Auniform disk with mass m = 9.11 kg and radius r = 1.37 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 319 n at the edge of the disk on the +x-axis, 2) a force 319 n at the edge of the disk on the –y-axis, and 3) a force 319 n acts at the edge of the disk at an angle θ = 32° above the –x-axis. 1) what is the magnitude of the torque on the disk about the z axis due to f1? n-m submit 2) what is the magnitude of the torque on the disk about the z axis due to f2? n-m submit 3) what is the magnitude of the torque on the disk about the z axis due to f3? n-m submit 4) what is the x-component of the net torque about the z axis on the disk? n-m submit 5) what is the y-component of the net torque about the z axis on the disk? n-m submit 6) what is the z-component of the net torque about the z axis on the disk? n-m submit 7) what is the magnitude of the angular acceleration about the z axis of the disk? rad/s2 submit 8) if the disk starts from rest, what is the rotational energy of the disk after the forces have been applied for t = 1.5 s? j submit

1) τ = 437 N m, 2)  τ = 437 N m , 3) τ = 344.4 N m ,  7) α  = 142.5 rad/s²  and

8)  Em = 1.95 105 J

Explanation:

The torque is given by the expression

    τ = F x R

Bold indicates vectors, we can calculate the magnitude of this torque with the expression

     τ = F R sin  θ

The direction is given by perpendicular to the two vectors F and R

1) To calculate the torque due to the F1 force. The angle between F1 and R is 90°

    sin 90 = 1

    τ = F R

    τ = 319 1.37

    τ = 437 N m

In the positive Z direction

2) The torque due to F2 again the angle is 90 ° without 90 = 1

    τ = F R

    τ = (-319) 1.37 sin 270

    τ = 437 N m

In the positive z direction

3) The torque  F3 where θ  = 32 ° measured from the negative side of the x axis

     τ = F R sin θ  

Angle is measured from the positive side of the x-axis, so this angle is

    θ  = 180 - 32 = 128 °

    τ = 319 1.37 sin 128

    τ = 344.4 N m

4) The x components of this torque is

    F1 R + F3 R cos 128 = 437 + 319 1.37 cos 128 = 437 + 269

    τₓ = 706 N m

5) the component and

     F2 R +F3 R sin 128 = 437 + 344.4

     τ y =  781.4 N m

The component has to be perpendicular to the moment

7) angular acceleration

    τ = I α

The moment of inertia of a disk is

    I = ½ M R²

    I = ½ 9.11 1.37²

    I = 8.55 Kg m²

    α  = τ / I

    α  = (τ1 + τ2 + τ3) / I

    α   = (437 + 437 + 344.4) / 8.55

    α  = 142.5 rad / s²

8) the disk starts at rest, let's calculate the speed in the given time

    w = w₀ + α  t

    w = 0 + 142.5  1.5

   w = 213.75 rad / s

Energy is

   Em = K = ½ I w²

   Em = ½ 8.55 213.75²

   Em = 1.95 105 J

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