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Solid disk on inclined plane

  1. May 5, 2009 #1
    1. The problem statement, all variables and given/known data

    A solid disk of mass m = 2.2 kg and radius r = .20 m starts from rest at a height h = 3.2 m and rolls without slipping down an incline plane with an angle of 30 degrees above the horizontal.

    a What is the linear velocity of the disk at the bottom of the incline?

    b How many revolutions has the disk turned through when it reaches the bottom?

    c What is the magnitude total Kinetic Energy of the sphere at the bottom of the incline?


    2. Relevant equations
    I=.5mr^2
    Kinetic = .5*I*omega^2


    3. The attempt at a solution
    I started by solving for the hyp => sin 30 = 3.2 /x where x=6.4
    That's where I am stuck
     
  2. jcsd
  3. May 5, 2009 #2

    LowlyPion

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    It rolls without slipping ... an important condition. That means that the potential energy ... m*g*h ... is going to go into kinetic energy. But what kind of kinetic energy?

    Won't it be rotational energy at 1/2*I*ω2 ?

    Once you've found ω, it's all down hill from there, even though you are already at the bottom, since ω = v/r.
     
  4. May 5, 2009 #3
    If 1/2*I*ω2 = mgh, I solve for the KE at about 69 and omega at 56. But, using ω = v/r, my v = 56*.2, or 11.2, which is not one of the multiple choice answers on the problem set.

    I am also still uncertain about solving for the # revolutions the disk has turned through when it reaches the bottom.
     
  5. May 5, 2009 #4

    LowlyPion

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  6. May 5, 2009 #5
    Whoops, I did have a math error. I got omega = 39.6. But, again, using ω = v/r, my v = 39.6*.2, or 7.9, which is not one of the multiple choice answers on the problem set.
     
  7. May 5, 2009 #6
    You must either include translational KE, or use the
    moment of inertia about the point of contact.

    Equating total KE to mgh is by far the simplest way of
    finding v.

    David
     
    Last edited: May 5, 2009
  8. May 5, 2009 #7

    LowlyPion

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    Oops. Right you are. I left out the 1/2*m*v2 that is the translational energy. How silly.

    Haste makes waste.

    1/2*I*ω2 + 1/2*m*v2 = m*g*h

    V = ω*r so ...

    1/2*ω2*(I + r2) = g*h
     
  9. May 5, 2009 #8
    Sorry to nitpick but

    1/2*ω2*(I + m*r2) = m*g*h
    where I = (1/2)m*r^2

    (Sort of confirms the parallel axis theorem)

    David
     
    Last edited: May 5, 2009
  10. May 5, 2009 #9

    LowlyPion

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    Geez. No problem. You're absolutely right. Thus confirming that haste indeed makes waste in not converting I. Trying to do too many things is not good for concentration ... at least mine.
     
  11. May 12, 2009 #10
    Here's what I did but I am still not getting the answer:
    mgh=0.5mv^2 (KE) + 0.5Iw^2

    34.5 = .5Iomega^2
    mgh*cos30 = 59.7

    PLugging in:
    59.7 = 1.1v^2 + 34.5
    22.9 = v^2
    4.8 = v

    Not a possible answer on our sheet...
     
  12. May 12, 2009 #11
    You haven't even read how we got this, let alone tried it, have you?
     
  13. May 12, 2009 #12
    Yes, I did. I thought that was the theorem. My variable, v, is not in that equation you posted. Based on that equation, what would I be solving for? I went back and used LowlyPion's equation.
     
  14. May 12, 2009 #13
    solve for w (omega)
    v = wr

    BTW LowlyPion was having a bad hair day, as you might have gathered
    from our good-natured exchange.
    Sorry if I sounded a bit cross in my last post.

    David
     
  15. May 12, 2009 #14
    mgh=0.5mv^2 (KE) + 0.5Iw^2
    is correct: PE lost = total KE gained.
    What you did next was just crazy!

    Now substitute w=v/r and solve for v.
     
  16. May 12, 2009 #15
    Good lord, I think I got it.
    v=6.47
     
  17. May 12, 2009 #16
    Whoopee:smile:
     
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