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Find maximum kinect energy

  1. Jan 1, 2013 #1
    Hi,
    I am trying to solve a problem but I can't figure how to continue..
    There's an object moving in a circular path with radius 0.5m in an horizontal surface. The rope will break if the tension exceeds 16N. What is the maximum kinect energy?

    So r = 0.5m

    T = 16N

    K=[itex]\frac{1}{2}[/itex]m[itex]v^{2}[/itex]

    How can I find m and v? I tried with T = mg <=> 16 = m(9.8) <=> m = 1.63kg but I don't know if it is correct. But besides that I can't figure how to find v
     
  2. jcsd
  3. Jan 1, 2013 #2

    Doc Al

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    Staff: Mentor

    Hint: Make use of the fact that the motion is circular.
     
  4. Jan 1, 2013 #3
    I also tried that

    v = [itex]\sqrt{a*r}[/itex]

    But how to find a? Will it be g?
     
  5. Jan 1, 2013 #4
  6. Jan 1, 2013 #5
    It is solved :) Thanks!
     
  7. Jan 1, 2013 #6

    Doc Al

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    Just rearrange that to write a in terms of v and r. Then apply Newton's 2nd law.
    No.
     
  8. Jan 1, 2013 #7
    But with that I don't get the right answer. Just to confirm: is the mass correct?
     
  9. Jan 1, 2013 #8

    Doc Al

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    Show what you did with it.
    You mean the value of the mass you calculated in post #1? No, that's not correct. For some reason, you set the tension equal to the weight of the mass. Why would you do that?

    Hint: You don't need the mass to answer the question. It asks for kinetic energy, not mass.
     
  10. Jan 1, 2013 #9
    a = [itex]\frac{v^{2}}{r}[/itex] and then F = m*a <=> F = m*[itex]\frac{v^{2}}{r}[/itex]
    <=> 16 = 1.63 * [itex]\frac{v^{2}}{0.5}[/itex]

    But since 1.63 isn't right it will give wrong results

    I did that because I couldn't figure any other way to find the mass.. trial and error

    But K=[itex]\frac{1}{2}[/itex]m[itex]v^{2}[/itex] so I need the mass
     
  11. Jan 1, 2013 #10

    Doc Al

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    Perfect so far.
    Yes, your value for mass is wrong. (Hint: There's not enough information to determine the mass. But you don't need it!)
    You do not need the mass, you need the kinetic energy. So you just need to solve for 1/2mv2. Go back and stare at the first equation in this post.
     
  12. Jan 1, 2013 #11
    Ok, finally I figured that :P

    So, i did this:
    F = ma <=> a = [itex]\frac{F}{m}[/itex]

    a = [itex]\frac{v^{2}}{r}[/itex] <=> [itex]v^{2}=a*r[/itex] <=> [itex]v^{2}=\frac{F}{m}*r[/itex]

    K = [itex]\frac{1}{2}*m*\frac{F*r}{m} = \frac{1}{2}*F*r[/itex]

    Thanks!
     
  13. Jan 1, 2013 #12

    Doc Al

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    Cool.

    A short cut would be to recognize mv2 when you see it:

    [tex]F = \frac{m v^2}{r}[/tex]
    Thus:
    [tex]m v^2 = F r[/tex]
    [tex]\frac{1}{2}m v^2 = \frac{1}{2}F r[/tex]
     
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