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Conservation of Energy & Circular motion

  1. Dec 3, 2005 #1
    Hey there!

    I'm having trouble with the following problem and I will appreciate it very much if you could help me out.

    A rollercoaster with mass m slides frictionless off a path with initial height h. At a certain point it makes a looping with radius r.

    [​IMG]

    a) Give an expression for the height h if the rollercoaster just makes the looping.

    b) For safety reasons the height is chosen in such a way, so that it is 3/2 of the size of the minimum height. What's the velocity of the rollercoaster in the highest point of the looping?

    c) In which direction is the acceleration pointing in the highest point of the looping when the track is given the restrictions of question b? And what's it's magnitude?
     
  2. jcsd
  3. Dec 3, 2005 #2

    siddharth

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    Where exactly are you having trouble in this problem?
     
    Last edited: Dec 3, 2005
  4. Dec 3, 2005 #3
    Actually I need a start with all of the questions :$.....
     
  5. Dec 3, 2005 #4

    siddharth

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    Ok, let's take the first question. It asks for the height if the roller coaster just completes looping. So what do you understand by this?

    Let the roller coaster be at any point on the loop. What forces are acting on the roller coaster? What forces are acting on the roller coaster when it is just falling off? From this, can you figure out what condition is satisfied when the roller coaster starts to fall off?

    Next, if the roller coaster just completes the loop, what should the velocity be at the top of the loop? (Hint: Use the condition you obtained in the previous part)

    Express this velocity in terms of the initial height (Using the Conservation of Energy). Once you do this, you can find the value of h for the roller coaster to just complete the loop
     
  6. Dec 3, 2005 #5

    daniel_i_l

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    About question c)
    in what direction is a mass accelerating if it is in circular motion?
     
  7. Dec 3, 2005 #6

    BobG

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    a) The centripetal acceleration has to equal gravitational acceleration. Since 'r' is pretty much unchangeable, that will drive your value for v. This problem requires you to express the velocity at the top of the loop in terms of 'r'. The velocity (in terms of 'r') can then be used in your equation for the work-energy theorem.

    After you've gotten part a, I think you'll have the general idea and be able to figure out the last two on your own.

    (By the way, interesting that they would say 'for safety reasons' in part b. If you calculate the forces on the person at the bottom of the loop, you'll see that a person would be extremely uncomfortable - which is why real roller coasters always have more of a corkscrew shape than a true loop.)
     
  8. Dec 6, 2005 #7
    For part a:

    The centripetal force is the gravitation force when the rollercoaster just makes the looping. Therefore ac= g and g= v2/r
    Therefore v2= g r

    Before the rollercoaster makes the looping it has only gravitational energy and in the looping it has gravitational energy + kinetic energy.

    Therfore

    mgh= mg(diameter looping)+ 0.5mv2
    gh= g2r+ 0.5 v2
    gh= g 2r + 0.5 g r
    h= 2r + 0.5 r = 2.5 r

    Correct?
     
  9. Dec 6, 2005 #8
    For part b:

    The height is now 3/2 h= 3/2 * 2.5 r= 7.5 r

    mgh= mg 2r + 1/2 mv2
    gh= g 2 r + 1/2 v2
    g 7.5 r= g 2 r+ 1/2 v2
    g 14 r= g 4 r + v2
    v2= g 10 r
    v= sqrt (g 10 r)

    Correct?
     
  10. Dec 6, 2005 #9
    C is very simple indeed:

    The acceleration points to the center of the looping, because the rollercoaster does a circular motion. ac= v2/r

    Calculated in b is v= sqrt (g 10 r), therefore
    ac= g 10 r/ r = 10 g

    That's right huh?
     
  11. Dec 6, 2005 #10

    BobG

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    Double check your arithmetic. 3/2 * 2.5 r = 3.75 r.

    Other than that, you've got the idea.
     
  12. Dec 7, 2005 #11
    Whoops! Of course :$ Thanks for checking & helping me out :D
     
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