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Using a stone to hunt down an animal with Centripetal Forces

  1. Mar 20, 2013 #1
    1. The problem statement, all variables and given/known data
    A bushman wants to hunt down a kangaroo so he uses a vine which is 2.0m long with a rock tied to the end of it. The bushman holds the other hand above his head which is 2.0m above the ground. The kangaroo observes that when the angle with the vine and vertical reaches 60°, the vine breaks. So, what's the minimum distance for the rock not to hit the kangaroo?

    θ = 60°
    vine length = 2.0m

    Y-component initial = 2.0m
    Y-component final = 0.0m

    2. Relevant equations
    ac = V2/r


    3. The attempt at a solution
    I drew a diagram of it and put in the data I know so far.
    tumblr_mjxzwoGQhc1qe908uo1_500.jpg
    I thought of this question for a bit and I feel like this question requires the concept of projectile motion (when the vine breaks and the rock flies to the kangaroo) and centripetal forces (because the rock is being spun tied to a vine).

    The thing is, if I try to find acceleration here, it would be the perpendicular acceleration but when the vine breaks, wouldn't the acceleration that would be affecting the rock by then is a parallel acceleration because it does not go in a circle anymore?
     
    Last edited: Mar 20, 2013
  2. jcsd
  3. Mar 20, 2013 #2
    Additional info: the stone is being spun horizontally
     
  4. Mar 20, 2013 #3

    haruspex

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    Despite the detailed diagram, it's not clear what plane the rock is spinning in.
     
  5. Mar 25, 2013 #4
    Alright so after a few days of thinking about this. I learned that my diagram is wrong because it would be impossible for the angle to angle up. So the 60° angles down making it look like it's spinning in the shape of a cone.

    Since this question involves the concept of centripetal forces and projectile motion. This is was I found.

    Y-initial=2-2cos60°
    =1m
    So when the vine breaks off, it starts off 1m above the ground.

    Since the rock spins horizontally, it has no acceleration in the vertical component. Force of tension will have X and Y-components.
     
  6. Mar 25, 2013 #5
    So this is my corrected diagram. The changes made:
    60° angles down, rather than up
    Initial position of Y-component when it breaks changes to 1m
     

    Attached Files:

  7. Mar 25, 2013 #6
    And this is the fbd of the rock :)
     

    Attached Files:

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