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Rock Problem

  1. Jun 19, 2009 #1
    1. The problem statement, all variables and given/known data
    A person standing at the top of a hemispherical rock of radius R kicks a ball (initially at rest) to give it a horizontal velocity vi.
    a. What must be the rock’s minimum speed if the ball is never to hit the rock after it is kicked?
    b. With this initial speed, how far from the base of the rock does the ball hit the ground?


    2. Relevant equations
    Projectile Motion


    3. The attempt at a solution
    I tried my way a couple times but the answer I get is wrong. Ok I tried to find the function of the curve of the rock. Setting y=0 at the top of the rock, I get y=-(gt²)/2. And vt=R. Solving for t when y=-R t=√(2r/g) and v=(√Rg)/2. Is this not vmin? I even checked it by putting vmin back into the y as function of x. Then I put something bigger back into that same function and y is greater for vmin. This means that the ball is always higher than the rock if kicked greater than (√Rg)/2. The answer key says vmin=(√Rg). What did I do wrong?
     
  2. jcsd
  3. Jun 19, 2009 #2

    LowlyPion

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    To not touch the rock doesn't the initial velocity need to satisfy

    m*v2/R > m*g

    Armed with that velocity which is horizontal, and the time to fall determined by R, then you know how far away.
     
  4. Jun 20, 2009 #3
    Um don't know about m yet. Can you explain what I did wrong?
     
  5. Jun 20, 2009 #4

    LowlyPion

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    m cancels out if you might notice.

    That expresses the contact relationship, which is that it will lose contact when the centripetal acceleration is greater than gravity.
     
  6. Jun 20, 2009 #5
    LowlyPion, losing contact isn't the question here, the ball is kicked and loses contact at t=0
    Forces are completely irrelevant. Though your approach looks interesting, I can't quite make heads or tails of it.
    Ooh, I see now. My approach is much more time-consuming and yours gives the solution straight away. @@

    The approach I'd choose here would be kinematics and analytic geometry.
    Do you know how to construct the y=f(x) equation representing a ballistic trajectory, and the y=f(x) equation representing a hemisphere?

    The latter requires a tiny bit of knowledge to develop on your own, so if you get stuck, here's a spoiler, but knowing how to develop the former is critical to all such problems.
    The equation representing a circle of radius R is:
    R²=x²+y²
    Can you transform this into a function of the form y=f(x)?
     
    Last edited: Jun 20, 2009
  7. Jun 20, 2009 #6

    LowlyPion

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    Consider a ball on a string that is cut exactly at the top of the sphere. If the speed v at t=0 meets the condition that v2/R > g, then it will not touch the sphere now will it. If the speed is less than that, then were it to continue, since the tension is not sufficient to keep it taught, it would necessarily come in contact, in fact be in contact, with the sphere at the top.
     
  8. Jun 20, 2009 #7
    Heh, 100% correct, the exact result my geometry got me. Could you please elaborate on why this is correct, please?
     
  9. Jun 20, 2009 #8

    Redbelly98

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    LowlyPion is using the centripetal force for circular motion:

    [The force on a mass constrained to move on the sphere's surface at constant velocity v]​

    is greater than

    [The actual force mg]​

    Note: if the OP is now learning about uniformly accelerated motion, then centripetal force & acceleration has probably not been covered in class yet.
     
  10. Jun 20, 2009 #9

    Redbelly98

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    When x=vt=R, y is not -R.

    That would be true if the ball were constrained to move along the sphere's surface, but it does not do that. It will be above the surface, so y is somewhere between 0 and -R when x=R.
     
  11. Jun 20, 2009 #10
    ....
     
  12. Jun 20, 2009 #11

    LowlyPion

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    As a side note, this is another problem in the series he was dealing with earlier.
    https://www.physicsforums.com/showthread.php?t=320363

    I should hope he is learning about centripetal acceleration by now.
     
  13. Jun 20, 2009 #12

    LowlyPion

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    That can only be a typo.
     
  14. Jun 20, 2009 #13

    Redbelly98

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    Okay, yes, that is clearly a centripetal acceleration problem so it has been covered.
     
  15. Jun 20, 2009 #14
    I can see how when:
    v²/R > g
    it follows that the body loses contact with the hemisphere (From the accelerated frame, the net force points up).

    But how does that show that it won't return to the hemisphere? I know that v²/R > g is the correct answer, from my own calculations, but I don't quite see how the above explanation of losing contact works for the rest of the motion.

    Thanks Redbelly98, I'll have to read up on curvatures to see what you mean, exactly. Hopefully the wiki article will be enough. :s
     
    Last edited: Jun 20, 2009
  16. Jun 20, 2009 #15

    Redbelly98

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    For a parabola, the path's curvature gets straighter with time. Whereas the circular path maintains a constant curvature.

    If the curvatures were equal initially, the parabola always has less curvature for t>0. Therefore it can never meet the circle again.
     
  17. Jun 20, 2009 #16

    LowlyPion

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    The first part is not necessarily true. g = v²/R is merely the condition under which the ball exerts at most 0 contact force on the surface at any point around the sphere. The top of course is where gravity acts normally into the surface, so like New York, NY, if it won't touch it there, it won't touch it anywhere.
     
  18. Jun 20, 2009 #17
    I am sorry I am not following along with any of the force or mass stuff. I know I am close but what did I do wrong?
     
  19. Jun 20, 2009 #18
    I prefer the problem where the rock is moving.
     
  20. Jun 20, 2009 #19
    Ok you if you want the rock to be moving go ahead. Still what did I do wrong? I am going crazy over this problem!
     
  21. Jun 21, 2009 #20

    ideasrule

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    "And vt=R. Solving for t when y=-R t=√(2r/g) and v=(√Rg)/2. Is this not vmin?"

    That's the problem: by using vt=R, you're assuming that at vmin, the ball touches the hemisphere when it gets to the ground. It can't; it must land a certain distance away from the hemisphere, and part b) of your question asks you to find this distance.
     
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