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Loop gravitational roller caoster- height, velocity, breaking force?

  1. Nov 4, 2009 #1
    1. The problem statement, all variables and given/known data

    Gravitational roller coaster makes a circular loop of radius 8 m.
    - What should the minimum height be for the roller coaster to make the loop without falling at the top?
    - What is the velocity of the cart in the point B, after it finishes the loop? (See picture)
    - What is the braking force of the roller coaster after it finishes the loop for it to stop after 10 m? The mass of the roller coaster is 2000 kg.


    2. Relevant equations

    PE(gravitational)= mgh
    KE= ½ mv²

    3. The attempt at a solution

    First part: What should the minimum height be for the roller coaster to make the loop without falling at the top?

    Energy at point A
    PE(gravitational)= mgh
    KE= 0

    Energy at point B
    PE(gravitational)= mg2r
    KE= ½ mv²

    Equlilibrium:

    mgh= mg2r + ½ mv²

    Substitution: v²= gr

    mgh= mg2r + ½ mgr → h= 2r + r/2= 5r/2
    h= 20 m

    Second part: What is the velocity of the cart in the point B, after it finishes the loop?

    PE(gravitational-at top)= KE(at bottom)
    mgr= ½ mv² → gr= ½v²
    v= sqrt(2gr)
    v= 12.65 m/s


    Are my calculations correct?

    I really need a hint for solving the third part:What is the braking force of the roller coaster after it finishes the loop for it to stop after 10 m? The mass of the roller coaster is 2000 kg. [/SUB]
     
    Last edited: Nov 4, 2009
  2. jcsd
  3. Nov 4, 2009 #2
    I forgot to attach the picture.

    http://www.slide.com/s/ULHvFJZS4z8EtEhA1ODWVdhI7p2Jkwm3?referrer=hlnk [Broken]
     

    Attached Files:

    Last edited by a moderator: May 4, 2017
  4. Nov 4, 2009 #3
    I came to some conclusions and I really hope they are correct.

    Third part: What is the braking force of the roller coaster after it finishes the loop for it to stop after 10 m? The mass of the roller coaster is 2000 kg.

    v(initial)= 12.65 m/s
    v(final)= 0 m/s

    a= v(final)-v(initial) / t

    s= v(initial)*t → t= s/v(initial)

    Substitution:

    a= v(final)-v(initial) / s/v(initial)
    a= -16 m/s

    F=ma= 2000 kg*(-16 m/s²)= -32005 N

    Are my conclusions correct?
     
  5. Nov 5, 2009 #4
    Can someone please take a look into this problem? I would really like to know if I solved it correctly?:confused:
    Thank you!:smile:
     
  6. Nov 5, 2009 #5

    rl.bhat

    User Avatar
    Homework Helper

    Your last part is not correct.
    The initial velocity is given. Final velocity is zero.
    Displacement is given.
    Using vf^2 - vi^2 = 2*a*s, find a.
     
  7. Nov 5, 2009 #6
    Thank you!

    v(final)² - v(initial)²= 2as
    a= v(initial) ²/ 2s
    a= 8 m/s²

    F= ma
    F= 2000 kg* 8 m/s²= 16, 000 N

    Is this now correct?
    What about the first two parts? Are my calculations correct?
    Thank you for helping!
     
  8. Nov 5, 2009 #7

    rl.bhat

    User Avatar
    Homework Helper

    Yes. Other two calculations are correct.
     
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