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Homework Help: Dart strikes lead sphere

  1. Feb 26, 2009 #1
    1. The problem statement, all variables and given/known data

    A 16.00 lead sphere is hanging from a hook by a thin wire 3.20 long, and is free to swing in a complete circle. Suddenly it is struck horizontally by a 5.50 steel dart that embeds itself in the lead sphere.

    What must be the minimum initial speed of the dart so that the combination makes a complete circular loop after the collision?

    2. Relevant equations

    3. The attempt at a solution

    1) i've try equating T = mv^2 / r = mg when the sphere reaches the top of the swing.

    v=5.6 as r = 3.2 and m cancels out.

    2) then initial p = final p
    5.5 v = (16+5.5) (5.6)
    i'll get initial v of 21.9m/s for the dart but its wrong.

    can anyone tell me what's wrong? or should i just the mgh PE at the top approach?
  2. jcsd
  3. Feb 26, 2009 #2


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    One issue is that just getting to the top will not make the sphere travel in a complete circle. The lead sphere must reach the top with enough velocity so that its centrifugal force overcomes gravity. Otherwise the thin wire will buckle and the lead sphere drop before you get to the top.

    Even though it has enough kinetic energy to reach the top not all of that will convert to potential energy and the path will look like the letter 'G' on its back .

    There is a difference between the mass on a wire and a mass on a light rigid rod.
  4. Feb 26, 2009 #3
    ok, but any idea how i can approach this problem??
  5. Feb 26, 2009 #4

    Doc Al

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    Good! Be careful how you write it though. While mv^2/r = mg, that's not the tension. (I assume T stands for tension; T = 0.)

    That's the minimum speed of the "sphere+dart" at the top of the swing. Now figure out the speed at the bottom of the swing, where the collision takes place. (What's conserved as the sphere swings up?)

    It's wrong because you're using the speed at the top instead of at the bottom.

    This part was done correctly in the first step. (Except for the "T =" part.)
  6. Feb 26, 2009 #5
    hi thanks thanks so much for the reply. i understand that for the first part, it should be T+mg = mv^2/r , but T= 0 as assumed.

    but for the 2nd part,
    i'm using v=5.6 at the top for final p. i don't get what you mean by that i should use the speed at the bottom instead. isn't that what i'm supposed to find(speed at the bottom)?
  7. Feb 26, 2009 #6

    Doc Al

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    You are supposed to find the speed of the dart before the collision. To do that, you need the speed of the "sphere + dart" after the collision. You can find that post-collision speed from the speed at the top of the swing.

    What happens to the speed of the object as it swings up?
  8. Feb 26, 2009 #7
    oh!! i kind of get what you meant. so now, i have post collision at top = 5.6m/s

    1) at the top, it has k.e. of 0.5(16+5.5)(5.6)^2 and p.e. of (16+5.5)(9.8)(3.2*2)
    2) solve for v using k.e. at bottom of 0.5(16+5.5)v^2, p.e. =0
    3) i'll get v at bottom of 12.52m/s
    1) using p(initial) = p(final)
    5.5 * v = 12.52(16+5.5)
    v = 48.95 for initial speed of dart

    hope i'm finally correct?? :)
    :p you're so kind to offer your help. really appreciate your precise and prompt replies.
  9. Feb 26, 2009 #8

    Doc Al

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    Perfectly correct! :approve:

    (And you are most welcome.)
  10. Feb 26, 2009 #9
    thank you so much once again. :)
  11. Mar 4, 2009 #10
    Why the initial momentum of the lead sphere= 0?
  12. Mar 4, 2009 #11

    Doc Al

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    Presumably it's just hanging there, initially at rest.
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