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Homework Help: Another centripetal force problem

  1. Jun 9, 2014 #1
    A stuntman swings from the end of a 4 m long rope along the arc of a circle. If his mass is 70 kg, find the tension in the rope required to make him follow his circular path at (a) the beginning of his motion, assuming he starts when the rope is horizontal, (b) at a height of 1.5 above the bottom of the circular arc, and (c) at the bottom of the arc.

    (Equation) Fc = mv^2/r
    KE= 1/2mv^2
    PE = mgy

    (Attempt at solution)
    Find the total energy of the system
    4 m (9.8)(70 kg) = 2744 J
    Find the kinetic energy at 1.5 m to find the velocity
    70(9.8)(1.5) = 1029
    2744 - 1029 = 1715 J
    1/2mv^2 = 1715
    V= 7 m/s
    Find the centripetal force using Fc= mv^2/r
    70(49/4)
    Fc = 857.5
    The correct answer is 1.29 x 10^3 N
     
    Last edited: Jun 9, 2014
  2. jcsd
  3. Jun 9, 2014 #2

    vela

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    You were asked to find the tension in the rope; you found the centripetal force. They're not the same thing. I suggest you draw a free-body diagram to try figure this problem out.
     
  4. Jun 10, 2014 #3
    What makes them different? The tension in the rope must match the centripetal force in order for the man to move in a circle.
     
  5. Jun 10, 2014 #4

    vela

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    That's not true. You should drop the notion of centripetal force and instead think in terms of a centripetal acceleration: objects that follow a circular path of radius ##r## experience a centripetal acceleration of magnitude ##v^2/r##. When you plug this into Newton's second law (for the radial direction), you get
    $$\sum F_\text{r} = ma = m\frac{v^2}{r}.$$ You're getting the lefthand side of the equation wrong. Did you draw a free-body diagram yet?
     
  6. Jun 10, 2014 #5

    Nathanael

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    The tension in the rope must cause the net centripetal force, but that does not mean that the tension in the force must match the centripetal force. (Because there are other forces involved.)

    (If you're stuck, draw a FBD)
     
  7. Jun 11, 2014 #6
    Here is my free-body diagram. The T stands for tension, CF for centripetal force, TF for tangential velocity, and w for weight.
     

    Attached Files:

  8. Jun 11, 2014 #7
    Do you see it now?
     
  9. Jun 12, 2014 #8
    No I don't. What should I be seeing?
     
  10. Jun 12, 2014 #9

    Nathanael

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    Ok, so we have a weight swinging around on a rope in circular motion. Let's look at the time when the rope is at the bottom of it's swing (so it's completely vertical) just to simplify things.


    Suppose, like you said, that the tension in the rope is equal to the centripetal force.


    What would be the net force on the weight (or person)? The tesnison is directed upwards (towards the center of the circle) but gravity is directed downwards, so what would be the Net Force on the weight? Would that Net Force be equal to the centripetal force?


    For an object to move in a circle there must be a NET force equal to [itex]\frac{mv^2}{2}[/itex]

    If the tension was equal to [itex]\frac{mv^2}{2}[/itex] then would the Net Force be equal to [itex]\frac{mv^2}{2}[/itex]?

    What must the tension be in order for the Net Force to equal [itex]\frac{mv^2}{2}[/itex]?

    (Hint: The tension is not constant throughout the swing)
     
  11. Jun 12, 2014 #10
    So the centripetal force doesn't have to be the same as the tension because the tension is reliant on the weight and the acceleration due to gravity while the centripetal force relies on the mass, the velocity , and the radius. So do I even have to use centripetal force to solve this problem? Can I just use trigonometry and the weight vector to find the tension?
     
  12. Jun 12, 2014 #11

    Nathanael

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    I don't think this is a good interpretation.

    You do need the centripetal force still. You'll need both methods.

