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Derivation of angular acceleration

  1. Nov 7, 2007 #1
    Derive the expression [tex]\alpha[/tex]=[tex]\frac{mgr}{I+mr^{2}}[/tex]for the acceleration a=[tex]\alpha[/tex]r of the falling mass m



    I don't even know what this question is asking...



    Any ideas on how to approach this wonderful little puzzle?
     
  2. jcsd
  3. Nov 7, 2007 #2
    anybody?
     
  4. Nov 7, 2007 #3

    Astronuc

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    That's all the information given? I presume this is a mass in freefall, in which case the translational acceleration is g, but it would appear that it is rotating as well.
     
    Last edited: Nov 7, 2007
  5. Nov 7, 2007 #4
    Unfortunately, yes. Can you figure out even which one is supposed to be derived or am I supposed to get one from the other? I'm really confused.
     
  6. Nov 7, 2007 #5

    Doc Al

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    Please post the full problem! I can't believe that that's all you were given. There must be some context. You have the rotational inertia and mass of what? :wink:

    Just by looking at the equation, I could make up a problem for which that would be the answer. But I'd rather see the actual problem.
     
  7. Nov 7, 2007 #6

    Astronuc

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    Look at - http://hyperphysics.phy-astr.gsu.edu/hbase/rotq.html#rq,

    one sees that the definition of tangential acceleration is at = [itex]\alpha[/itex]r.

    Now I am wonder if this problem is about a mass 'falling' in a circular arc, so it starts at a height of r, which gives a gravitational potential energy of mgr for a height of r above the bottom of the arc.

    At the moment, that's all I can think of off the top of my head.
     
    Last edited: Nov 7, 2007
  8. Nov 7, 2007 #7

    Doc Al

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    Here's one idea: An object (rotational inertia I) rotates freely about a horizontal axis through its center of mass. A mass m is attached such that it is a horizontal distance r from the axis. What's the acceleration of the "falling" mass?
     
  9. Nov 7, 2007 #8
    OK I can clarify this much.

    We did a lab. There was a frictionless wheel spinning in the horizontal plane. there was a string attached to the wheel and we timed different masses falling from a specified height.

    The question involves no numbers, they just want us to derive the acceleration...

    Sorry this is really stupid and I'm just not getting it.
     
  10. Nov 7, 2007 #9

    Doc Al

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    That's better. Just apply Newton's 2nd law to both the wheel and the falling mass. Combine the resulting equations and you can solve for the acceleration. What forces act on each?
     
  11. Nov 7, 2007 #10

    Astronuc

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    That's a vital piece of information! The wheel is spinning in the horizontal plane and the masses were falling, so presumably there is a frictionless (or nearl so) pulley over which the string is pulled?
     
  12. Nov 7, 2007 #11

    Doc Al

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    I suspect that the wheel is the pulley. They wound a string around the wheel, attached a mass, and let it drop.
     
  13. Nov 7, 2007 #12

    Astronuc

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    I would have thought so if the wheel was spinning in the vertical plane, but the new information states the wheel is spinning horizontally, so I imagined the string to be traveling horizontally which would require another pulley somewhere if the string is tied to a falling mass. I was trying to visualize the geometry.
     
  14. Nov 7, 2007 #13

    Doc Al

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    Oops, you're right it does say "horizontal plane". I just assumed that he meant to write horizontal axis. But either way will work.
     
  15. Nov 7, 2007 #14

    Astronuc

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    Yeah - it's the same thing, but it was a vital piece missing in the OP. The falling mass at r makes all the difference.
     
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