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Find the radius

  1. Feb 13, 2012 #1
    1. The problem statement, all variables and given/known data

    A block of mass m1=2kg is attached to a cord. The cord goes down through a hole in the table and is attached to mass m2=4 kg hanging below the table. The 2 kg mass moves on the table in a circle at a speed of 3.5 m/s the table top is friction less and there is no friction between the cord and the side of the hole. What is the radius of the circle if the 4 kg mass remains at rest?

    2. Relevant equations

    Sum of force x direction m1: T=mv^2/r

    Sum of forces y direction m1: N-mg=mv^2/r

    Sum of forces x direction m2: Mg-t= Mv^2/r

    3. The attempt at a solution

    I have tried to use this equations and i cannot come up wit the solution. Do i have my equation correctly in this problem?
     
  2. jcsd
  3. Feb 13, 2012 #2

    PeterO

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    The sum of forces in the y direction for the mass m1, on the table is zero! It is not moving in the vertical direction.

    Mass m2 is not moving in any direction at all so sum of forces on it is also zero.

    The common force to m1 and m2 is the tension in the cord - which pulls in the x direction on m1 and in the y direction on m2.
     
  4. Feb 13, 2012 #3
    In such case if I try to find the radius when the 4 kg mass is at rest (no tension in both of the masses, I believe) than i have this equations for both of the masses r=mv^2 and when i sum both of my radius's i don't get the right radius. the radius I'm looking for when the 4kg mass is at rest is 62.5 cm
     
  5. Feb 13, 2012 #4

    PeterO

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    You believe wrong!

    If the tension was zero, the 4kg mass would accelerate towards the floor.

    If the tension was zero, the 2 kg mass would not follow a circular path

    Not sure how you got two radius's [it is actually radii] to add. the 4kg mass is stationary, so doesn't have a radius associated with it ?
     
  6. Feb 13, 2012 #5

    PeterO

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    from your original post

    Sum of force x direction m1: T=mv^2/r

    Sum of forces y direction m1: N-mg=mv^2/r

    Sum of forces x direction m2: Mg-t= Mv^2/r

    I have already suggested they should read

    Sum of force x direction m1: T = mv^2/r

    Sum of forces y direction m1: N - mg = 0

    Sum of forces y direction m2: Mg - T = 0

    But clarified that the T value in the y direction for m2 is the same size as the T value for the x direction of m1.
     
  7. Feb 13, 2012 #6
    Than dude can you give me a hint? cause I really don't understand what it mean "remain at rest". If the 4kg mass is hanging, isn;t it already at rest? O_O if is not how can a 4kg mass hanging from a hole remain at rest?? I do not understand this question... really... please help?
     
  8. Feb 13, 2012 #7

    PeterO

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    For m1 to moving in a circle, something has to pull it towards the centre - the cord is that thing.
    Your first equation effectively said that T = mv2/R

    For m2 to be stationary, something has to apply an upward force to balance the force of gravity [its weight] - again it is the cord (the Tension in the chord).

    From the hanging mass, it is easy to calculate the required tension.

    Using that tension figure, you can calculate the Radius for m1, since you already know the m and v values.
     
  9. Feb 13, 2012 #8
    Got it!!! What i did it was just like you told me. I took the m2 equation mg-t=0 and found the tension than i used m1x equation t=mv^2/r than I solved for r to find the radius and I got r= (2kg)(3.5m/s)^2/39.2N = 0.625 m that I multiplied by 100 to convert it in cm and got 62.5 cm . Thanks man!! You rock!
     
  10. Feb 13, 2012 #9

    PeterO

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    It is possible to replicate a similar experiment at home.

    Note: If you have a simple BIC ball pont pen , or similar, you can remove the inner parts to create a tube.
    Thread a cotton through the tube and tie one washer to one end, and several washers to the other.
    Hold the tube above your head, with the several washers against the tube, and the one washer hanging down (you will need to hold it down). When you release the single washer, the bunch if washers immediately accelerates down, and the single washer accelerates up until that single washer jams against the top of the tube.

    Now repeat, but this time hold onto the bunch of washers and begin to rotate the single washer.
    As you increase the speed of that washer, you will find that you need less and less force to support the bunch of washers.

    With a bit of practice and acquired skill you will be able to twirl the single washer while having the bunch of washers hanging stationary below the tube - them being supported by the tension in the cotton - the exact same tension that keeps the twirling washer in circular motion.

    That set-up is very similar to the situation you have been presented with.

    I really recommend you try this experiment - but don't expect to get it to work first try!!
     
  11. Feb 13, 2012 #10
    I'll try it! I'll let you know my results if i had in the first try :D
     
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