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Homework Help: Gravitational potential energy of ball

  1. Dec 19, 2007 #1
    1. The problem statement, all variables and given/known data
    A 7.26 kg ball hangs from end of 2.5 m rope. The ball is pulled back until rope makes 45 degree angle with vertical. What is the gravitation potential energy?

    2. Relevant equations


    3. The attempt at a solution
    I thought I would just multiply mass times gravity times height, but the answer is 52 J.
    What do I do with the angle I'm given?
  2. jcsd
  3. Dec 19, 2007 #2
    What are the maximum and minimum vertical heights of the ball?
  4. Dec 19, 2007 #3
    0 m and 2.5 m....?

    I provided the entire problem.
  5. Dec 19, 2007 #4
    In the back of the book it says to find height do
    h=2.5m * (1-cos(45))
    Then plug that into Ug=mgh.
    Why 1-cos(45)?
  6. Dec 19, 2007 #5
    No. The 2.5 m corresponds to the total height of the rope. The ball is not lifted so that it's level with the top of the rope- it's pulled along like a pendulum such that the rope is at 45 degrees to vertical.

    Edit: Think of swinging your arm such that it's 45 degrees to vertical. How far has your hand moved vertically?
  7. Dec 19, 2007 #6
    Ok, I think I understand. It's going to be hard to explain, but in other words, when the ball is in the air and has that 45 degree angle, it's going to "lose" height; it's going to be "shorter" than hanging straight down. So that's why 1 is being subtracted from cos(45).
  8. Dec 19, 2007 #7
    I'm not sure I understand your explanation. At 45 degrees the ball is higher than completely vertical- so I'm not sure why you claim it loses height.

    Try answering my question. If you swing your arm by 45 degrees from vertical how far has your hand moved vertically (upwards)? (You could take the length of your arm to be e.g. 1.1 meters.)
  9. Dec 19, 2007 #8
    i don't know. i'm flustered and confused.
  10. Dec 19, 2007 #9
    OK, calm down!

    This problem involves a little trig. It's best to draw a diagram.

    Draw a vertical line of length, e.g. 2.5m between points A and B (with A at the top, B at the bottom.)

    Next- draw a line between points A and C which is of the same length, but at 45 degrees to the AB line.

    Tell me once you've done this.
  11. Dec 19, 2007 #10


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    Just a comment: to be exact, the question does not make sense because the gravitational potential energy depends on the position of the origin so all answers are possible depending on where you set your origin. Did they mean to calculate the change of potential energy as teh ball is pulled back?
  12. Dec 19, 2007 #11
    kdv: The gravitational potential energy refers to the potential energy that the ball can convert into kinetic energy without cutting the rope. If we allow the rope to be cut- then yes we would have to specify at what height we take as the zero.
  13. Dec 19, 2007 #12


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    But it is still arbitrary even if the rope is not cut. One may define the gravitational potential energy to be zero at th elowest point of the swing or at the ceiling or at any other point. The value mgy never appears isolated in the equations, it's only the combination [itex]mg(y_i-y_f)[/itex] (the work done by gravity) that is relevant. So if the question is worded exactly the way given in the OP and without specifying what origin must be used, then the question is actually meaningless. My bet is that they meant to ask what is the change of potential energy as the ball is pulled from its initial position to its final position. That would make sense as a question because a difference of potential energy is the same no matter where the origin is chosen.
  14. Dec 19, 2007 #13
    This is nitpicking and isn't helping the thread. The potential in this case is clearly the amount that can be completely converted to kinetic energy at the vertical. We're all aware that energy has to be defined with respect to a reference point- but it's absolutely clear in this case what the reference point is! If you want to debate this further then message me.
  15. Dec 19, 2007 #14


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    Sorry. Ten years of teaching college physics has forced me to become careful about making sure things are clear in the students mind before they do a calculation. It did not seem nitpicking to me since it is the very definition of potential energy that it may be defined with respect to any origin. I thought explaining this to the OP would give him/her a better understanding of the concepts (and that tehre is not a unique answer here) but if it's nitpicking, I apologize.
  16. Dec 19, 2007 #15
    Sorry if I was a little harsh.
  17. Dec 19, 2007 #16
    christian-as you explained is how i drew my diagram in the first place, so it's done.
  18. Dec 19, 2007 #17
    OK, good. Now for the evil next step.

    Draw a line horizontally from C so that it intersects the AB line. Tell me when you're done.
  19. Dec 19, 2007 #18
    Your diagram should look like a triangular flag on a post. Does it?
  20. Dec 19, 2007 #19
    forgot to mention, i drew the horizontal line previously also.

    now, i'll apply trig-->
    2.5(cos45)=1.7678 and that's the height from A to the horizontal line that intersects AB
  21. Dec 19, 2007 #20
    OK. label 'D' the intersection of the horizontal with AB

    So AD=2.5 cos(45)

    But what is the vertical height the ball has been raised equal to on the diagram?
  22. Dec 19, 2007 #21
    .....2.5 m...?
  23. Dec 19, 2007 #22
    No. The ball started off at B and is now at C. You can think of the total displacement from B to C as a vertical movement of y meters followed by a horizontal movement of x meters. What are x and y on your diagram?
  24. Dec 19, 2007 #23
    y=D, x=C...? yikes!!!!!!!!!!
  25. Dec 20, 2007 #24
    I think you've got it- but the length of a line between (e.g.) E and F is called EF.

    so y=??
    where ?? stands for 2 letters.
  26. Dec 20, 2007 #25
    oh ok-->
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