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Circular Motion: 2-step System (very confusing!)

  1. Oct 28, 2008 #1
    1. http://paer.rutgers.edu/PT3/experiment.php?topicid=5&exptid=177
    The three black stoppers that are whirling in a circle have a combined mass of 58.7g. The length of man's arm from elbow to wrist is about 30 cm.

    2. Find the mass of the other object that is tied at the bottom.

    3. I solved for time by first counting the number of frames it takes to complete a full circle, which was approximately 18. Then since the video is 30 frames per second, I set up a proportion (18/30 = t/1), where t=time. I then assumed that the radius was 30 cm. which I am very unsure on. I plugged in radius into C=2(pie)(r) to solve for the circumference. I divided the circumference by the time to find velocity (does this even make sense?) Then since F=m(v2/r), I plugged in mass (58.7), velocity, and radius to solve for force. I was also able to figure out acceleration since F = ma, but honestly I'm not sure if my steps so far are correct. And how would I go about solving the mass of the other object tied at the bottom?
     
  2. jcsd
  3. Oct 28, 2008 #2
    Well the mass of the other object on the bottom is the Ft and Ftx is what is proving the centripetal force. So if you have T and you have r and m you can caluclate Fc .

    Now for the small mass draw its Force diagram

    Fg is point down.
    Fty is pointing up same magnitude of Fg
    Ftx is the Fc (towards center)
    and Ft is diagonal and it is equal to Fgof mass on bottom

    So if you get Fc which is Ftx then you can use pythagorus Fg(of mass thats spinning)^2 + Ftx(which is Fc)^2 = mg^2
     
  4. Oct 28, 2008 #3
    I'm sort of unsure on what you mean by Ftx and Fty and then using the pythagorus on Fg.

    Do you think it's possible for you to paste a diagram? That would be amazing
     
  5. Oct 28, 2008 #4
    I attached a diagram made in word.
     

    Attached Files:

  6. Oct 28, 2008 #5
    O thanks alot! I'm waiting for it to be approved by the admins...
     
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