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Uniform Pully with a single block, do not understand where mass relates?

  1. Oct 26, 2008 #1
    OK, my largest issue with dealing with uniform cylindrical pulleys and a single block (object) is understanding where mass of the block relates in any of the equations.

    Specifically, I was doing some problems earlier with no issues at all. Then I came across one asking about the mass of the block hanging from the pulley. The book and my notes do not speak of this.
    Known:
    mass of the pulley
    radius,
    Tension,

    This seems like a pretty straight-forward topic, yet I cannot seem to locate much on it.
    Can anyone please explain the topic?
    Thank you in advance, I have already found much help from these forums, and without them would probably not pass calculus based physics!
     
  2. jcsd
  3. Oct 26, 2008 #2
    I tried solving for acceleration then using:

    ma = mg - T, rearranged as T = m(g-a) but this did not give me a correct solution either. hmmmmm.
     
  4. Oct 27, 2008 #3

    Doc Al

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    Staff: Mentor

    Are you talking about a cylinder that has a massless rope wrapped around it from which a block is hanging? And you're supposed to find the acceleration of the block?
    Generally the tension is not known, but you can figure it out.

    That looks like a reasonable equation for the block. Note that you have two unknowns, tension and acceleration. You need a second equation--one for the cylinder--to solve for the acceleration.
     
  5. Oct 27, 2008 #4
    Ok, thank you. I will look for some info on this. I am not too familiar with the equations of a pulley.
    To clarify the problem, I am looking for the mass of a block hanging from a single pulley.
    From the problem I am given the mass of the pulley, the radius of the pulley, and the tension in the cable.
    Do you know, specifically, which equation I am missing for the cylinder?
    Thank you again for the help. I need to know this subject in and out...
     
  6. Oct 27, 2008 #5

    Doc Al

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    Staff: Mentor

    OK. Instead of the tension being unknown, the mass is unknown. You still have two unknowns and need two equations.
    Yes, the rotational equivalent of Newton's 2nd law: Net Torque = Iα
     
  7. Oct 27, 2008 #6
    you have PM.
     
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