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Can someone please help me with this problem?

  1. Aug 14, 2005 #1
    Can Someone please help me with this problem! I need some hints how to solve this problem, i am already stuck on part A :(.
    This is my practice assignment for the final coming up next 2 weeks. Unfortunately, the professor does not provide solution.

    THE PICTURE OF THIS PROBLEM IS ILLUSTRATED BY THE LINK BELOW.

    Starting from the initial position (the origin of the coordinate system) a cart with a mass of 2000 kg is accelerated from the initial velocity zero by acceleration a1 which is uniform over a segment of 40 m. Following this segment, the cart proceeds through a turn (20 m radius), is then accelerated again over a segment of 10 m length, coasts toward the top of a hill where its velocity is near zero. The cart then rolls down to ground level, through a circular valley (radius 10 m) and along a straight segment of 2 m length, before leaving the track in free fight, initially inclined at 45 degree. In order to ensure a safe landing, an angular impulse is exerted on the cart just before leaving the track, causing the cart to rotate by a constant angular velocity w, during the flight. This rotation is necessary to have the cart axis aligned with the track at the landing point. Departure and landing occur at the same vertical position. On a short segment of 5 m the cart is decelerated by a3, before coasting through the second turn (20 m radius) and it is further decelerated uniformly (a4) over the final 30 m to come to a full stop at the origin of the coordinate system.

    Determine the following design parameters:
    a) Appropriate accelerations a1, a2, to meet the following requirements:
    _ The cart must come to a stop on top of the hill.
    _ The magnitude of the acceleration through the first turn must not be larger than 3g.
    b) The coordinates of the point where the cart departs into free flight.
    c) The velocity of the cart at the point of departure.
    d) The coordinates of the landing point and the flight path angle (_2) at that point.
    e) The duration of the flight, and the angular velocity w which is required to have the cart aligned with the track at landing.
    f) The velocity of the cart as it reaches ground level (just before the 5m deceleration segment).
    g) Appropriate accelerations a3, a4, to meet the following requirements:
    _ The cart must come to a full stop at the origin of the coordinate system.
    _ The magnitude of the acceleration through the second turn must not be larger than 3g.
    h) The duration of the entire ride.
    *PS. This problem will help you with the final. GOOD LUCK!


    [​IMG]
     
  2. jcsd
  3. Aug 14, 2005 #2

    HallsofIvy

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    What have you done? What have you tried yourself?
     
  4. Aug 14, 2005 #3
    For part A.
    Knowing V0=0, X-X0=40=X1, X2=10
    I have (V1)^2=(V0)^2 + 2*(a1)(X-X0)
    (V1)^2= 2*(a1)*(X1)

    and on top of the loop V = 0

    SO m*g*h = 1/2*m*(V3)^2
    (V3)^2 = 2*g*h
    (V3)^2 = V2^2 + 2(a2)*(X2)

    V1 = V2 (since, it's circular uniform)
    Ac = V^2/R

    I am stuck here, i dont know how to find the a2 and a1, and V2. I think, im missing something, i have 2 equations with 3 unknows..but i tried for many hours and still couldn't figure it out.

    Please help me out, thank you so much.
    Thank you
     
    Last edited by a moderator: Aug 14, 2005
  5. Aug 14, 2005 #4

    Fermat

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    Your two eqns in 3 unknowns reduce down to one eqn in two unknowns - a1 and a2.
    So, theorectically, you can have an inifinite number of solutions!

    Practically, however there are a number of constraints. One of them being that the centripetal accln must be no greater than 3g.
    This puts a limit on possible values for a1
    Since the number of choices for a1 and a2 are still infinite, your problem now is to select two values appropriate to your design of an amusement park joy-ride.
     
    Last edited: Aug 14, 2005
  6. Aug 14, 2005 #5
    But, is my concept correct for part A? I feel like i didn't do it right. B/c, i do think that Ac= V^2/R still have not use yet. I believe it has something to do with the V2 somehow.
     
  7. Aug 14, 2005 #6
    Oh, i have thought about the 3g think.
    so the acceleration through the first turn must not be larger than 3g.
    SO

    A = 3g =< sqrt(At ^2 + Ac^2)

    but At = (V1)^2 / (2*X1)
    Ac = (V1)^2/ R

    solve for V1, and V1=V2

    Does this make sense?
     
  8. Aug 14, 2005 #7

    Fermat

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    Sorry, but not yet

    I imagine that At is the accln called a1 in your attachment, for the motion along the 1st 40 m. But the eqn doesnt seem to make sense!

