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Simple angular acceleration problemhelp

  1. Nov 14, 2005 #1

    Nm

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    Two masses (mA= 2 kg, mB= 4 kg) are attached to a (massless) meter stick, at the 0 and 75 cm marks. The system is then hung from a string, so that it stays horizontal. Now, if mass B was removed, and no additional force was supplied, calculate the size of the angular acceleration of the meter stick at that instant.

    I calculated the center of mass to be at the 50 cm mark, but I don't know how to get the angular acceleration. Do I just use +rF/mr^2 = angular acceleration? I got the wrong answer form that. This seems like a simple problem, but maybe I'm thinking too much..
     
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  3. Nov 14, 2005 #2

    Tide

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    I think what you're looking for is the relationship between torque and angular acceleration. Here it is: torque = moment of inertia X angular acceleration.
     
  4. Nov 14, 2005 #3

    Nm

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    I used that relationship to get inertia of a point mass to equal mr^2 and the net torque to equal rFsin90. From the two equations I got angular acceleration to equal F/mr. Am I really far off because I don't know the force...?
     
  5. Nov 14, 2005 #4

    Tide

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    The force is the weight of the suspended object!
     
  6. Nov 14, 2005 #5

    Nm

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    If the weight is mg, the angular acceleration would be 9.8m/s^2/(2kg)(0.75m)?
     
  7. Nov 14, 2005 #6

    Tide

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    Don't forget the force holding the meter stick in place!
     
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