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Toy car loop physics problem

  1. Feb 25, 2007 #1
    A toy car rolls down a hill at with a height of h and goes through a loop with a radius r. What is the height the toy car has to start off from rest at, if everything is frictionless and is able to make it throught the loop.

    I believe the answer it 2r because of the energy involved. If everything is frictionless, then the toy car must start from the height of the top of the loop. My professor wants equations, but i don't know how to break this one down. Thanks for the help.
     
  2. jcsd
  3. Feb 25, 2007 #2

    Doc Al

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    That would give the car zero speed at the top of the loop--which is not enough for it to maintain contact. (It would have fallen away from the track before it made it to the top.) Hint: Consider the forces acting on the car at the top of the loop and the acceleration of the car.
     
  4. Feb 25, 2007 #3
    Ok, that makes sense. so h>2r? So how would i achieve that answer by using equations. Gravity is acting on the car at the top of the loop, and a normal force.
     
  5. Feb 25, 2007 #4

    Doc Al

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    Apply Newton's 2nd law.
     
  6. Feb 25, 2007 #5
    Fnet=ma???
    ok....i know the formula. How do i apply it though. I don't have mass, i don't have acceleration. What are you talking about. I'm dealing with potential energy.
     
  7. Feb 25, 2007 #6

    Doc Al

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    Energy methods alone will not solve this problem. Another hint: What kind of motion is the car undergoing as it loops the loop?
     
  8. Feb 25, 2007 #7
    + and - motion?? i don't understand what your asking.
     
  9. Feb 25, 2007 #8

    Doc Al

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    Yet another hint: It's going in a circle!
     
  10. Feb 25, 2007 #9

    yeah...ok. I know it's going in a circle which is why i'm given a radius=R. I don't know where you are going with this.
     
  11. Feb 25, 2007 #10
    The sum of the forces must keep the car in a circle. But what is the minimum force acting at the top of the loop?
     
  12. Feb 25, 2007 #11
    The normal force
     
  13. Feb 25, 2007 #12
    how do you get that normal force?
     
  14. Feb 25, 2007 #13
    By the gravitational force or weight. But in this case. Wouldn't N be just barely over zero or zero because both of the forces are pointed downward?
     
  15. Feb 25, 2007 #14
    well, if you had the car upright on a flat surface, yes the normal force would be the one that's caused by gravity

    but in this case, the car is upside-down, if gravity wants to pull it off the track...what is resisting that?

    certainly not the weight of the car, right?

    you can't have N = 0 if you want the car to stay on the track, otherwise it's just in freefall, right? if gravity is the only thing acting on it...it'll do what it wants

    so, something is causing a force here...read through the previous posts (doc al's hints) carefully

    and keep Newton's second law in mind
     
    Last edited: Feb 25, 2007
  16. Feb 25, 2007 #15
    I'm doing this same problem, and have approached it like this-

    At the beginning, the car has potential energy = mgh. At the bottom, this potential has been converted into kinetic energy, 1/2mv^2.

    1/2mv^2=mgh, or, 2mgh=mv^2.


    During the loop, the car will undergo circular motion, where the net forces are equal to = (mv^2)/r. At the very top of the loop, the car will only feel two forces- normal force and the force of gravity.

    So...

    N + mg = mv^2/R

    Subsititue 2mgh in for mv^2, which gives us

    N + mg = 2mgh/R

    N=2mgh/R - mg

    N=(mg)(2h/R - 1)

    Now, this is where I am not positive what I'm doing is legal- I rewrite normal as mg

    mg = (mg)(2h/R-1)

    1 = 2h/R - 1
    2= 2h /R
    2R=2h
    r=h


    Which would mean that the height only needs to be as tall as the radius of the loop? Am I to assume, then, that inertia is what carries the car through the rest of the loop? Or have I made a silly mistake?
     
  17. Feb 25, 2007 #16
    I'm fairly positive you have made a silly mistake. I know the answer is 5/2R but i'm trying to figure out why. The answer is not just R. If the height was only half of the loop, do you actually think it would make it though the whole loop?? doubt it.
     
  18. Feb 25, 2007 #17
    what is the N in this equation?
     
  19. Feb 25, 2007 #18
    N was meant to stand for the normal force.
     
  20. Feb 25, 2007 #19
    The way you set up the problem, h is the height above the top of the loop, which makes the total height 3R. Youre wrong, though, and not with the algebra. The minimum centripetal force is not N + mg
     
  21. Feb 25, 2007 #20
    I guess I don't see what other forces are acting on the car at the very top of the loop. It would seem at the top, the point I am looking at, there would only be gravity and the normal force. What other forces would be acting on it at this point?
     
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