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Centripetal Acceleration Problem

  1. Feb 16, 2006 #1
    Hi Everyone! I've been working on this Physics problem for the last week and I still can not figure it out! I'm sure I'm missing some very basic concept, but I'm not sure how to proceed:

    An Australian bushman hunts kangaroos with the following weapon - a heavy rock is tied to one end of a light vine of length 2 m, he holds the other end above his head, at a point 2 m above ground level, and swings the rock in a horizontal circle, about his body. The cunning kangaroo has observed that the vine always breaks when the angle, measured between the vertical and the string, is 60°. At what minimum distance from the hunter can the kangaroo stand with no danger of a direct hit?

    :surprised

    I understand that this problem is made up of both centripetal acceleration concepts and projectile motion concepts. I can figure out the projectile motion using the formula d = v1(t) + 1/2(a)(t)^2 and sine/cosine/tan. Because the radius of the circle and the length of the vine form a triangle with the "extra" vertical displacement being the opposite side, I used the angles to determine that when the hypotenuse (length of vine) was 2, the radian was cos(30)(2) = 1.73 and the vertical displacement was sin(30)(2) = 1 so the total vertical displacement is 3 metres. This means that the time the projectile is in flight is 0.782 seconds and I just need the horizontal velocity to solve for the horizontal displacement.

    Ok, so after that mess :smile: , I know that all I need is the horizontal velocity of the rock the instant that the vine breaks and it is travelling as a projectile. This is where I get confused. The only two forces at work on this system are the force of tension (both vertical and horizontal components) and the force of gravity.

    Fgravity = mg
    Ftension = ma = mv^2/r

    I'm just not sure how to proceed to solve for v...I would really appreciate any suggestions. I don't mind doing the "work", but I just don't know where to start....Thank-you!
     
  2. jcsd
  3. Feb 16, 2006 #2

    Doc Al

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    Staff: Mentor

    Sanity check: Does the vine angle up from where its held or does it angle down?

    As you said yourself, the force of tension has two components, horizontal and vertical. Vertically, there's no acceleration, so the vertical forces add to zero. Horizontally there's centripetal acceleration; so set the horizontal force equal to ma.

    You'll get two equations (horizontal and vertical). Solve them together and you'll find the speed at which the vine breaks.
     
  4. Feb 16, 2006 #3
    Thank-you!

    Thanks you for the "Sanity Check" :rofl:

    Nothing like a good and :blushing: humiliating assumption problem, eh?

    Anyways, the problem was actually incredibly simple once I changed my point of view. Thanx!
     
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