to F1?
449.55
2)What is the magnitude of the torque on the disk about the z axis due to F2?
0
3)What is the magnitude of the torque on the disk about the z axis due to F3?
349.365697
4)What is the x-component of the net torque about the z axis on the disk?
0
5)What is the y-component of the net torque about the z axis on the disk?
0
6) What is the z-component of the net torque about the z axis on the disk?
100.184303
7)What is the magnitude of the angular acceleration about the z axis of the disk?
12.87372622
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? (disk is starting from stop)
All above answers are correct I'm just having trouble figuring out 8
Please help! thanks!
The rotational energy of the disk after the forces have been applied for t = 1.6 s is 8.845 J.
To find the rotational energy of the disk, we need to use the equation:
Rotational Energy = (1/2) * moment of inertia * angular velocity^2
The moment of inertia of a uniform disk rotating about its central axis is given by:
moment of inertia = (1/2) * mass * radius^2
Let's calculate the moment of inertia first:
moment of inertia = (1/2) * (8.54 kg) * (1.35 m)^2
moment of inertia = 8.1166 kg·m^2
Since the disk starts from rest, the initial angular velocity is zero (ω0 = 0). We can find the final angular velocity (ω) using the equation:
torque = moment of inertia * angular acceleration
From question 7, we already know the magnitude of the angular acceleration:
angular acceleration = 12.87372622 rad/s^2
We can rearrange the equation to solve for angular acceleration:
torque = moment of inertia * angular acceleration
Let's calculate the torque due to F3 using:
torque on disk = force * lever arm
The lever arm is the perpendicular distance from the force to the axis of rotation (in this case, the z-axis). The force F3 is given as 333 N, and the angle θ is given as 39°. We can use trigonometry to find the lever arm:
lever arm = force * sin(θ)
lever arm = (333 N) * sin(39°)
lever arm = 203.21775 N
Now, let's calculate the torque using:
torque on disk = lever arm * force
torque on disk = (203.21775 N) * (333 N)
torque on disk = 67,675.25275 N·m
Now, we can substitute the values into the equation:
torque = moment of inertia * angular acceleration
torque = (8.1166 kg·m^2) * (12.87372622 rad/s^2)
torque = 104.5779 N·m
Now, we can use this torque value to find the final angular velocity (ω):
torque = moment of inertia * angular acceleration
torque = moment of inertia * (final angular velocity - initial angular velocity)
104.5779 N·m = (8.1166 kg·m^2) * (ω - 0)
ω = 104.5779 N·m / 8.1166 kg·m^2
ω ≈ 12.879 rad/s
Now, we can find the rotational energy:
Rotational Energy = (1/2) * moment of inertia * angular velocity^2
Rotational Energy = (1/2) * (8.1166 kg·m^2) * (12.879 rad/s)^2
Rotational Energy ≈ 678.535 J
Therefore, the rotational energy of the disk after the forces have been applied for t = 1.6 s is approximately 678.535 J.
To find the rotational energy of the disk after the forces have been applied for t = 1.6 s, you need to use the equation for rotational kinetic energy:
Rotational Kinetic Energy (KE) = (1/2) * I * ω^2
where I is the moment of inertia of the disk and ω is the angular velocity of the disk.
First, let's find the moment of inertia of the disk. The moment of inertia of a uniform disk rotating about its axis is given by:
I = (1/2) * m * R^2
where m is the mass of the disk and R is the radius of the disk.
Given:
m = 8.54 kg
R = 1.35 m
I = (1/2) * 8.54 kg * (1.35 m)^2
I = 8.54 kg * 0.91125 m^2
I ≈ 7.785 m^2 kg
Now, you need to find the angular velocity ω. To do that, you can use the equation:
ω = α * t
where α is the angular acceleration of the disk and t is the time.
Given:
α = 12.87372622 rad/s^2
t = 1.6 s
ω = 12.87372622 rad/s^2 * 1.6 s
ω = 20.59876195 rad/s
Now, you have both I and ω, so you can calculate the rotational kinetic energy:
KE = (1/2) * I * ω^2
KE = (1/2) * 7.785 m^2 kg * (20.59876195 rad/s)^2
KE ≈ 645.946 Joules
Therefore, the rotational energy of the disk after the forces have been applied for t = 1.6 s is approximately 645.946 Joules.