Respuesta :
Answer:
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
The magnitude of the torque on the disk about the z-axis due to Force 1 (F1) is equal to 437.03 Newton.
Given the following data:
- Mass of disk = 9.11 kg.
- Radius of disk = 1.37 m.
- Force 1 = 319 Newton.
- Force 2 = 319 Newton.
- Force 3 = 319 Newton.
- Angle of inclination = 32°.
- Time = 1.5 seconds.
How to determine the torque on the disk about the z.
Mathematically, the magnitude of the torque on the disk about the z-axis due to Force 1 (F1) is given by:
[tex]\tau_1 =F_1r_1 sin\theta_1\\\\\tau_1 = 319 \times 1.37 \times sin 90\\\\\tau_1 = 437.03\;Newton[/tex]
For the tension due to Force 2 (F2), we have:
[tex]\tau_2 =F_2r_2sin\theta_2\\\\\tau_2 = 319 \times 1.37 \times sin 180\\\\\tau_2 = 0\;Newton[/tex]
For the tension due to Force 3 (F3), we have:
[tex]\tau_3 =F_3r_3sin\theta_3\\\\\tau_3 = 319 \times 1.37 \times sin (90-32)\\\\\tau_3 = 370.62\;Newton[/tex]
Since the torque generated by Force 2 (F2) has a magnitude of zero and torque 2 and 3 are in the z-direction, there is no x-component of the net torque about the z-axis.
Similarly, there is no y-component of the net torque about the z-axis.
How to calculate the z-component of the net torque.
[tex]\tau_{n}=\tau_1 - \tau_3\\\\\tau_{n}=437.03-370.62\\\\\tau_{n}=66.41\;Newton[/tex]
How to calculate the magnitude of the angular acceleration.
First of all, we would determine the moment of inertia of the disk by using this formula:
[tex]I=\frac{1}{2} mr^2\\\\I=\frac{1}{2} \times 9.11 \times 1.37^2\\\\I=8.55\;kgm^2[/tex]
For the angular acceleration, we have:
[tex]\alpha =\frac{\tau_n}{I} \\\\\alpha =\frac{66.41}{8.55} \\\\\alpha =7.77\;rad/s^2[/tex]
How to calculate the rotational energy of the disk.
First of all, we would determine the angular velocity by using this formula:
[tex]\omega = \alpha t\\\\\omega =7.77 \times 1.5\\\\\omega =11.66\;rad/s[/tex]
For the rotational energy, we have:
[tex]K.E=\frac{1}{2} I\omega^2\\\\K.E=\frac{1}{2} \times 8.55 \times 11.66^2\\\\K.E =581.21\;Joules[/tex]
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