1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Loop the loop

  1. Feb 7, 2006 #1
    A steel ball rolls down an incline into a loop-the-loop of radius R. What is the minimum speed the ball must have at the top of the loop in order to stay on the track? (the ball has a radius r and starts at a height of h)
    This is a similar image to the one in my book:
    [​IMG]

    I'm not sure how to solve this question. But I think it has something to do with conservation of energy. I tried doing that and got:
    v = square root of (v02 - 20/7 gR)
    but the answer is the square root of (gR), which is must simpler than my answer...so I must be doing something wrong.
     
  2. jcsd
  3. Feb 8, 2006 #2

    Fermat

    User Avatar
    Homework Helper

    It has to do with centripetal force, rather than energy conservation.
    The ball has to not fall off the top of the track, which means it must maintain a minimum amount of circular motion.
    It is centripetal force that provides this circular motion.
    When travelling around a track, it is the normal reaction of the track against the ball which provides the centripetal force.
    When the motion is in a vertical plane, this reaction is modified by any radial component of the ball's weight.
    At the top of the track, the radial component of the ball's weight is mg.
    If there is no reaction force from the track then mg is the minimum amount of centriptal force.
    If mg maintains circular motion (no falling off the top of the track) then mg = mw²r
     
  4. Feb 8, 2006 #3
    This 2nd last sentence I don't understand.
     
  5. Feb 8, 2006 #4

    Doc Al

    User Avatar

    Staff: Mentor

    There are two forces acting on the ball at the top of the loop: gravity and the normal force (the track pushing on the ball). These forces add to produce the centripetal force on the ball. The faster the ball goes, the greater will be the normal force. To find the minimum speed such that the ball barely maintains contact with the track, set that normal force to zero. (Then apply Newton's 2nd law.)
     
  6. Feb 8, 2006 #5
    I think I'm beginning to understand.
    But why does the normal force have to be greater if the speed is greater? Is it because there needs to be a greater force to keep the ball in the loop? (Rather than fly off tangentially)

    so I use Fcentripetal = mg, and get the answer v = square root of (gR)
     
  7. Feb 8, 2006 #6

    Doc Al

    User Avatar

    Staff: Mentor

    Exactly.

    Right.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Loop the loop
  1. Loop the Loop (Replies: 8)

  2. Loop the loop (Replies: 1)

  3. Loop the Loop (Replies: 10)

  4. Loop the loop (Replies: 3)

  5. Loop a loop (Replies: 4)

Loading...