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Ball mechanics problem

  1. May 9, 2010 #1

    bon

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    1. The problem statement, all variables and given/known data

    A solid ball of constant density, a hollow ball and a hoop have the same radius.
    They are simultaneously released to roll without slipping down the same ramp. Write
    the order in which they reach the bottom and justify your answer.


    2. Relevant equations



    3. The attempt at a solution

    The thing is, it's worth 7 marks..I have no idea what working i need to do!

    So as they fall, PE is converted into KE..

    KE is translational and rotational..

    so mgh = 1/2mv^2 + 1/2Iw^2 = 1/2mv^2 + 1/2 I (v/r)^2 where h is the height fallen

    so gh = 1/2v^2 + 1/2 I/m (v/r)^2...

    Now im guessing the I/m will be the most for the solid ball, then for the hollow ball and least for the hoop..so hoop arrives first, then hollow ball, then solid ball..

    is this right? How can i justify it more rigorously?
     
  2. jcsd
  3. May 9, 2010 #2

    Doc Al

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    Staff: Mentor

    Re: Mechanics

    Why are you guessing? Look it up! (Or just think about it a bit.)

    Other than that, you have the right idea.
     
  4. Jun 6, 2010 #3

    bon

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    Re: Mechanics

    Further to this (sorry it was ages ago...)

    I think the MOI of a solid sphere would be 2/5ma^2, of a hollow sphere ma^2, and of a hoop ma^2..

    So solid ball arrives first then hoop and hollow sphere arrive together
     
  5. Jun 6, 2010 #4

    bon

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    Re: Mechanics

    Is this right?
     
  6. Jun 6, 2010 #5

    Doc Al

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    Re: Mechanics

    You're almost there. You have the correct MOI for the solid sphere and for the hoop, but not for the hollow sphere.
     
  7. Jun 6, 2010 #6

    bon

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    Re: Mechanics


    But isnt all the mass situated on the outer shell so for each infinitesimal mi the distance to the central axis is ri^2? so MOI is ma^2?
     
  8. Jun 6, 2010 #7

    Doc Al

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    Re: Mechanics

    Yes, the mass is on the outer shell. But the distance of that shell to the axis varies from 0 to R. (The distance to the center is fixed, but not to the axis.)
     
  9. Jun 6, 2010 #8

    bon

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    Re: Mechanics

    Oh i see so 2/3ma^2. Great thanks.

    so order: solid sphere first, then hollow, then hoop...?
     
  10. Jun 6, 2010 #9

    Doc Al

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    Re: Mechanics

    Right! The more energy (per unit mass) required to turn the thing, the less energy is left for translational motion.
     
  11. Jun 6, 2010 #10

    bon

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    Re: Mechanics


    Yay...thanks for all your help!

    Sorry to corrupt this thread with a different question (but i guess it's sort of related)..

    A rigid body is free to rotate about a fixed axis k through the origin. A force F
    is applied to the body at a point R. State the relationship between the torque due to
    F and the angular momentum about the axis. How is this related to Newton’s Second
    Law applied to linear momentum?

    So i know torque due to F is the rate of change of angular momentum i.e. G = dJ/dt

    or here: R x F = d/dt(mR x R dot)

    But what about the second part? how is this related to NII for linear momentum..

    I know that if you start with J = m r x r dot and differentiate you see that dJ/dt = r x mr double dot..then you can replace mr double for F due to NII, but i dont think this answers the question...Thanks

    Also - when you have a point mass m at a radius r in a gravitational potential V(r), why is the general motion confined to a plane..I know that the angular momentum about z i.e. r x r dot is constant..but why does this => plane? thanks again!
     
  12. Jun 6, 2010 #11

    Doc Al

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    Re: Mechanics

    I really don't know what they are looking for with that second question. I would have done something similar to what you did. Start with the definition of angular momentum, differentiate and apply NII. Beats me!

    The initial plane of motion can be defined as that perpendicular to the vector ri X vi. But angular momentum conservation implies that that r X v doesn't change, so the plane is fixed.

    Look at it this way. The vectors r, v, and the acceleration start out in the same plane, right? So how can the mass leave that plane?
     
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