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Rubber band

  • Thread starter huskydc
  • Start date
  • #1
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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?
 

Answers and Replies

  • #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.
 
  • #3
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0.09 is also incorrect....
 
Last edited:
  • #4
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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:
  • #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:
  • #6
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=( , i tried them all, none of them works....
 
  • #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?

zwtipp05 said:
I found delta L to be .002 m as well.
Here's what I did.
...
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
[itex]\omega = 2 \pi[/itex], not 1. (It's measured in radians.)
 
  • #8
78
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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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