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!

Circular motion of a loop-the-loop machine

  1. Sep 6, 2006 #1
    A loop-the-loop machine has radius R of 18m.
    a)What is the minimum speed at which a cart must travel so that it will safely loop the loop?

    I am unsure which formula to use as I know it has something to do with KE and GPE. I was thinking it might be :

    v = square root ____(2GM)______
    r

    but it cant be as we don't the mass.

    any ideas???
     
  2. jcsd
  3. Sep 6, 2006 #2

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Hi there busted and welcome to PF,

    Your an the right tracks with energy. Energy must be conserved, ignoring friction all the kinetic energy at the bottom of the loop will be converted into potential energy and the top of the loop. Therefore, at the minimum speed the initial kinetic energy must equal the potential energy at the top of the loop. Do you follow?
     
  4. Sep 6, 2006 #3
    The trick to this is to remember it needs to have some excess velocity (and therefore KE) to get round, it can't have v=0 at the top, else it would fall vertically. The constraint for the particle at the top of the turn is centripetal acc=g.

    Also, the mass cancels.
     
    Last edited: Sep 6, 2006
  5. Sep 6, 2006 #4
    thanks guys i'll hav a go working it out
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Circular motion of a loop-the-loop machine
Loading...