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Torque and Rotational Motion

  1. Mar 13, 2006 #1
    A uniform 2.9-kg cylinder can rotate about an axis through its center at O. The forces applied are: F1 = 2.4 N, F2 = 3.9 N, F3 = 6.1 N, and F4 = 4.6 N. Also, R1 = 11.2 cm and R3 = 4.0 cm. Find the magnitude and direction (+: counterclockwise; -: clockwise) of the angular acceleration of the cylinder. (see attachment)

    I'm having a hard time with this one. I know that:
    Torque (net) = Inertia * Angular Acceleration
    and
    Torque = Force * Radius

    I've got:
    T(F1)=0.2688 Nm
    T(F2)=-0.4368 Nm
    T(F3)=-0.0244 Nm
    T(F4)=0 Nm

    Tnet=-0.1924 Intertia=0.0181888
    and the angular accerleration= -10.58 rad/s^2 (INCORRECT)

    What am I doing wrong?
     

    Attached Files:

  2. jcsd
  3. Mar 13, 2006 #2

    nrqed

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    I can't see the figure yet so I can't tell. But the torque is F times r times sin theta. That may (or may not, depending on what direction the forces are acting) explain a discrepancy.

    Pat
     
  4. Mar 13, 2006 #3

    Doc Al

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    The diagram seems to show a single radius. What are R1 and R2? What's the radius of the cylinder?


    OK.
    Don't forget the angle that the force is applied.

    Torque = Force * Radius * sin(angle)
     
  5. Mar 13, 2006 #4
    All angles a 90deg, that would just be multiplying by 1, right? Also, is my F3 torque correct? I did .004 * 6.1
     
  6. Mar 13, 2006 #5

    nrqed

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    Your T(F3) is off by a factor of 10. That will probably do it.
     
  7. Mar 13, 2006 #6

    Doc Al

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    Perhaps you are defining things differently, but only F1 and F2 are applied at 90 degrees to the radial line through the point of application.
     
  8. Mar 13, 2006 #7

    UGH! That was it...I got it. Thanks!
     
  9. Jan 10, 2009 #8
    O k Doc I am just having trouble get my (T1), T(2)..., my numbers are off, so what is the formula for that?
     
  10. Jan 10, 2009 #9

    Doc Al

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    Formula for what? You just quoted the formula for finding the torque.

    Show what you did.
     
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