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Tarzan is swinging on a vine

  1. Nov 9, 2014 #1
    1. The problem statement, all variables and given/known data
    Tarzan is swinging on the end of a vine, the vine is 6m long and is attached to the top of a tree. Tarzan weighs 90kg. At the highest level, he is 2.25 meters above ground and the lowest level 0 meters above ground.

    Questions: At which point is the centripetal force/acceleration at its greatest and how big is it?

    2. Relevant equations
    F = ma
    a = v^2 / r

    3. The attempt at a solution
    I draw an angle and show his weight (mg), and have drawn relevant lines aswell. See the picture. rOb95JA.png
    Okay, mg is the force downwards. mg is a cathetus, and you can extend the centripetal force so that it is another cathetus and mg is the hypothenuse. Alright. I have marked two angles which are the same. So how do we calculate "centri"? We need to know an angle and one side/hypothenuse. We know one cathetus. So let's call the angle x.

    Tan x = mg / centri. Tan x * centri = mg.

    How do I continue from here? How do I know at which point it is the biggest and how big it is?
     
  2. jcsd
  3. Nov 9, 2014 #2

    Doc Al

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    Not quite sure what you are doing here.

    Start here.
     
  4. Nov 9, 2014 #3
    Hello! Could you give me another pointer? Will do my best to continue from that point.
     
  5. Nov 9, 2014 #4

    Doc Al

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    You are asked to find where the centripetal force and acceleration are greatest. Well, how would you determine the centripetal force?
     
  6. Nov 9, 2014 #5
    Hello, okay I tried it. But when the vine is straight (0m from the ground) isn't r = 0? Which makes it undefined for an acceleration? So it is at greatest when it is at the top because it will have a bigger radius? Am I on the right track to continue?
     
  7. Nov 9, 2014 #6

    Doc Al

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    No. r is the radius of the circular path that Tarzan traces out as he swings. It equals the length of the vine.
     
  8. Nov 10, 2014 #7
    So it is 6 meters at both the lowest and highest point? Since the vine is the radius.
     
  9. Nov 10, 2014 #8

    Doc Al

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    Yes.
     
  10. Nov 10, 2014 #9
    So at both points, the centripetal force is a = v^2 / 6?

    See the illustration. Am I understanding it correctly?
    hN8MCbb.png
    These are both the states. The first state, it is 2.25 m above ground, the vine is 6m and the radius that it spins around is 6m. In the second state, the vine is 6m. There is no radius except a "theoretical" one that points outwards, and it is 6m? How do I continue from here?
     
  11. Nov 10, 2014 #10

    Doc Al

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    Yes. Of course, the v changes as Tarzan swings.

    I'm not sure what those horizontal 6m distances are.

    Tarzan swings in a circle. So draw a circle. Put the center at the top of the vine. The radius of that circle is the vine itself, which is 6 m at all points. (Nothing "theoretical" about it.) Trace out the circular path that Tarzan follows as he swings.
     
  12. Nov 10, 2014 #11
    So.. like this?
    j6ajZK7.png
    But where does the "0 m above ground" and at most "2.25 m above ground" part come into play? How do I determine which has the greatest force?
     
  13. Nov 10, 2014 #12

    Doc Al

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    Better, but he only travels a part of the circle. So mark off the starting point (the highest point that he reaches) and the lowest point. That will show you the path that he travels.

    The bottom of the circle is at ground level. He starts at a higher point than that.

    Use the formula:
     
  14. Nov 10, 2014 #13
    JJ5RLiO.png
    Is that better?

    EDIT: Forgot to thank you for your patience with me.
     
  15. Nov 10, 2014 #14

    Doc Al

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    Much better.

    Now, at what point is the speed (v) greatest?

    You are welcome. Keep going. You'll get there!
     
  16. Nov 10, 2014 #15
    It is bigger on the way down? But how do you know if he began from the top and not the bottom? Because if he started at the bottom and then went upwards there would be a need for a greater velocity/acceleration than on the way down, if I'm not thinking wrong.
     
  17. Nov 10, 2014 #16

    Doc Al

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    Yes, the lower he gets, the faster he gets.

    If he started from the bottom he'd have a heck of a time getting going! From the top, he'd just be able to hang on and start swinging.

    If he started at the bottom, he'd already have to be moving.

    But no matter how he started, he is now swinging back and forth. So at what point is his speed greatest?
     
  18. Nov 10, 2014 #17
    It would be on the way down, and then as he goes back up he will reach a lower height and so on.

    Alright! So now I need to figure out the force when he goes down? But there are two unknowns here, v and a. I know r and m. How do I figure out either of the variables? Any pointers?
     
  19. Nov 10, 2014 #18

    Doc Al

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    You need to find the speed at the lowest point. Hint: Make use of the fact that the maximum height is given.
     
  20. Nov 10, 2014 #19
    Can't I use conservation of energy laws? All potential energy is converted to kinetic energy at the bottom.
    m * g * h = 1/2 m * v2

    I know m, g, h, m and not v. I get v, enter it into a=v^2 / r and I am done. Is it correct?
     
  21. Nov 10, 2014 #20

    Doc Al

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    Excellent!

    Yes. That will give you the centripetal acceleration. Then you can find the centripetal force, if you need to.
     
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