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Homework Help: Rubber band

  1. Aug 11, 2005 #1
    A 100 g ball is attached to a rubber tube and is spun around in a circle at a rate of one revolution every second. A force of 0.5 N is required to stretch the tube 1.0 cm.

    If the original length is L = 1.0 m, what will be the change in length of the rubber tube when the ball is revolving?

    It was given as a hint that centripetal force is equal to elastic force:

    F_c=mass *length*omega^2,

    where length is now the stretched length, that is,
    length= L+delta L.

    The elastic force is equal to
    F_el=k*delta L.

    first, I found k to be 50. but I plugged in k to the above equation and tried to find delta L. as .002 , but it's incorrect, hints?
     
  2. jcsd
  3. Aug 11, 2005 #2

    Gokul43201

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    I get delta L = about 0.09m.

    I see nothing wrong with your approach. You must have a calculation error.
     
  4. Aug 11, 2005 #3
    0.09 is also incorrect....
     
    Last edited: Aug 11, 2005
  5. Aug 11, 2005 #4
    I found delta L to be .002 m as well.
    Here's what I did.

    Fc=m*(dL+1)*w^2
    Fs=k*dL

    .5 N=k*.01 m
    k=50 N/m

    Fc=Fs
    m*(dL+1)*w^2=k*dL
    m*w^2=k*dL - m*w^2*dL
    m*w^2=dL*(k-m*w^2)
    dL=(m*w^2)/(k-m*w^2)

    When you plug in the values (m=.1 kg; w=1 rps; k=50 N/m) the answer is dL=.1/49.9
    So dL = .020 m
     
    Last edited: Aug 11, 2005
  6. Aug 11, 2005 #5

    Galileo

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    I also get 0.09 (0.086) m...

    [tex]\Delta L = \frac{mL \omega^2}{k-m\omega^2}\approx 0.09 m[/tex]
     
    Last edited: Aug 11, 2005
  7. Aug 11, 2005 #6
    =( , i tried them all, none of them works....
     
  8. Aug 11, 2005 #7

    Doc Al

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    I agree with Gokul; the answer is about 0.09m. What makes you think that this answer is wrong?

    [itex]\omega = 2 \pi[/itex], not 1. (It's measured in radians.)
     
  9. Aug 11, 2005 #8
    the answer turns out to be .0857m, I guess we'er all too caught up in the big picture and totally forgot about the omega being 2pi, thanks
     
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