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Math Help - Swing Problem

  1. #1
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    Swing Problem

    A 16 kg child on a playground swing moves
    with a speed of 9.5 m/s when the 2 m long swing is at its lowest point.

    The acceleration of gravity is 9.81 m/s^2

    What is the angle that the swing makes
    with the vertical when the swing is at its
    highest point? Answer in units of degrees.



    For this problem, i solved for KE first. KE = 1/2mv^2, so plugging in some numbers I get 76 joules. Then I plug this into the PE equation (PE = mgh), so when 76 = (16kg)(9.81)(h) the height I get is .4841997m.

    From this point on, I think I get stuck. I understand that energy is conserved, but how do I find out the angle that the swim makes? I tried using pythagorean's theorem, but that doesn't seem to work because the swing is always 2m long (which gives it a round opposite).
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  2. #2
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    Think this way: sin opposite hypotenuse, cos adjacent hypotenuse,
    tan opposite adjacent

    SOH CAH TOA

    You have enough information to find theta (the angle).

    -Andy
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  3. #3
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    The opposite side is curved, not straight, so when I tried using SOH CAH TOA the answer I got was wrong
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  4. #4
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    I'm not really sure where to go with this one. I tried subtracting the height I got from 2m to create a "new" adjacent angle, and then tried using Cos^-1(1.52m/2m) and I get about 40 degrees or something (i forget) but that answer is incorrect.
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by vesperka View Post
    I'm not really sure where to go with this one. I tried subtracting the height I got from 2m to create a "new" adjacent angle, and then tried using Cos^-1(1.52m/2m) and I get about 40 degrees or something (i forget) but that answer is incorrect.
    You have the right method. (I could have sworn I posted my own diagram here. Ah well.)

    Check your height again. I'm getting h = 4.60 m.

    -Dan

    Edit: I see what went wrong. You didn't square the 9.5 in your kinetic energy equation.
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  6. #6
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    Oh yeah I screwed that up big time. Still stuck though. So now with the new height, how do I go about setting it up? The hypotenuse is still 2m, but if the adjacent side's length is negative it messes up my calculator (maybe I'm typing it wrong :\ )
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by vesperka View Post
    Oh yeah I screwed that up big time. Still stuck though. So now with the new height, how do I go about setting it up? The hypotenuse is still 2m, but if the adjacent side's length is negative it messes up my calculator (maybe I'm typing it wrong :\ )
    I didn't notice that. It does kind of screw the problem up, doesn't it. I'd say this one is FUBARed. Talk to your instructor.

    -Dan
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  8. #8
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    Heh thanks anyways, I'll be sure to give you an update tomorrow once I figure out how to get this solved
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