Torque: Turning Heavy Door with "Steady" Force

  • Thread starter JFlash
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[SOLVED] Torque: Turning Heavy Door with "Steady" Force

Homework Statement



Hello. The question reads, "...This is the world's heaviest hinged door. The door has a mass of 44,000kg, a rotational inertia about an axis through its hinges of 8.7X10^4 kg*m^2, and a width of 2.4m. Neglecting friction, what steady force, applied at its outer edge and perpendicular to the plane of the door, can move it from rest through an angle of 90 degrees in 30s? Assume no friction acts on the hinges."

Homework Equations



The work I did below followed 4 steps:
1. Use this equation: [tex]\Theta = (1/2)(\omega initial + \omega)t[/tex] to find the angular velocity at 30s.
2. Find the average angular acceleration over the 30s period of applied force.
3. Set the torque equal to the rotational inertia of the door times its angular acceleration.
4. Set torque equal to r X F, and solve for F.

The Attempt at a Solution



My answer came to 40.27N, which is about a third of the actual answer: 130N. As I look over my work, I realize that I never used the mass of the door, which is probably a reason why my answer is wrong. Anyway, here's what I did:

[tex]\Theta = \pi/2 = (1/2)(0 + \omega)30s... \omega = \pi/30 rad/s[/tex]
[tex]\alpha = (\pi/30 rad/s)/(30s-0s) = \pi/900 rad/s^2 [/tex]
[tex]\tau = 87000kgm^2 * \pi/900 rad/s^2 = 96.67[/tex]
[tex]\tau = rF sin \phi = 2.4m * F * sin(90 degrees) = 40.27N[/tex]

Does anyone know where I went wrong?
 
Last edited:

Answers and Replies

  • #2
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you forgot to multipy by pi in: [tex]\tau = 87000kgm^2 * \pi/900 rad/s^2 = 96.67[/tex]
 
  • #3
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Thanks man.
 

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