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Physics Rotating Disk, Moment of Inertia, Angular Momentum, Angular Velocity

  1. Sep 11, 2008 #1
    1. The problem statement, all variables and given/known data

    Here's the problem statement, uploaded onto google docs:
    http://docs.google.com/View?docid=d277r7r_48f97jqsgs

    2. Relevant equations



    3. The attempt at a solution

    i)
    L=r*p
    L=R*m*v
    I think that's the answer, b/c they don't give us any constants to work with, but it seems too easy, so I don't trust myself on that.

    ii)
    I=[M(R^2)]/2

    iii) Here's where I get really confused. I thought the final moment of inertia would be I+(L/2*pi) because that would be like taking the effect of the ball, and dividing it by the entire circumference of the circle. Does that make sense?

    iv) The final angular velocity, this is where I'm totally tripped up. I thought it would be v/(answer in iii), because it's like taking v/m in straight kinematics, but translating it to rotational kinematics. Does that make sense again?
     
  2. jcsd
  3. Sep 11, 2008 #2

    Doc Al

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    Staff: Mentor

    No. What's the moment of inertia of a point mass?

    If you like, you can imagine the mass spread out over the circumference of the circle. What shape would that be and what is its moment of inertia?
     
  4. Sep 11, 2008 #3

    Aaah, I see. I guess the moment of inertia would be that of a hoop, which is mr^2. But, since the mass is now spread out over the entire circumference, would it be like (m/2*pi*R)*r^2, which simplifies to (mr)/(2pi) Is that right now?
     
  5. Sep 11, 2008 #4

    Doc Al

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    Exactly. (Now stop while your ahead. :wink:)
    Nope. Looks like you're trying to apply the linear mass density of the hoop for some reason. But you already know the moment of inertia of a hoop, which is in terms of its total mass.
     
  6. Sep 11, 2008 #5
    Oh, wow! It's a lot simpler than I thought. I don't know why I was complicating it so much. Okay, so we know the moment of Inertia for the ball is mR^2, and the moment of Inertia for the disk is .5*MR^2, so is the moment of inertia for the final assembly just the other moments of inertia added up? That is to say,

    I (final) = R^2(m+.5M)
     
  7. Sep 11, 2008 #6

    Doc Al

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    You got it.
     
  8. Sep 11, 2008 #7
    Sweet! You rock, Doc Al. Does that mean my answer for #1 is right too? And how do I solve for iv?
     
  9. Sep 11, 2008 #8

    Doc Al

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    Yes.
    What's conserved during the collision?
     
  10. Sep 11, 2008 #9
    Energy is conserved, I guess in this case there would be no translational kinetic energy, so all the energy from the ball, the 1/2mv^2, would have to be changed into energy of rotation. So, I could set up this equation:

    .5*m*v^2 = .5*I*w^2, and solve for w.

    My answer is that w=the square root of (mv^2)/I
     
  11. Sep 11, 2008 #10

    Doc Al

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    Kinetic energy is not conserved. This is a inelastic collision--the mass sticks to the disk.

    But something else is conserved. Hint: It was the first thing they asked about.
     
  12. Sep 11, 2008 #11
    Aaah, I see. The ball sticks, soo kinetic energy isn't conserved, okay. But, just like in 'normal' kinematics, momentum is conserved, so in this case Angular momentum is conserved. So, would that mean that I set up an equation that:

    R*m*v=I*w
    Thus, w=(Rmv)/I
     
  13. Sep 11, 2008 #12

    Doc Al

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    Exactly. (Where I is the total moment of inertia, of course.)
     
  14. Sep 11, 2008 #13
    All riiight. Sweeeet. Many thanks to you, Doc Al. Marvellous way of teaching without giving me the answer and without being too hard either. Really magnificent.
     
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