Calculate Angular Acceleration of Rotating Door

[nutster]

More questions!
A rotating door is made from four rectangular glass panes, as shown in the drawing. The mass of each pane is 91 kg. A person pushes on the outer edge of one pane with a force of F = 58 N that is directed perpendicular to the pane. Determine the magnitude of the door's angular acceleration.

OK, so equations...

F sub T = mass * alpha sub T,
and alpha sub T = radius * angular accel

So if I try to apply 58N as the tangential force and 364kg as the mass, and solve for alpha sub T, why can't I divide alpha sub T by the radius of the door and get the answer?

 

[Pyrrhus]

could you be more clear with your notation?

 

[nutster]

Sorry, I'll give LaTeX a try.
$$F_T=m\alpha_T$$, and $$\alpha_T=ra$$, so why doesn't $$F_T=mra$$?

If you were to do this, you'd get...

58N=364kg*1.2m*a, right? That's .133, but it's wrong

 

[Pyrrhus]

$$a = r \alpha$$

$$\sum_{i=1}^{n} F_{i} = ma$$

 

[nutster]

Something without integrals, please :) It's trig calc.

 

[nutster]

Sorry, trig physics.

 

[Pyrrhus]

That's the sigma notation for summation, that's not an integral...

 

[nutster]

Sorry, blonde moment. OK, so the only force being applied is the tangential force of 58N...right? So,

$$a=\frac{58N}{364kg}$$, or 0.159. Applying this value to a, and applying to the other equation, $$0.159=1.2m\alpha$$;$$\alpha=0.1325$$...but that isn't the answer. What have I missed?

 

[Pyrrhus]

I won't know what's wrong without the image. So far your original equations were wrong, now it seems you're using "correctly" the right ones.

 

[nutster]

Image: http://www.webassign.net/CJ/09_34.gif

 

[nutster]

bump, please :)

 

[whozum]

I would tackle this from a torque perspective. You have a force of 58N acting at a radius of 1.2M. Find the center of mass of the panes and find that point's moment of inertia and used Newton's 2nd Law for torques to find alpha..