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Projectile motion on a hemisphere

  1. Mar 11, 2012 #1
    1. The problem statement, all variables and given/known data
    A person standing on the top of a hemispherical rock of radius R kicks a ball (initially at rest on the top of the rock) to give it horizontal velocity [itex]v_i[/itex]
    What must be it's minimum initial speed if the ball is never to hit the rock after it is kicked?


    2. Relevant equations
    [itex]v=v_i+at,\\ v^2=v_i ^2 + 2ar, \\ r=v_i t +\frac{1}{2}at^2[/itex]


    3. The attempt at a solution

    I'm not sure how to do this one as a parabola and a semicircle are 2 different shaped curves, a little nudge in the right direction would be helpful, thanks guys.
     
  2. jcsd
  3. Mar 11, 2012 #2

    tiny-tim

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    Hi Bread18! :wink:
    Write out the two equations, and see where they intersect …

    what do you get? :smile:
     
  4. Mar 11, 2012 #3
    Hmm ok...
    Well the eqn of the semicircle is [itex]y=\sqrt{R^2 - x^2}[/itex]

    The eqn of motion is [itex]y = \frac{1}{2}at^2 \\ x = v_it \\ y = \frac{1}{2}a(\frac{x}{v_i})^2 \\ 2yv_i^2 = a x^2?[/itex]
     
  5. Mar 11, 2012 #4

    tiny-tim

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    = a(R2 - y2) ? :wink:
     
  6. Mar 11, 2012 #5
    [itex]ay^2 + 2yv_i^2 - aR^2 = 0 \\ 4v_i^4 +4a^2R^2 < 0 \\ v_i^4 < -a^2R^2[/itex]

    and that's not right...
     
    Last edited: Mar 11, 2012
  7. Mar 11, 2012 #6

    tiny-tim

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    after that, you've lost me :redface:
     
  8. Mar 11, 2012 #7
    Solve the quadratic, they don't touch so the discriminate needs to be < 0 (typo in the other post, I'll fix it)
     
    Last edited: Mar 11, 2012
  9. Mar 11, 2012 #8

    tiny-tim

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    ah, sorry, i'd forgotten what the question asked for! :redface:
    yes, that's fine :smile:
    remember, your "a" was negative! :wink:
     
  10. Mar 11, 2012 #9
    Yes but it's a^2, so it cancels out the negative, leaving me with -g^2R^2?
     
  11. Mar 11, 2012 #10

    tiny-tim

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    hmm … you're right! :redface:

    ok, let's go back and check your original equations …
    ah! should be y = R + 1/2 at2 ! :smile:

    (no wonder it couldn't avoid hitting the circle!! :biggrin:)
     
  12. Mar 11, 2012 #11
    Haha well spotted :smile:

    So, now with that fix, I get [itex]2v_i^2(y-R)=ax^2 \\ 2v_i^2(y-R)=a(R^2 - y^2) \\ 2v_i^2 = -a(R+y) \\ v_i^2 = \frac{g}{2}(R+y)[/itex]
     
  13. Mar 11, 2012 #12

    tiny-tim

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    looks good! :smile:

    (but on my recent performance, i could be missing something! :blushing: :rolleyes:)
     
  14. Mar 11, 2012 #13
    Haha yeah, we've all been missing simple things..:rolleyes:
    I don't see how this ties in with it not hitting the semi circle though.
     
  15. Mar 11, 2012 #14

    tiny-tim

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    because …
    gives you the y-coordinate where it hits the circle

    for example, if v = 0, the ball drops straight down, and hits the circle at y = -R !

    ok, now what happens to y as you increase v ? :wink:

    (btw, the other solution, y = R for any value of v was eliminated from the equation when you divided it by (y - R) :wink:)
     
  16. Mar 11, 2012 #15
    So I want y = 0?
     
  17. Mar 11, 2012 #16

    tiny-tim

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    in a moment, i'm going out for an hour or so

    draw yourself some diagrams of a parabola touching and cutting a circle, and see what happens as the parabola changes shape :wink:

    (also, think how many times can a parabola touch and cut a circle?)
     
  18. Mar 11, 2012 #17
    Ok, thanks for your help, I think I'm going to go to bed now (2am here). Hopefully When I wake up it'll all become obvious...
     
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