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Centripetal Acceleration of Stone Problem

  1. Mar 2, 2009 #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 1.20 m at a constant speed v0 = 1.50 m/s. The center of the sling is 1.50 m above the ground. What is the range of the stone if it is released when the sling is inclined at 30 degrees with the horizontal (a) at A? (b) at B? What is the acceleration of the (c) stone just before it is released at A? (d) just after it is released at A?

    http://www.fen.bilkent.edu.tr/~mb/phys101/CH4.pdf [Broken]
    The figure is on page 7. Its problem #57


    3. The attempt at a solution

    I'm having trouble just thinking of what a set up would be like for this problem. I wish I could give some work here but I'm tottally stumped
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 2, 2009 #2

    lanedance

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    Hi

    probably need to recoginise the answer to c) and d) before attempting a) & b)

    for c) and d) think about the type of motion, for c) this is the circular motion, so what is the acceleration. for d) what forces act on the stone after release?

    for a) and b) you know initial velocity & position so should be able to calculate range...
     
  4. Mar 3, 2009 #3
    I'm not sure where to begin. First off I don't know what its really asking. I have an idea that the stone at the end of point A will fling at a higher range than point B, but other than that I don't see how to get there

    So I think the appropriate eq is xf = x0 + v0t --> 1.20 + 1.50cos(30 deg)t
    Solving for t in the y direction 1.50 = 1/2(9.8)t^2 --> t = .55328
    Then, xf = 1.20 + 1.50(cos(30 deg))(.55328) = 1.9187 m

    I'm not sure what to do with point B. Don't point A and point B both have the same data? But that wouldn't really work out because point B is supposed to have a smaller range than point A
     
  5. Mar 3, 2009 #4

    lanedance

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    look like you're on the right track...

    this questions consiet of two main parts
    - finding vector components
    - using constant acceleration formula

    first you need to find the vector components of the initial position (on the circle) and velocity

    Once you have these you find use the initial hieght & y velocity to findt he time the stone is in the air for, as i think you have done.

    Then use the time to find the distance travelled in the x direction

    i think in your calulations you need to be a bit more careful defining your starting points... (ie initial position & velcoity) for example the y velocity is not included in your calculation of time
     
    Last edited: Mar 3, 2009
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