    You should have two eqns of motion - one for the 40m stretch and one for the 10m stretch. Have you got those?
     
  9. Aug 14, 2005 #8

    Fermat

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    Ah yes, of course you have, looking back at post #3.
    What you need now is.

    Ac = V^2/R

    what you need now is a value for Ac, then eliminate V2 (or v1, depending on how you used V1 = V2)

    I's up to you to choose a suitable value for Ac.
     
  10. Aug 14, 2005 #9
    So you're saying this does not make sense at all?

    A = 3g <= sqrt(At ^2 + Ac^2)

    but At = (V1)^2 / (2*X1)
    Ac = (V1)^2/ R

    then substitute At, and Ac back into
    A = 3g <= sqrt(At ^2 + Ac^2)

    solve for V1, we know that V1=V2

    SO on top of the hill,
    m*g*h = 1/2*m*(V3)^2
    (V3)^2 = 2*g*h
    (V3)^2 = V2^2 + 2(a2)*(X2)

    Solve for a2, and a1 can be found using (V1)^2 = 2*(a1)*(x1)
     
  11. Aug 14, 2005 #10

    Fermat

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    Are you saying that At is the same as a1 ?
     
  12. Aug 14, 2005 #11
    well that is my assumption, a1 = at (the tangential acceleration when it enters the curve)

    But, if the cart is decreasing at a constant speed. Then, can it be that the tangential acceleration = 0 (for the whole turn)?

    So then a = Ac=V1^2/R

    and with the givin "a" must not be greater than 3g

    so 3g >= V1^2R

    Solve for V1, and V1=V2.

    :( i dont know, i'm confused now.
    If anyone know how to solve this part please help out.
     
    Last edited by a moderator: Aug 14, 2005
  13. Aug 14, 2005 #12

    Fermat

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    The cart accelerates through the 1st 40m at a1.
    It then travels at constant velocity around the turn.
    It then accelerates at a2 along the 10m stretch.

    I don't think, from reading your very first post about the problem that any (tranverse) accln is involved in travelling around the turn.

    I would say that only radial accln, Ac, is involved, with Ac = V2/r².

    Also, since we are asuming that V2 = V1, then there can't be any tranverse accln.
     
  14. Aug 14, 2005 #13
    I agree,
    V1 = V2 assumption
    So then, a = Ac = V2^2/r

    From here, i must pic a value for Ac ( must be < 3g)

    then solve for V2,

    is this right?
     
    Last edited by a moderator: Aug 14, 2005
  15. Aug 14, 2005 #14

    Fermat

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    substitute for V3, then get a2 in terms of V2, yes.
     
  16. Aug 14, 2005 #15
    Hmm, can you type that out Fermat?

    I am a bit confused here
     
  17. Aug 14, 2005 #16

    Fermat

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    v1² = 2a1.x1 ---------------------(1)

    Ac = v2²/r -----------------------(2)

    v3² = v2² + 2a2.x2 ---------------(3)

    v3² = 2gh (h=30m) ---------------(4)

    v1 = v2 --------------------------(5)

    use (3) and (4) to eliminate V3. Call this eqn (6)

    Substitute for v2² = Ac.r in (6), call this (7)

    Set Ac = 2g, say and solve (7) for a2.

    Solve for a1 and v1 and v2.

    Adjust the value of Ac to give appropriate/good values for the other design parameters. viz a1,a2,v1,v2
     
    Last edited: Aug 14, 2005
  18. Aug 14, 2005 #17
    Thank you for your kindess, Fermat.

    I will continue work on the other parts. If i have questions, is it possible for me to ask you again? I am sorry for asking too many questions
     
  19. Aug 14, 2005 #18

    Fermat

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    no problem.
     
  20. Aug 16, 2005 #19
    Can you guys help me double check part B, C. I don't if this is correct.

    B)

    L1= 2*cos45 + 10*sin45
    h= 10(1-cos45) + 2*sin45

    C)

    T1=0
    T2=1/2*m*Vdepart^2

    Work: U1-2 = mgH'
    H' = 30 - h

    T1 + U1-2 = T2

    mgH' = 1/2*m*Vdepart^2

    Vdepart = Squrt(2gH')


    Thanks
     
  21. Aug 17, 2005 #20

    Fermat

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    Yes. Those results, for L1 and Vdepart are correct.

    I assume you're using the symbols T1 and T2 to mean the kinectic energy at the top of the hill and at the take-off point.
     
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