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Youngs Modulus and Centripetal Acceleration

  1. Nov 27, 2012 #1
    Question #1)
    In a new bungee sport a 70 kg rider is projected from the shore, out into a lake for a splash with a human scale slingshot. The stretched slingshot is made with a set of 50 cylindrical rubber tubes 1) stretched at a 45 degree angle from the horizontal. The tubes are cylindrical, having a 1cm outside diameter and a 0.5cm inside diameter. They are stretched from a 10 meter loose length back by an additional 15 meters and released accelerating the rider. Y of rubber = (1.0x10^6 pa) . Consider the accelerated rider.

    a) Find the net force on the rider upon release.

    b) Find the maximum height of the rider in his flight out into the water.

    Question #2)
    A 14 Kg door is pulled shut using tension in a rope. The rope is fixed to the door, 10cm from the hinge. The rope passes over a pulley and extends down a 2Kg hanging mass. The mass falls at a nearly constant speed, falling a total of 10cm during the 5 seconds to pull the door through a 90 degree rotation from fully open to fully shut. Find the angular acceleration of the door when the rope is at a 45 degree angle from the doors surface. You may consider the shutting door as a mass of 14Kg pivoting 60 cm from the hinge
     
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  3. Nov 27, 2012 #2
    Sorry for the repost.
    This is what I've done for question 1a)

    F/A = (y x delta L) / L

    delta L = 25-10 = 15 meters
    A = A(outer) - A (inner)
    A = 50pi(7.5x10^-5)
    =0.0118 m^2

    F = ((0.0118)(1x10^-6)(15))/(10)
    F = 17,700 N

    For Part B)
    Im not exactly sure how to do this but I know that K.E initial = P.E final.
    We need to calculate the velocity as he leaves the slingshot but I have no idea how.

    I thought maybe this could work. F=ma
    a= 17700/70
    a= 253 m/s^2

    v^2/r = a
    v = sqrt (253x25)
    v = 80 m/s

    The I did 0.5mv^2= mgh
    h = 322.6 meters

    I have no clue how to do question number 2.
     
  4. Nov 28, 2012 #3

    haruspex

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    Where did you get that equation from? I believe it to be right - just don't know on what basis you justify it.
    The usual way is conservation of energy. How much energy is stored in a spring?
    So all the KE goes into PE?
     
  5. Nov 28, 2012 #4
    Well I assumed that the net force is the one we calculate using the young's modulus equation. Once i get acceleration I use a= v^2/r. Assuming that a is the centripetal acceleration. The radius is 25 because he's being stretched 25m from the slingshot at a 45 degree angle, hence creating a circle.

    It says what is the maximum height reached, and this is only when the K.E energy is 0, and P.E is maximum.
    Yes, I assumed that all of it gets transferred into P.E. We can ignore non-conservative forces like air resistance etc.
     
  6. Nov 28, 2012 #5

    haruspex

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    We seem to be interpreting the set-up quite differently. You mention centripetal acceleration and a circle... what circle? The slingshot is what in England I would have called a catapult. The hand-held version is a Y-shaped stick with elastic stretched across the prongs. You put the payload in the middle of the elastic, pull and release.
    In the present case, the initial trajectory is at 45 degrees to horizontal (thereby maximising range). The trajectory as a whole will be a parabola.
     
  7. Nov 28, 2012 #6
    Oh, now I see the problem. It's not accelerating in a circle, so there is no centripetal acceleration. But then how do I calculate the velocity with which it leaves the slingshot?
     
  8. Nov 28, 2012 #7

    haruspex

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    Conservation of energy.
     
  9. Nov 28, 2012 #8
    for the first question a) how did you calculate the area?
     
  10. Nov 28, 2012 #9

    haruspex

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    Forget about the area. That was based on a misreading of the problem.
    (Edit: ignore that comment)
     
    Last edited: Nov 28, 2012
  11. Nov 28, 2012 #10
    so why is he multiplying by 50. is it because of the number of tubes. I have the same question and i a little confused with the first one . i got my force to be 88.35N and if i multiply it by 50 i get 4417.5N. i don't know if im correct.
     
  12. Nov 28, 2012 #11
    Yeh sorry, you're right. The answer if 4417.5 N. I was using the diameter, not the radius lol. Still dont know what to do for the second part.
     
  13. Nov 28, 2012 #12
    yea cody pretty much killed us this time
     
  14. Nov 28, 2012 #13
    I dont really like this system of taking tests back home and bringing them next class.
    He should just make the questions on the tests easier, so we dont have to take them home.
     
  15. Nov 28, 2012 #14
    exactly, i sent him an email complaining on how we can't even do the questions at home with all the resources, because they are too damn hard.
     
  16. Nov 28, 2012 #15
    A = A(outer) - A (inner)
    A = 50pi(7.5x10^-5)
    =0.0118 m^2

    I dont get that part, how did you get the actual area and where did you get 7.5x10^-5 from?
    could you plase explain that part to me
     
  17. Nov 29, 2012 #16
    Yes can someone please explain how they are getting there area. Because when i do my area, i am not getting the same value for my F as you guys are. My value is extremely low and im confused on what im doing wrong. my radius i was using was 7.5x10^-03.
     
  18. Nov 29, 2012 #17
    A= 50pi( (5x10^-3)^2-(2.5x10^-3)^2)
    = 2.94x10^-3 m^2
     
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