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Loop-the-Loop problem help

  1. Apr 7, 2004 #1
    A solid marble of mass m = 40 g and radius r = 2 cm will roll without slipping along the loop-the-loop track if it is released from rest somewhere on the straight section of track. From what minimum height h above the bottom of the track must the marble be released to ensure that it does not leave the track at the top of the loop? The radius of the loop-the-loop is R = 0.95 m.

    So this problem has to do with Rotation and Inertia and Conservation of energy I think.

    I found the Inertia i believe with I=(2/5)mr^2

    So the top of the loop-the-loop is 2 times the radius which is .95, so it'd be 1.90 m.

    Conservation of mechanical energy says
    (1/2)mv_f^2 + mgy_f = (1/2)mv_i^2 + mgy_i ?

    How do I arrive to that formula?
    Last edited: Apr 7, 2004
  2. jcsd
  3. Apr 7, 2004 #2


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    Yes, it surely does have to do with "conservation of energy".

    The last time I tried to answer a problem like this Doc Al jumped all over me. He's much better with physics problems than I am so: Over to you Doc Al!
  4. Apr 7, 2004 #3
    Due in 2 hours. You know where he answered a similar question?
  5. Apr 7, 2004 #4
    He answered them here. But the search feature doesn't want to find them.

    Anyway, Doc Al's point (if I rememer correctly, anyway...) was that the marble (roller coaster in the other example, actually) must have a certain amount of kinetic energy in the form of horizontal velocity in order to stay on the track at the top of the loop. If it has zero kinetic energy at the top, it's not moving at all in either the vertical or horizontal directions, and therefore it's going to fall straight down!

    So how fast does the ball have to be moving at the top to stay on the track? Here's a hint: think circular motion...

    Last edited: Apr 7, 2004
  6. Apr 7, 2004 #5
    You don't know how to go about arriving at Conservation of energy formula?
  7. Apr 7, 2004 #6
    Conservation of Energy can be derived from Newton's Second Law. There is no external force, so therefore there will be no change in energy.

  8. Apr 7, 2004 #7
    So first i need to find w..?
    Can i do that without time?
  9. Apr 7, 2004 #8
    w, as in omega ([itex]\omega[/itex]), the angular velocity?

    At the top of the loop, the centripetal force will be provided by gravity. So if F is known to be mg, do we have another equation that relates angular velocity and F? Do we have an equation that relates angular velocity and linear velocity?

  10. Apr 8, 2004 #9
    Isn't it just h > 2R? For the marble to stay on track, its speed at the top of the loop-the-loop must be greater than zero. I don't believe these is only one speed with which it can approach the top, since there is also the normal force N which will balance mg so that:
    Fc = mv2/R
    That is why you are asked to find the "minimum height h" rather than the "only possible height h" or "range of heights h". If there is no friction, energy is conserved so as long as the marble is released from a height greater than 2R, it will reach the top of the loop with speed.
    Last edited: Apr 8, 2004
  11. Apr 8, 2004 #10

    Doc Al

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    Staff: Mentor

  12. Apr 8, 2004 #11

    Doc Al

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    Staff: Mentor

    No. See the link I refer to in my previous post.
    If you release from a height of 2R, you won't make it around the loop. A minimum speed (greater than zero!) is needed to maintain contact with the track at the top of the loop. Anything less and you will fall off.

    In talking about the normal force and centripetal acceleration, you guys are on the right track ( :biggrin: ), now just follow through correctly.
  13. Apr 8, 2004 #12
    I think I see what my error was. I thought the marble was traveling along the outer circle of the loop, i.e "outside" the loop, rather than "inside" the loop. When it travels outside the loop, the normal force is on the opposite direction of the centripetal force of the marble, so any speed (greater than zero) will allow it to continue without falling (partially because it has nowhere to fall to :wink:). But you are right, since it travels inside the loop and the normal force contributes to the centripetal force, there exists a minimum speed with which the marble must approach the top of, or any other point of, the loop.
  14. Apr 8, 2004 #13
    Now what I want to know is why that post didn't come up when I do a search for it. Searching for "roller coaster" or "roller" or "coaster" or "loop" all fail to return that thread.

    Think of all the effort that could be saved if we could just throw links at people once we got all the common problems solved!

  15. Apr 8, 2004 #14
    I think it's called a FAQ. :wink:
  16. Apr 8, 2004 #15
    Like anybody actually reads that thing!

  17. Apr 8, 2004 #16
    If anyone did read it, it wouldn't be a FAQ anymore. It could be a knowledge base maybe, but not a FAQ which isn't meant to be read by definition. :wink:
  18. Apr 8, 2004 #17
    Maybe we should just write up some solutions and throw them up top. Gather some problems and divy them up, make it a group effort.

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