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Pendulum hanging from a roller coaster

  1. May 9, 2010 #1
    1. The problem statement, all variables and given/known data

    In one roller coaster car, a small 0.10 kilogram ball is suspended from a safety bar by a short length of light, inextensible string. The car is then accelerated horizontally, goes up a 30 degree incline, goes down a 30 degree incline, and then goes around a vertical circular loop of radius 25 meters. For each of the four situations described in parts (b) and (e), do all three of the following. In each situation, assume that the ball has stopped swinging back and forth.

    1. Determine the horizontal component Th of the tension of the string in newtons.
    2. Determine the vertical component Tv of the tension in the string in newtons.

    (b) The car is at point B (flat surface) moving horizontally with an acceleration of 5.0 m/s squared.

    (c) The car is at point C and is being pulled up the 30 degree incline with a constant speed of 30 m/s.

    (d) The car is at point D moving down to 30 degree incline with an acceleration of 5.0 m/s squared.

    (e) The car is a tpoint E moving upside down with an instantaneous speed of 25 m/s and no tangential acceleration at the top of the vertical loop of radius 25 meters.

    2. Relevant equations

    F = ma
    At equilibrium, F1 + F2 + F3 = 0
    A squared plus B squared = C squared

    3. The attempt at a solution


    Fg = Tv
    Fg = ma
    Fg = 0.10 * 9.8 m/s^2
    Fg = 0.98 N
    Tv = 0.98 N

    Fh = 0.10 kg * 5.0 m/s^2
    Fh = 0.5 N
    Fh = Th
    Th= 0.5 N

    (c) This is where it gets complicated because there's an angle involved.

    See: http://img52.imageshack.us/img52/9761/physics0001.jpg [Broken]

    Here there's no acceleration, so there's equilibrium. Therefore, the only tension is vertical against gravity. Right?

    Tv= 0.98 N
    Th = 0 N

    (d) Here, there's an acceleration and an angle. I solved it as if there was no angle- does the angle matter?

    Tv= 0.98 N
    Th= 0.5 N

    (e) No acceleration and no angle, but the car is on a circle.

    Tv= 0.98 N
    Th = 0N

    I can't believe that the angle has absolutely nothing to do with anything. Where have I gone wrong?
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 9, 2010 #2

    Doc Al

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    Staff: Mentor



    Sure the angle matters! Apply Newton's 2nd law to the horizontal and vertical directions. What are the vertical and horizontal components of the acceleration?

    No acceleration? :yuck: (It's moving in a circle. What kind of acceleration does it experience? What's the direction and magnitude of the acceleration?)

    (Note: Please resize your images!)
  4. May 9, 2010 #3
    (d) I drew a vector diagram to help. So I've got a right triangle with hypotenuse 5.0 m/s^2 and a 30 degree angle.

    Sin 30 = x / 5.0 m/(s^2) = .5 (NOTE: Calculator set to degree mode)
    x (opposite side) = 2.5 m/(s^2)

    Cos 30 = y / 5.0 m/(s^2) = .866
    y (adjacent side) = 4.33 m/(s^2)

    Now it's just another force calculation, right?

    F = ma
    Fv = 0.10 kg * 2.5 m/(s^2)
    Fv = 0.25 N
    Tv = 0.25 N

    F = ma
    Fh = 0.10 kg * 4.33 m/(s^2)
    Fh = 0.43 N
    Th = 0.43 N

    (e) Oops, forgot about centripetal acceleration! Since the cart is at the very top of the circle, all centripetal acceleration provides vertical tension on the string.

    Fc = m(v^2)/r
    Fc = 0.10 kg * 625 m/s squared / 25 m
    Fc = 2.5 N
    Tv = Fc + Fg
    Tv = 3.48 N

    Thanks for the tips and the resize, Doc Al- it's a big help. Have I gotten it this time around?
  5. May 9, 2010 #4

    Doc Al

    User Avatar

    Staff: Mentor


    Sure, but you must include all forces.

    What about gravity?


    Careful! Fc is the net force on the ball. Set that equal to ΣF.

    You're almost there.
  6. May 9, 2010 #5
    Ahh, that's right, I missed the tension created by Fg in part D. Thanks!

    I'm lost on the net force bit. The cart is at the very top of the track upside down and the only acceleration is centripetal. Since centripetal acceleration is always to the center of the orbit circle and the direction from the anchor point of the ball to the center of the orbit circle is straight down, doesn't it all add to the tension on the string?
  7. May 10, 2010 #6

    Doc Al

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    Staff: Mentor

    You are correct that the only acceleration is centripetal and that it points straight down.
    Just apply Newton's 2nd law and see:

    ΣF = ma

    What forces act on the ball?
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