A uniform disk with mass m = 9.42 kg and radius R = 1.31 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 315 N at the edge of the disk on the +x-axis, 2) a force 315 N at the edge of the disk on the –y-axis, and 3) a force 315 N acts at the edge of the disk at an angle θ = 35° above the –x-axis.
√ 1) What is the magnitude of the torque on the disk about the z axis due to F1?
412.65 N-m √
√ 2) What is the magnitude of the torque on the disk about the z axis due to F2?
0 N-m √
√ 3) What is the magnitude of the torque on the disk about the z axis due to F3?
338N-m √
√ 4) What is the x-component of the net torque about the z axis on the disk?
0 N-m √
√ 5) What is the y-component of the net torque about the z axis on the disk?
0 N-m √
# HELP IN 6,7 & 8 PLEASE!!
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 = 1.7 s?
J
6) The z-component of the net torque about the z-axis on the disk can be found by summing the torques due to each individual force. Since the forces in the +y and -y directions have equal magnitudes and opposite directions, their torques cancel each other out. Therefore, the z-component of the net torque is simply the torque due to F3.
7) The magnitude of the angular acceleration about the z-axis of the disk can be calculated using the equation:
τ = Iα
Where τ is the net torque, I is the moment of inertia of the disk, and α is the angular acceleration. Rearranging the equation gives:
α = τ / I
Plug in the value of the net torque and the moment of inertia of the disk to find the angular acceleration.
8) To calculate the rotational energy of the disk after the forces have been applied for a certain time, you need to use the equation:
Rotational Energy = (1/2) I ω^2
Where I is the moment of inertia of the disk and ω is the angular velocity. Since the disk starts from rest, the initial angular velocity is zero. To find ω, you can use the equation:
τ = Iα
Rearranging the equation gives:
α = τ / I
Once you have found the angular acceleration, you can use the kinematic equation:
ω = α t
Where t is the time. Plug in the values to find ω, and then use it to calculate the rotational energy.
To answer questions 6, 7, and 8, we need to calculate the net torque and use it to find the angular acceleration and rotational energy.
6) To find the z-component of the net torque about the z-axis, we need to sum the torques due to each force. The torque due to a force is given by the cross product of the force and its position vector from the origin.
The position vectors for each force are:
1) F1: (R, 0, 0)
2) F2: (0, -R, 0)
3) F3: (-Rcos(θ), Rsin(θ), 0)
The torques are given by:
τ1 = r1 x F1 = (R, 0, 0) x (315, 0, 0) = (0, 0, 315R)
τ2 = r2 x F2 = (0, -R, 0) x (0, -315, 0) = (0, 0, 0)
τ3 = r3 x F3 = (-Rcos(θ), Rsin(θ), 0) x (315cos(θ), 315sin(θ), 0)
= (0, 0, -315Rcos(θ)sin(θ))
The z-component of the net torque is the sum of these torques:
τz_net = τ1z + τ2z + τ3z = 315R - 315Rcos(θ)sin(θ)
7) To find the magnitude of the angular acceleration about the z-axis, we can use Newton's second law for rotational motion, which states that the net torque is equal to the moment of inertia multiplied by the angular acceleration:
τz_net = I * α
Since the disk is a uniform disk, the moment of inertia can be calculated as:
I = (1/2) * m * R^2
Therefore, the angular acceleration is:
α = τz_net / I = (315R - 315Rcos(θ)sin(θ)) / ((1/2) * m * R^2)
8) To find the rotational energy of the disk after the forces have been applied for t = 1.7 s, we can use the formula for rotational kinetic energy:
Rotational Energy = (1/2) * I * ω^2
Since the disk starts from rest, the initial angular velocity ω is 0. We can use the angular acceleration calculated in question 7 to find the final angular velocity ω_f:
ω_f = α * t
Then, we can calculate the rotational energy:
Rotational Energy = (1/2) * I * ω_f^2
To solve questions 6, 7, and 8, we need to use the concept of torque and rotational motion.
6) The z-component of the net torque about the z-axis can be calculated by considering the individual torques from each force. Since forces 1 and 2 do not have any z-component, only force 3 contributes to the z-component of the net torque. The torque due to force 3 is given by:
τ3 = r3 * F3 * sin(θ)
Where r3 is the distance from the origin to the point where force 3 is applied, F3 is the magnitude of force 3, and θ is the angle between force 3 and the negative x-axis.
Since the disk lies in the x-y plane and is centered at the origin, the distance r3 is equal to R (the radius of the disk). In this case, r3 = R = 1.31 m.
Substituting the given values:
τ3 = 1.31 m * 315 N * sin(35°)
Calculate this expression to find the z-component of the net torque about the z-axis.
7) The magnitude of the angular acceleration about the z-axis of the disk can be found using the equation:
τnet = I * α
Where τnet is the net torque about the z-axis, I is the moment of inertia of the disk, and α is the angular acceleration.
Since the disk is a uniform disk, the moment of inertia can be calculated using the formula:
I = (1/2) * m * R^2
Where m is the mass of the disk and R is the radius of the disk.
Calculate the net torque τnet from the given values and substitute it in the equation to find the magnitude of the angular acceleration α.
8) The rotational energy of the disk is given by the formula:
Rotational energy = (1/2) * I * ω^2
Where I is the moment of inertia of the disk and ω is the angular velocity.
Since the disk starts from rest, the initial angular velocity ω is zero. Therefore, the rotational energy of the disk after the forces have been applied for time t can be calculated using the formula:
Rotational energy = (1/2) * I * ω^2 = (1/2) * I * (α * t)^2
Substitute the given values of I, α, and t to find the rotational energy of the disk after time t.
6. Add the three torques, but as I see it, trorque 3 is in the clockwise direction, so it should subtract from the sum of the other two. sketch out that to confirm.
7. A= net torque/I
8. net torque*time=(momentinertia) *angularvelocityfinal
solve for final angular velociyt, so rotational energy=1/2 (momentinertia) wf^2