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Conservation of Energy Physics Olympiad

  1. Jan 8, 2012 #1
    1. The problem statement, all variables and given/known data
    A ball of mass M and radius R has a moment of inertia of I =2/5MR. The ball is released from rest and rolls
    down the ramp with no frictional loss of energy. The ball is projected vertically upward o a ramp as shown in
    the diagram, reaching a maximum height ymax above the point where it leaves the ramp. Determine the maximum
    height of the projectile ymax in terms of h
    2rwrmud.png

    The image is cut off but h is the height from the top of the ramp to the ball
    2. Relevant equations
    PE=1/2mv^2
    Rotational KE= 1/2 I omega^2
    KE= 1/2mv^2


    3. The attempt at a solution
    Using x as the distance between the bottom and top of the ramp:
    mg(h+x) = 1/2mv^2+1/2I omega^2 =1/2mv^2 + 1/2(2/5mr^2)omega^2 = 1/2mv^2 + 1/5mv^2 = 7/10mv^2

    7/10mv^2=mgx + 1/2mv'^2

    1/2mv'^2 = ymax

    im stuck here D: any hints?
     
  2. jcsd
  3. Jan 8, 2012 #2

    jedishrfu

    Staff: Mentor

    Okay so physically the ball rolls down and then exits the ramp spinning and going upward a certain height.

    It's energy initially is mgh right? And it's energy afterward will be in the spin and in its motion so if you can determine its rotational energy and subtract it from the mgh you'd have a new mgh right from which to get the final height.

    Does that sound right?
     
  4. Jan 8, 2012 #3
    its energy initally should be mg(h+x), because mgh is just the energy from the top of the ramp to the inital position of the ball
     
  5. Jan 8, 2012 #4

    NascentOxygen

    User Avatar

    Staff: Mentor

    I'm not sure why you introduced an x. It rolls from an initial vertical height h, and rises to a height ymax.
     
  6. Jan 8, 2012 #5

    jedishrfu

    Staff: Mentor

    Don't you need an equation to relate the omega to the ball rolling down with no friction and no slipping? Also do you need the angle of the ramp or does that fall out somewhere? If not I suppose you could choose say 60 degrees or better yet use a variable for it.
     
  7. Jan 8, 2012 #6

    NascentOxygen

    User Avatar

    Staff: Mentor

    The clue that is the key to problems like this is to recognize that, by descending without slippage, the rotational speed of the ball is directly related to its linear velocity. When you think about it, it seems obvious, though it may still help if it's spelled out: the ball rotates once in the same time that it rolls a distance 2 Pi R metres.

    While you may think the angle of the ramp is needed, it seems not. During the descent, the gravitational potential energy of the ball is totally converted into linear and rotational energy.
     
  8. Jan 8, 2012 #7
    Sure, however, that still leaves many unknowns. Including the velocity at the bottom and the max y.
     
  9. Jan 12, 2012 #8

    NascentOxygen

    User Avatar

    Staff: Mentor

    By equating PE at start to KE + rotational energy on the level, I was able to then obtain a direct proportionality between ymax and h. Recognize that as it rises, the KE is converted into PE while the rotational energy remains as rotational energy.
     
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