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Bug on a Band


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#1 elfmotat

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Posted 4 March 2013 - 05:57 AM

Came across this interesting problem on the Feynman Lectures site. The answer is there, so no peeking!

 

An infinitely stretchable rubber band has one end nailed to a wall, while the other end is pulled away from the wall at the rate of 1 m/s;  initially the band is 1 meter long. A bug on the rubber band, initially near the wall end, is crawling toward the other end at the rate of 0.001 cm/s. Will the bug ever reach the other end? If so, when?

 

 

bug.gif


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\delta S=0


#2 michel123456

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Posted 4 March 2013 - 09:26 AM

No peeking?

 

As a remark to the problem: if the bug is not traveling ON the rubber band but NEXT to the rubber band (on the ground for example), it will never catch the end of the band.



--------------

Remark#2

 

If I understand correctly, the bug on the band, as observed by an observer at rest, is accelerating. That is not obvious from the (provided) maths.


Edited by michel123456, 4 March 2013 - 09:08 AM.

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Michel what have you done?


#3 adjarklo2000

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Posted 15 March 2013 - 06:34 PM

let length be L, velocity be U , acceleration be A and time be T

 

Li=1m, Lf=?

Ui=1m/s, Uf=?

A=?

 

Uf=Ui+AT

Uf=Ui, T=0

 

A=Uf/1= 1m/Square sec

 

Lf=1/1=1m

 

Lt= Li+Lf

Lt=2m

 

t=2/0.00001

t=200000s

 

conclusion: the bug will reach the end in 200000 seconds by crawling a distance of 2 meters.


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#4 Daedalus

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Posted 19 March 2013 - 10:33 PM

This is from a previous post of mine that not only solves the problem, but also provides an explanation.
 
Spoiler

 

If I understand correctly, the bug on the band, as observed by an observer at rest, is accelerating. That is not obvious from the (provided) maths.

 
You are correct michel123456, the bug on the rubber rope is observed to be accelerating by an observer at rest relative to the rope. Given the position function
 
x(t) \ = \ \left(v_r \, t + x_1\right) \frac{v_a}{v_r} \, \text{ln} \left(\frac{v_r \, t \, + \, x_1}{x_1}\right)

the acceleration can be found by taking the second derivative

\frac{d^2x}{dt^2}\ x(t) \ = \ \frac{v_a \ v_r}{v_r \, t+x_1}

We can see that the acceleration approaches zero as time approaches infinity. However, the acceleration would be very different if a nonzero net force were to act upon the bug. In that case, the bug would be accelerating relative to the rope instead of just having a constant velocity.

Edited by Daedalus, 19 March 2013 - 10:34 PM.

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#5 ydoaPs

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Posted 19 March 2013 - 10:49 PM

Came across this interesting problem on the Feynman Lectures site. The answer is there, so no peeking!
 
 
 
bug.gif

Assuming uniform expansion of the band (it's stretching everywhere at once), I'd say yes, but I can't be bothered to Faf about with the math atm. Otherwise, probably not.
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#6 Daedalus

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Posted 19 March 2013 - 11:17 PM

Assuming uniform expansion of the band (it's stretching everywhere at once), I'd say yes, but I can't be bothered to Faf about with the math atm. Otherwise, probably not.

 

I already showed the math. Check the spoiler tag in my previous post.


Edited by Daedalus, 19 March 2013 - 11:20 PM.

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#7 mooeypoo

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Posted 21 March 2013 - 09:39 PM

!

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Off topic posts split to new topic: http://www.sciencefo...-bug-on-a-band/

Enjoy!


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No trees were harmed in the creation of this post.
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#8 King Thando Mathe

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Posted 3 April 2013 - 10:09 AM

It will never reach the end

let length be L, velocity be U , acceleration be A and time be T
 
Li=1m, Lf=?
Ui=1m/s, Uf=?
A=?
 
Uf=Ui+AT
Uf=Ui, T=0
 
A=Uf/1= 1m/Square sec
 
Lf=1/1=1m
 
Lt= Li+Lf
Lt=2m
 
t=2/0.00001
t=200000s
 
conclusion: the bug will reach the end in 200000 seconds by crawling a distance of 2 meters.



Uf=Ui=1m/s for the band's end because we assume constant rates
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