Calculating Rotational Motion: Solving for Time in Uniform Door Rotation

[pinky2468]

I am having a very hard time with this problem! Even my teacher had trouble with it(of course we are still expected to have it done on our homework!)

2 doors are uniform and identical. Door A rotates about an axis to its left edge and door B rotates about an axis through the center. The same force F is applied perpendicular to each door at its right edge and the force remains perpendicular to as the door turns. Starting from rest, door A rotates through a certain angle in 3.00s. How long does it take door B to rotate through the same angle.

So I got as far as 1/3ML^2(alpha)=1/12M(L/2)^2(alpha) and then I get stuck on how o make time fit in. I have been looking at all the equations for rotational motion, but I don't know which one?

 

[Doc Al]

Use \tau = I \alpha to find the \alpha for each case. Then apply the kinematic equation that gives \theta as a function of time.

 

[pinky2468]

This is what I am coming up with:
1/3ML^2(2\theta/t^2)= 1/12ML/2^2(2\theta/t^2)

ML^2 cancel out and so does the 2\theta
and I am left with (1/3)(1/t^2)=(1/12)(1/4)(1/t^2)

Is that right? Where do I go from there?

 

[Doc Al]

This is what I am coming up with:
1/3ML^2(2\theta/t^2)= 1/12ML/2^2(2\theta/t^2)

The torque is different in each case.

 

[pinky2468]

But the same force is applied to each door in the same spot?? If the torque is different then I am not sure what to do b/c the only known value is time.

 

[krab]

A: torque is FL
B: torque is FL/2

 

[pinky2468]

That is what I had, but can't I set them equal to each other?

 

[Doc Al]

That is what I had, but can't I set them equal to each other?

No, since they are not equal! You can easily relate them though.

 

[CartoonKid]

You should use the information given to find out the angle for door A. Treat door A and B as separate cases, door A is there just to tell you the angle. Then you find out the \alpha from the \tau and I that you already have. If you can find out the angle of door A, then you shouldn't have problem working out time for door B because it's sort of working backward with the method you used for A.