Angular acceleration problem - masses on massless rod

A 3 kg mass and a 4 kg mass are attached to either end of a 3 m long massless rod. If the system is rotated about the center of mass by a force of 7 N acting on the 4 kg mass, (perpendicular to the rod), what will the size of the angular acceleration of the system be?

I used angular acceleration = rF/mr^2, and got 1.3608 rad/s^2, but got the answer wrong. Anyone have any idea why?
 

nrqed

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FlipStyle1308 said:
A 3 kg mass and a 4 kg mass are attached to either end of a 3 m long massless rod. If the system is rotated about the center of mass by a force of 7 N acting on the 4 kg mass, (perpendicular to the rod), what will the size of the angular acceleration of the system be?

I used angular acceleration = rF/mr^2, and got 1.3608 rad/s^2, but got the answer wrong. Anyone have any idea why?
What did you use for r? You must first find where the center of mass is. Did you do this?
 
For r I used 1.286. I got the center of mass is 15.4286 kgm^2, which is correct, according to WebAssign.
 

nrqed

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FlipStyle1308 said:
For r I used 1.286. I got the center of mass is 15.4286 kgm^2, which is correct, according to WebAssign.
Ok (btw, the value you give is the moment of inertia, not the center of mass).

I think you just made a mistake using your calculator. I used your values and got a different alpha (0.5835 rad/s^2)
 
Whoops, you're right, I gave the rotational inertia for the center of the rod lol. Center of mass is 1.714 m, from the 3 kg mass. What did you plug in for that 0.5835?
 

nrqed

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FlipStyle1308 said:
Whoops, you're right, I gave the rotational inertia for the center of the rod lol. Center of mass is 1.714 m, from the 3 kg mass. What did you plug in for that 0.5835?
rF/I = 1.286m * 7 N/15.428 kg m^2
 
Oh okay, thanks! So I was supposed to use the inertia of the center of mass. Interesting. Thanks!
 

nrqed

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FlipStyle1308 said:
Oh okay, thanks! So I was supposed to use the inertia of the center of mass. Interesting. Thanks!
I am not sure what you mean. But the r you use there is the distance between where the force is applied and the axis of rotation, as a general principle. And the I is always calculated with respect to where the axis of rotation is located.
 
Makes sense...I never thought of that. Thanks!
 

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