# Rotation about a fixed axis

1. Nov 16, 2004

### 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????

2. Nov 16, 2004

### Staff: Mentor

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

3. Nov 16, 2004

### 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?

4. Nov 16, 2004

### Staff: Mentor

The torque is different in each case.

5. Nov 16, 2004

### 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.

6. Nov 16, 2004

### krab

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

7. Nov 16, 2004

### pinky2468

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

8. Nov 16, 2004

### Staff: Mentor

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

9. Nov 16, 2004

### 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.

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