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Angular acceleration problem - masses on massless rod

  1. Apr 21, 2006 #1
    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?
     
  2. jcsd
  3. Apr 21, 2006 #2

    nrqed

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    What did you use for r? You must first find where the center of mass is. Did you do this?
     
  4. Apr 21, 2006 #3
    For r I used 1.286. I got the center of mass is 15.4286 kgm^2, which is correct, according to WebAssign.
     
  5. Apr 21, 2006 #4

    nrqed

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    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)
     
  6. Apr 21, 2006 #5
    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?
     
  7. Apr 21, 2006 #6

    nrqed

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    rF/I = 1.286m * 7 N/15.428 kg m^2
     
  8. Apr 21, 2006 #7
    Oh okay, thanks! So I was supposed to use the inertia of the center of mass. Interesting. Thanks!
     
  9. Apr 21, 2006 #8

    nrqed

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    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.
     
  10. Apr 21, 2006 #9
    Makes sense...I never thought of that. Thanks!
     
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