    If you're at the bottom of the swing (completely vertical) then you have the equation:

    [itex]F_{centripetal}=F_{net}=F_{tension}-F_{gravity}[/itex]

    Therefore:
    [itex]F_t=F_c+F_g=\frac{mv^2}{2}+mg[/itex]


    The bottom of the swing is the easiest place to analyze it (because no angles are involved)


    I'll let you figure it out for other parts of the swing (that do involve angles) but do you understand the logical principle I'm applying?

    The logic is that the only two forces are gravity and tension and they must combine to give a Net Force of [itex]\frac{mv^2}{2}[/itex] (towards the center)
     
  13. Jun 12, 2014 #12
    OK I see it now. Thanks!
     
  14. Jun 12, 2014 #13

    Nathanael

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    No problem
     
  15. Jun 12, 2014 #14
    Alright now I am having trouble with the angle. I know that at 1.5 m above the ground the force of gravity is perpendicular to the ground and in order to find the amount of that force that effects the tension I have to use trigonometry. So T = 1/2mv2 + mg sin θ (or cos θ). The problem I have is that I don't know whether to use sin or cosine or how to find the angle. Do I use radians to find the angle?
     
  16. Jun 12, 2014 #15

    Nathanael

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    First step is to find the angle. How? Draw a picture.


    Can you find the θ in my picture? (Maybe I shouldn't have labeled the 2.5m side haha)


    If you're stuck on whether to use cosine or sine and whatnot or something else I want to see you draw a picture to try to figure it out
     

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  17. Jun 13, 2014 #16
    ImageUploadedByPhysics Forums1402688824.676259.jpg

    I found theta by doing the inverse cos of 2.5/4 = .89566. I found out whether to use sin or cos by drawing the picture. So
    T = 70(7)/2 + mg/ cos .89566
    T = 1715 + 686 / cos .89567
    T = 2812.59
    This is not the correct answer
    Ignore the caption under the photo
     
    Last edited: Jun 13, 2014
  18. Jun 13, 2014 #17

    Nathanael

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    Don't you want g (or mg) to be the hypotenuse? Because you're trying to break up the force of gravity into it's components (specifically it's tangent and perpendicular components)

    Your drawing is wrong, gravity should be the hypotenuse.


    EDIT:
    Can you explain your calculations?

    Also in your drawing you put 0.89566 degrees, but it would actually be radians (I don't think this effected your math though)


    Why didn't your calculation involve the centripetal equation?
     
    Last edited: Jun 13, 2014
  19. Jun 13, 2014 #18
    I thought the force of gravity always was a straight line down?
     
  20. Jun 13, 2014 #19

    Nathanael

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    It is, but you can sill make it the hypotenuse. Look at the drawing I've attatched, that is how it should be.


    But why didn't your calculations involve the centripetal force equation?
     

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  21. Jun 13, 2014 #20
    Ok I did that but I'm still not getting the right answer. I did
    Mg cos .88566 = 428.75
    Then I plugged that into the equation
    Mv^2/r + 428.76 = T
    70(7)/2 + 428.75 = 2143.75
    Which is not the right answer
    I got the velocity 7 from an earlier calculation attempt.
     
  22. Jun 13, 2014 #21

    haruspex

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    You run the risk of accumulating rounding errors by substituting in numbers too soon. Keep everything symbolic until the final step. This also helps simplify the process by finding cancellations of terms, helps others follow your logic and helps you spot algebraic errors. It's a very good habit to get into.
    In the present case, you would have avoided taking any square roots, since it's the value of v2 that you need. You would also have avoided figuring out the angle since the information provided gives you the trig function of it that you need (cos).
    If you still don't get the right answer, please repost your working in purely symbolic form.
     
  23. Jun 13, 2014 #22
    I don't think it has to do with my rounding. My answer is double the correct answer..
     
  24. Jun 13, 2014 #23

    Nathanael

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    Edit:
    nevermind
     
  25. Jun 13, 2014 #24

    The answer is 1.29 x 10^3
     
  26. Jun 13, 2014 #25

    Nathanael

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    I was just thrown off because you put 70(7)/2 instead of 70(7^2)/2

    Honestly I don't know why your answer is wrong, I'll let you know if I think of anything
     
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