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Classic Sled on a Hill Problem, Need a Little Help

  1. Aug 4, 2015 #1
    1. The problem statement, all variables and given/known data
    We are trying to solve a problem where, we'll call him John, is being pushed down the hill too fast by Mary and when he goes back up the next hill he ends up flying up off the ground. We are trying to solve where, on the graph (coordinates) he lands. Here is what I know about the problem.
    He starts at the origin.
    Mary is pushing him at a velocity of 5 m/s.
    He starts at a standstill.
    Friction is to be ignored.
    His mass is only 1 kg and the sled is to be considered mass-less. Apparently for simplicity's sake?
    It's on earth so gravity is of course 9.81 m/s squared.

    2. Relevant equations
    Equation of the hill is y(t)= -sin(x(t)).

    3. The attempt at a solution
    I'm kind of lost, and I don't pretend to know much about math and physics, but this problem has piqued my interest and I'd like to know how to solve it.

    Thanks!
     
  2. jcsd
  3. Aug 4, 2015 #2
    Also, I don't know if this belongs here D: Sorry if it doesn't!
     
  4. Aug 4, 2015 #3
    Also, I don't know if this belongs here D: Sorry if it doesn't!
     
  5. Aug 4, 2015 #4

    billy_joule

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    Those two statements cannot both be true.
    John cannot both be at a standstill and be going 5 m/s.
    Maybe Mary pushes with a constant force of 5 N?

    Without doing any math, I assume, if John does become airborne, he'll do so at the next inflection point, when x = pi.
    From there, standard projectile motion applies.
     
  6. Aug 4, 2015 #5
    I think the standstill thing wasn't right. Thanks for responding! Uhm, how do i know the velocity of the sled at the inflection point? How would i calculate that?
     
  7. Aug 4, 2015 #6
    Plus she also stops pushing at the bottom of the hill
     
  8. Aug 4, 2015 #7

    billy_joule

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    Conservation of energy is probably the easiest method.

    If John starts from a standstill and isn't pushed then his velocity will be zero at the inflection point. Like a pendulum, it'll can't swing higher than it's starting point.
    If you want to know the velocity with pushing, then you need to specify the push force. The additional energy from the work done by the push force will be equal to the kinetic energy at the inflection point.

    Then Wpush = 1/2 mv2
    and
    Wpush = Fd

    Where
    d is the arc length of sin from 0 to pi/2 (the distance over which Mary pushes)
    F is the push force
    m is the sled mass
    v is the velocity at (pi, 0) with direction tangential to -sin(pi).

    I would wait for someone to confirm though, I'm not 100% confident on this approach :wink:
     
  9. Aug 4, 2015 #8
    woah that's a lot haha... uh what is the correct equation for arc length? i know i'm close with ds = intrgl sqrt(1 + (dy/dx)) ^2 dt, but i can't quite finish haha.

    then how do i find the v in there? I'm really confused sorry xD
     
  10. Aug 5, 2015 #9

    haruspex

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    There's a problem with the profile equation. y = sin(x) doesn't mean anything here unless the units for x and y are specified. Can we assume metres?
    Your integral for arc length should be wrt x, not t. Given y as a function of x, what is dy/dx here? Plug that into your integral. The solution is difficult. Cheat: https://en.m.wikipedia.org/wiki/Sine#Arc_length
    It won't leave the ground at the inflexion point, but some point later. You'll need to calculate where that is. What forces and accelerations will apply at that point?
     
  11. Aug 5, 2015 #10
    Yeah meters. All in all what i'm trying to figure out is this. You have the equation y(t)= -sin(x(t)), and there is a sled at the origin with John on it being pushed at 5 m/s by Mary. There is no friction, he weighs 1 kg and the sled weighs nothing, and she stops pushing at the bottom of the hill. What i want to know is, after he is done being pushed and starts going up the next "hill", and the sled flies into the air because of said forces, where does it then reconverge and "land" back on the y(t)= -sin(x(t)) graph.
     
  12. Aug 5, 2015 #11

    haruspex

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    Ok, but as has been pointed out, it doesn't make sense to say that he is pushed at a steady 5m/s given that he is also sliding down hill.
    And I repeat, before you can determine where he lands, you have to find out where he takes off. What can you say about the forces and accelerations at that point?
     
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