- #1

JFlash

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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.

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) F = 40.27N[/tex]

Does anyone know where I went wrong?