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Centripetal/Cons. of Energy Problem

  1. Jun 14, 2009 #1
    1. The problem statement, all variables and given/known data
    Here's the question: You favourite pyhsics teach who is late for class attempts to swing from the roof of a 24m high building to the bottom of an identical building using a 24m rope. She starts from rest with the rope horizontal, but the rope will break if the tension force in it is twice the weight of the teacher. How high is the swinging physicist above level when the rope brakes.
    g= 9.8m/s[down]

    2. Relevant equations

    Ac= V^2/r
    Ep = mgh
    Ek = 1/2mv^2
    Fg = mg
    3. The attempt at a solution

    So far...
    2Fg = T
    2mg = T
    Figured T was the cent. force
    2mg = (v^2/r)m
    2g = (v^2/r)
    Solved for velocity...
    [2gr^(1/2)] = V

    I assumed enery is conserved so...
    mgh = 1/2mv^2 +mgz - where z is the height at which the rope breaks and h is the height of the building.

    mgh = 1/2m(2gr^(1/2))^2 + mgz
    mgh = 1/2(2mgr)+mgz
    h = r+z

    the radius = height so I end up getting 0 = z...the answer is 8.0 m
     
  2. jcsd
  3. Jun 14, 2009 #2
    The approach I would recommend here is to first find the velocity she would need to be traveling at for the rope to break through dynamics, and to then find the height she would have to drop in order to gain that velocity through energy.

    Start off drawing a free body diagram for the teacher, but with the rope bent at an angle Θ from the horizontal, so as to cover the general case. A good idea might be to look at the problem from the teacher's point of view (Accelerated system).

    An interesting thing to note is that the problem is equivalent in some aspects to this problem: https://www.physicsforums.com/showthread.php?t=319294
     
    Last edited: Jun 14, 2009
  4. Jun 14, 2009 #3
    Alright, well after looking at that i'm a little confused as to how to solve for the angle. From what I'm getting it seems that I should break up the Fg into a component in the direction of tension F = cosθFg = cosθ(2mg).

    So 2cosθmg = [(v^2)/r]m
    2cosθg = (v^2)/r

    cosθT = 2mg
    cosθ((V^2)/r)m = 2mg
    cosθ = 2g / ((V^2)/r)
    Sub into 1

    2(2g / ((V^2)/r) )g = (v^2)/r
    which eventually becomes...
    v=(2gr)^1/2
    which ends up being the same thing as I got before
     
  5. Jun 14, 2009 #4

    Doc Al

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    Staff: Mentor

    Please draw a free body diagram of the person when the rope makes an angle θ below the horizontal. What forces act on her? What's the net force in the radial direction? Then apply Newton's 2nd law.

    Note: The centripetal force is not just the tension; it's the net force in the radial direction.
     
  6. Jun 14, 2009 #5
    I've drawn the fbd but I only see two forces acting, the tension and gravity. So the net force in the radical direction is the tension plus the radical component of gravitational force?
     
  7. Jun 14, 2009 #6

    Doc Al

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    Staff: Mentor

    Good.
    Yes, but be careful with signs. (The two forces point in different directions.)
     
  8. Jun 14, 2009 #7
    So that means that the tension equals the component of gravity (2mg) in the radical direction?
     
  9. Jun 14, 2009 #8

    Doc Al

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    Staff: Mentor

    No. How did you deduce that?

    Apply Newton's 2nd law: Set the net force in the radial direction equal to mac.
     
  10. Jun 14, 2009 #9
    so m(V^2/r) = T + (-cosθ(2mg)) ?

    If thats true it seems I have too many unknowns to solve for.
     
  11. Jun 14, 2009 #10

    Doc Al

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    I don't understand that last term. It should be the radial component of the weight. (And is θ the angle the rope makes with the horizontal?)
    You'll need to combine this with conservation of energy. Also, T is given.
     
  12. Jun 14, 2009 #11
    Well the θ is the angle with the horizontal minus 90 degrees, because I made a triangle with a vertical line going through the object. Then I used the z pattern to find the angle the Fg makes with the radical component of Fg. This seems to be my biggest problem, I'm not sure how to solve for this angle as none are given... So far all the concepts involving dynamics seem to make sense and I overlooked the whole T = 2mg, but I'm kinda confused as to how to use the whole angle with the horizontal concept to help solve the problem.
     
  13. Jun 14, 2009 #12

    Doc Al

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    Draw a diagram showing the rope making an angle θ with the horizontal. Then draw a line through the person that is perpendicular to the rope. That line makes an angle θ with the vertical (the direction of gravity). Then it's just getting the trig right when finding the radial component of gravity.

    But if you want to use θ to be the angle the rope makes with the vertical, that's OK too. Just pick one definition and go with it. (The only difference is that one will give you a sinθ, the other a cosθ.) Obviously your final answer for the height will be the same.
     
  14. Jun 14, 2009 #13
    well I end up getting cosθmg for the component, but once again not sure how an unknown angle will help
     
  15. Jun 14, 2009 #14
    The major point here is that once you know Θ from the horizontal or vertical (It does not matter) at which T=2mg, you can find the height of the teacher through simple trigonometry.

    What you've already done is draw a diagram of the teacher with all the forces acting on her. You've already concluded that they are gravity, and the tension in the rope.

    In your diagram, you choose to define an angle Θ so you can measure the relation between gravity and the tension. What I recommend you do, is break both forces down into their radial and tangential components (Note that the tension is in the radial direction for any angle Θ).

    Then apply Newton's second law to the radial components of both forces, ΣF = mar
    Since this is circular motion,
    ar = mv²/R

    Then look at your specific case, where T=2mg.

    From this point on, you can extract an expression for the angle Θ. Once you know that angle, you can find the height of the teacher above the ground/below the starting point through simple trigonometry.

    To eliminate the variable v², use conservation of energy.
     
  16. Jun 14, 2009 #15
    Thanks alot for the help guys... this is what I've managed to come up with thanks to all your input... I feel i'm getting close.

    sinθT = mg
    sinθ(2mg) = mg
    sinθ = 1/2
    special triangle so... θ = 30

    (next part is the part I feel is probably incorrect)

    Let @ be the other degree in the right angle triangle (60)
    Let Fr be the radical component of Fg

    cos@mg = Fr
    Fnet = T + (-Fr)
    = T - Fr
    = 2mg - cos@mg

    Fnet = Fc = m(V^2/r)
    m(V^2/r)=2mg - cos@mg
    V^2/r = 2g - cos@g
    V = (2gr-cos@gr)^1/2

    Sub into cons. of energy formula
    Let z = height of disconnect
    Let r= radius/height

    mgr = (1/2)m((2gr-cos@gr)^1/2)^2) + mgz
    r = r - (cos@r/2) + z
    cos@r/2 = z
    cos(60)(24)/2 = 6

    So I ended up getting the wrong answer, and I feel it's probably due to my geometry; but other then that does it look good?
     
  17. Jun 14, 2009 #16

    Doc Al

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    No problem. You're using the angle between the rope and the vertical.
    You'll solve for the angle. Once you have the angle, you can figure out the height.

    This is incorrect. If you could solve for the angle so easily, you'd be done!

    You assumed that the net vertical force equals zero, but that's not true. Just stick to the radial component.

    You're on the right track. One mistake is thinking you already had the angle, so forget that. When you redo it, try this: Express everything in terms of the unknown angle. That should be your only variable, so you can solve for it. Once you have that angle, use it to find the height.
     
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