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Centripetal forces

  1. Nov 14, 2004 #1
    Lets say there was a poll in the middle of the ground and a 2m rope attached to it. Attached to the rope was a cart that sat a 80kg person. The person had a constant force being exerted on them so they the person went in a circle. There is a spring scale attached to the rope and poll to show the spring scale force. What would be the relationship between this spring scale force and the centripetal force?
  2. jcsd
  3. Nov 14, 2004 #2
    Is the cart massless? Is there an angle between the ground and the rope?
  4. Nov 14, 2004 #3
    massless cart, parallel between ground and rope
  5. Nov 14, 2004 #4
    actually, I think I am supposed to find the relationship between the centripetal force and v^2/r
  6. Nov 14, 2004 #5
    Well, if that's the case, then you know [tex]v^2/r[/tex] is equal to the centripetal acceleration. By Newton's Second Law, you know that [tex]F=ma[/tex]. That should be enough for you to finish the problem on your own.
  7. Nov 14, 2004 #6
    I need to plot a graph of (v^2/r) vs Centripetal force. There is a linear relationship, why is that?
  8. Nov 14, 2004 #7
    maybe it's just me, but I understand all these concepts already. However, it might just be the way the question is worded. All that's different between v^2/r and the centripetal force is that the centripetal force has a mass multiplied into it. The problem asks for is the "functional relationship." Cf=v^2/r times the mass. I thought it was asking for something more than that
  9. Nov 14, 2004 #8
    Its the same as asking you to plot F vs a, a is just determined differently in this case.
  10. Nov 14, 2004 #9


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    You know that C = m v^2/r. Let C = y and v^2/r = x, then the law becomes

    y = mx

    This is a line, of course, with slope m. The centripetal force is linear in (v^2/r).
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