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Uniform Circular Motion of a stone

  1. Feb 28, 2008 #1
    1. The problem statement, all variables and given/known data
    A stone at the end of a sling is whirled in a vertical circle of radius 1.00 m at a constant speed vi = 2.40 m/s as in Figure P4.57. The center of the string is 1.50 m above the ground.

    [​IMG]

    I solved these:

    (a) What is the range of the stone if it is released when the sling is inclined at 30.0° with the horizontal at A?

    1.05577 m

    (c) What is the acceleration of the stone just before it is released at A?
    Magnitude

    5.76 m/s^2

    (d) What is the acceleration of the stone just after it is released at A?
    Magnitude

    9.81 m/s^2




    2. Relevant equations

    (b) What is the range of the stone if it is released when the sling is inclined at 30.0° with the horizontal at B?


    3. The attempt at a solution

    I've tried doing it the same way I did part a. I used

    x(t) = x_o+v_o(t)

    and

    y(t) = y_o+v_o(t)+(.5)(9.8)(t^2)

    it didn't work and I've tried everything that I had in my notes.
     
    Last edited: Feb 28, 2008
  2. jcsd
  3. Feb 28, 2008 #2

    Doc Al

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    Include the diagram, or at least describe the difference between points A & B.
     
  4. Feb 28, 2008 #3
    Sorry that I didn't include one. I'm trying to find a host. It'll be up shortly.
     
  5. Feb 28, 2008 #4

    Doc Al

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    What did you use for your initial values of position and speed? The acceleration should be negative.

    When they ask for "range", what are they measuring with respect to? Position of the stone when released? Center of the circle?
     
  6. Feb 28, 2008 #5
    It doesn't specify but I believe it's center of the circle. For X-Component I used 1.2m/s and I used 2.0784m/s for the Y-Component. I'm really confused by this, so I don't really know what to do.
     
  7. Feb 28, 2008 #6

    Doc Al

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    OK.
    Those are the velocity components. What are their signs?

    What about initial position?


    I'd start by figuring out how long the stone is in the air before hitting the ground. Then figure out how far it got horizontally.
     
  8. Feb 28, 2008 #7
    I solved for time which is .68912 seconds. What do you mean by how far it would go horizontally? I don't know about initial position either.
     
  9. Feb 28, 2008 #8

    Doc Al

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    How did you solve for the time?

    That's what "range" means.

    What's the position of point B?

    How did you solve for the range in part (a)?
     
  10. Feb 28, 2008 #9
    Part a I did it wrong because I subtracted time that I got from 1.5 (height) and for some reason it gave me the correct answer.

    Here's how I solved part A.

    2.4COS(30) = 2.0784
    2.4SIN(30) = 1.2

    t =( - 1.2 + sqrt(1.2^(2) + 4(4.9)(1.5) ) / 9.8

    t = .444222

    1.5 - t = 1.06

    I don't really know how to get A either. :( I only have 2 tries left on the site and I feel really stupid right now
     
  11. Feb 28, 2008 #10

    Doc Al

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    Those are the y and x components of the initial velocity. The sign of the x-component should be negative.

    Where does this come from?

    What you are trying to find is the distance the stone lands (from the center of the circle, I presume). You need the initial position and velocity components. Calculate the time it takes to hit the ground (by analyzing vertical motion). Then analyze the horizontal motion.
     
  12. Feb 28, 2008 #11
    I know that the height is 1.5 m. I have the velocity components but I do not know how to find the initial position components
     
  13. Feb 28, 2008 #12

    Doc Al

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    The height of the center of the circle is 1.5 m. Use a little trig to find the initial position components of the stone (which is at the end of the string).
     
  14. Feb 28, 2008 #13
    1.5cos(30) = 1.299m
    1.5sin(30) = .75m

    now what do I do? I don't know how to use the information that I have.
     
  15. Feb 28, 2008 #14

    Doc Al

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    The length of the string is only 1 m. Once you find the components of the string length, combine that with the location of the center.
     
  16. Feb 28, 2008 #15
    Oh so

    1cos(30) + 1.5 = 2.36602 m
    1sin(30) + 1.5 = 2 m
     
  17. Feb 28, 2008 #16

    Doc Al

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    Almost. The location of the center is (0, 1.5m).

    Clearly indicate which is the x and which is the y component of the stone's position.
     
  18. Feb 28, 2008 #17
    I did this and still don't get the correct answer. One more try left

    the vertical velocity of the stone on release
    Vv = 2.4 sin 30 = 1.2m/s

    time to reach max height
    v = u+at (v=0 at max height, Vv = u)
    0 = 1.2 - 9.8t
    t = 0.122 sec

    max height above release reached
    v^2 = u^2 + 2as
    0 = 1.2^2 - 2 x 9.8 x s
    s = 0.073m

    so max height above the ground = 2.08 + 0.073 = 2.153m

    time taken to reach ground
    s = ut+ 1/2 at^2 (u = 0 as falling from max height)
    2.153 = 1/2 x 9.8t^2
    t = 0.663 secs

    total time in the air = 0.122 + 0.663 secs = 0.785s

    horizontal velocity of stone on release = 2.4 cos 30 = 2.08m/s

    range = Vh x t = 2.08 x 0.785 = 1.63m
     
  19. Feb 28, 2008 #18

    Doc Al

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    Where does 2.08 come from?


    You'll need to correct this part. Also, you found the horizontal distance from the release point, not its distance from the center. (Beats me what they want. Range is usually measured from the starting point, so you may be OK.)

    Why not find the time in one step? Using:
    Y = Y0 + Vy0*t -.5gt^2

    Solve for t where Y = 0.
     
  20. Feb 28, 2008 #19
    2.4COS(30)

    I'm not sure either I think I'm supposed to add 1.5 to whatever answer I get not really sure.
     
  21. Feb 28, 2008 #20

    Doc Al

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    That's speed, not height.
     
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