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A bead on loop

  1. Jul 29, 2011 #1
    A bead slides without friction around a loop-the-loop. The bead is released from rest at h=3,5R.

    a) What is the speed at point A?

    b) How large is the normal force at point A if its mass is m=5g?


    -------------------------------------------------------

    My question for:

    a) Do we consider the loop to have mass m=0 and the bead on it m=5g?

    b) If there is no friction at all, are we sure that the loop is not rotating?




    Fig: http://tinypic.com/view.php?pic=dr9tv&s=7
     
  2. jcsd
  3. Jul 29, 2011 #2
    The loop would presumably fixed, so it's mass shouldn't affect your calculations; use energy conservation to find the velocity at the top point. Same goes for friction. If there were any friction, the velocity of the bead at the top would be different.
     
  4. Jul 29, 2011 #3
    I assumed that the loop was fixed and used energy conservations. And i got the correct result.

    But the book reveals the answer and shows that the normal force is calculated as:

    N+mg=(v^2/R)*m


    But this means that the loop is rotating at the speed calculated in a.
     
  5. Jul 29, 2011 #4
    No, what the book is doing is writing the equation for the centripetal force, which keeps the bead moving in a circle. The equation is for the forces on the bead, not the loop.
     
  6. Jul 29, 2011 #5
    But R is the radius of the loop. Which implies that the loop is in rotation?
     
  7. Jul 29, 2011 #6
    no the loop is just to fix bead at a constant R. If there is no friction the loop does not affect the bead and if the loop has 0 mass it has no classical force
     
  8. Jul 29, 2011 #7
    The R refers to the radius with which the bead is executing circular motion. The formula [itex]\frac{mv^{2}}{R}[/itex] applies to circular motion of an object, and not what structure it is executing it on.
     
  9. Jul 30, 2011 #8
    Ok lets put this clear.

    The loop has speed v. And is not rotating. So the bead has the same speed and direction as the centre of the loop?

    How can there be centripetalacceleration in this case?
     
  10. Jul 30, 2011 #9
    The centripetal acceleration is the acceleration needed to change the direction of the bead. It is given by [itex]\frac{v^{2}}{R}[/itex] The direction of the bead's acceleration is radially inward to the circle's centre while the velocity is tangential to the circle at the point of contact.
     
  11. Jul 30, 2011 #10
    Okay, i think i got it now. Thanx!
     
  12. Jul 30, 2011 #11

    PeterO

    User Avatar
    Homework Helper

    A "loop-the-loop" is a common part of many modern Rollercoasters at amusement parks [though often the loop is not a perfect circle there].

    Also it is part of the track in some model car kits [Fast-Wheels?]

    I think you concentrated on the circular loop in the diagram, not noticing the "loop-the-loop" description in the thread of the question
     
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