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(Simple?) Pully problem

  1. Apr 7, 2005 #1
    M, a solid cylinder (M=1.39 kg, R=0.137 m) pivots on a thin, fixed, frictionless bearing. A string wrapped around the cylinder pulls downward with a force F which equals the weight of a 0.710 kg mass, i.e., F = 6.965 N. Calculate the angular acceleration of the cylinder.

    That I can do... But then I get...

    If instead of the force F an actual mass m = 0.710 kg is hung from the string, find the angular acceleration of the cylinder.

    Why would the second situation be differant than the first? And could you point me in the right direction? Thanks.
  2. jcsd
  3. Apr 7, 2005 #2
    The 0.71 kg mass provides a constant downward force only if it's not accelerating, itself.
  4. Apr 7, 2005 #3
    I get what you're saying, but I still have no idea where to go from here. Do you think you could give me some more hints? Thanks for the help. :confused:
  5. Apr 7, 2005 #4
    I have to leave this computer pretty soon, but here's a quick suggestion: The D.E. for the system will have a mass with a linear acceleration and a force, and a moment of inertia with an angular acceleration and a torque. The torque & ang. accel. can be converted to linear units of force & linear accel, and will equal the tension in the line and the acceleration of the mass. Gotta go...
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