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Friction in circular motion and torque transmission

  1. Aug 9, 2012 #1
    3 masses 120 degrees away from each other each one of 46g is spinning at 600RPM 58mm away from the center and are fixed to a guide pole. The masses are transmitting torque on a surface and the friction between the surfaces is 0.62 The area of contact is 192mm^2. How much torque is transmitted from one surface to the other.

    How can I make it to transmit as much force possible without changing the friction coefficient?

    Here is an image so you can have an idea hopefully you can help me out have no idea where to start.
     

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  2. jcsd
  3. Aug 9, 2012 #2

    Simon Bridge

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    This is the rotational equivalent of one block sliding over another isn't it?
    "0.62" what? Is this the actual friction force or is it the coefficient?
    How would you answer that question for linear friction and forces?

    Note: area of contact does not normally matter.
    https://www.physicsforums.com/showthread.php?t=106428
     
  4. Aug 9, 2012 #3
    the blocks slide on the blue part which is the one that is going to rotate when the friction between the blocks and the surface stick together

    The blocks are moving along a rotor's axis this makes the blocks move toward the blue part's wall that's when the blue part starts rotating. Like a centrifugal switch

    The friction coefficient is 0.62
     
  5. Aug 9, 2012 #4

    Simon Bridge

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    There is a lot of blue in that diagram - but I figured the blocks and arms were rotating inside a drum that the ends of the arms were brushing. I figured the blocks could move along the shafts too.

    It is the blocks that push - so, as the central axle turns faster, the blocks press harder against the drum and slow it down?

    The the force due to friction f is how hard the blocks press against the wall times the coefficient of kinetic friction ... f=μN just like you are used to... N would, in this case, be radial and outwards - what sort of force does that sound like? This produces a retarding torque τ=fr each.

    That the sort of thing you are looking for?
     
  6. Aug 11, 2012 #5
    That's right but I don't want to slow it down although it will I want the other part to also rotate
     
  7. Aug 11, 2012 #6
    I want to make the blue part rotate so the more it pushes the more force the more it is going to slow down and also transmit the motion to the blue part
     
  8. Aug 11, 2012 #7

    Simon Bridge

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    You are driving the middle 3-prong bit, and you want the blue drum to rotate as a result ... yep.

    This is the rotating version of those puzzles where you pull a block across the top of a bigger block and you want to get the acceleration of the bigger block. Work it out: you can now figure out the torque due to friction between the masses and the drum. You know the moment of inertia of the drum. [itex]\Sigma \tau = I\alpha[/itex].
     
  9. Aug 11, 2012 #8

    Simon Bridge

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    You are driving the middle 3-prong bit, and you want the blue drum to rotate as a result ... yep.

    This is the rotating version of those puzzles where you pull a block across the top of a bigger block and you want to get the acceleration of the bigger block. Work it out: you can now figure out the torque due to friction between the masses and the drum. You know the moment of inertia of the drum. [itex]\Sigma \tau = I\alpha[/itex].
     
  10. Aug 11, 2012 #9
    what do you mean, how do I calculate N as radial and outwards??
     
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