| Topic: |
Science > Physics |
| User: |
"David" |
| Date: |
23 Jun 2004 03:18:25 PM |
| Object: |
Tension in a circular rubber band |
I'm trying to derive the tension in a circular rubber band which is in
equilibrium and on which a constant force is being exherted on the
inside (e.g. a rubber band around a baloon).
This is what I do:
We consider half of the rubber band and put it in a coordinate system
going from -r to r. The curve describing the band would be y =
sqrt(r^2 - x^2).
At each point of the circle (band) a force f is being exherted
radially outwards. We take y-component of this force which would be
f*cos(theta), where theta is the angle between y-axis and the force
vector. cos(theta) = sqrt(r^2 - x^2)/r so the force in the y-direction
at the point (x, sqrt(r^2 - x^2)) is f*sqrt(r^2 - x^2)/r. If we sum
these up from -r to r we should get the total force in the y-direction
so we get:
T = int( -r, r, f*sqrt(r^2 - x^2)/r*dx) (T is tension)<->
T = f*int( -r, r, sqrt(r^2 - x^2)/r*dx) <->
T = f*int( -r, r, sqrt(1 - (x/r)^2)*dx) <->
set a = x/r
T = f*int( -1, 1, sqrt(1 - a^2)*da*r) <->
T = r*f*int( -1, 1, sqrt(1 - a^2)*da) <->
T = r*f*(pi/2) <->
T = f*r*pi/2
This should be the tension force in the points x = r and x = -r. But I
am pretty sure that's incorrect...
I know you could also do it by considering an infinitesimal increase
in radius of the band, but I can't figure out how to do that. Could
someone walk we through it?
Regards,
David
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| User: "Jim Black" |
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| Title: Re: Tension in a circular rubber band |
23 Jun 2004 11:12:29 PM |
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(David) wrote in message news:<3635fa5.0406231218.24210d22@posting.google.com>...
I'm trying to derive the tension in a circular rubber band which is in
equilibrium and on which a constant force is being exherted on the
inside (e.g. a rubber band around a baloon).
This is what I do:
We consider half of the rubber band and put it in a coordinate system
going from -r to r. The curve describing the band would be y =
sqrt(r^2 - x^2).
At each point of the circle (band) a force f is being exherted
radially outwards.
The force on a point is going to be zero. What you want is the force
per length of rubber band. This will have units of force per
distance. Calling this f, what you say next is essentially correct:
We take y-component of this force which would be
f*cos(theta), where theta is the angle between y-axis and the force
vector. cos(theta) = sqrt(r^2 - x^2)/r so the force in the y-direction
at the point (x, sqrt(r^2 - x^2)) is f*sqrt(r^2 - x^2)/r. If we sum
these up from -r to r we should get the total force in the y-direction
so we get:
T = int( -r, r, f*sqrt(r^2 - x^2)/r*dx) (T is tension)<->
The f*sqrt(r^2 - x^2)/r is right; that's the y-component of force per
rubber band length. The problem here is that the length of each piece
of this rubber band is not equal to dx, but to sqrt(dx^2+dy^2) =
dx/cos(theta). Needless to say, this correction makes the calculation
much simpler.
T = f*int( -r, r, sqrt(r^2 - x^2)/r*dx) <->
T = f*int( -r, r, sqrt(1 - (x/r)^2)*dx) <->
set a = x/r
T = f*int( -1, 1, sqrt(1 - a^2)*da*r) <->
T = r*f*int( -1, 1, sqrt(1 - a^2)*da) <->
T = r*f*(pi/2) <->
T = f*r*pi/2
This should be the tension force in the points x = r and x = -r. But I
am pretty sure that's incorrect...
Your approach is valid except for the mistake you made. Make sure you
remember to divide by two at the end, of course, because the force
between the top and bottom of the band is the sum of the tensions at
x=r and x=-r.
I know you could also do it by considering an infinitesimal increase
in radius of the band, but I can't figure out how to do that. Could
someone walk we through it?
Regards,
David
That approach is based on the conservation of energy. If the radius
of the band increases infintesimally, then the amount of work done by
the object stretching the rubber band is force times displacement =
(2*pi*r)*f*dr. The amount of work done on the rubber band is likewise
T*d(2*pi*r). These two energies will be equal, and one can then
simply solve this equation for the tension.
.
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| User: "Jack Ferman" |
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| Title: Re: Tension in a circular rubber band |
23 Jun 2004 06:24:40 PM |
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Might you look at the band as a segment of a round pressure vessel and do
the hoop stress from the internal pressure. This would be in any number of
strength of materials text books. I have donated all my old textbooks long
ago so can't quote a formula
In article <3635fa5.0406231218.24210d22@posting.google.com>,
mpe501@chello.se (David) wrote:
I'm trying to derive the tension in a circular rubber band which is in
equilibrium and on which a constant force is being exherted on the
inside (e.g. a rubber band around a baloon).
This is what I do:
We consider half of the rubber band and put it in a coordinate system
going from -r to r. The curve describing the band would be y =
sqrt(r^2 - x^2).
At each point of the circle (band) a force f is being exherted
radially outwards. We take y-component of this force which would be
f*cos(theta), where theta is the angle between y-axis and the force
vector. cos(theta) = sqrt(r^2 - x^2)/r so the force in the y-direction
at the point (x, sqrt(r^2 - x^2)) is f*sqrt(r^2 - x^2)/r. If we sum
these up from -r to r we should get the total force in the y-direction
so we get:
T = int( -r, r, f*sqrt(r^2 - x^2)/r*dx) (T is tension)<->
T = f*int( -r, r, sqrt(r^2 - x^2)/r*dx) <->
T = f*int( -r, r, sqrt(1 - (x/r)^2)*dx) <->
set a = x/r
T = f*int( -1, 1, sqrt(1 - a^2)*da*r) <->
T = r*f*int( -1, 1, sqrt(1 - a^2)*da) <->
T = r*f*(pi/2) <->
T = f*r*pi/2
This should be the tension force in the points x = r and x = -r. But I
am pretty sure that's incorrect...
I know you could also do it by considering an infinitesimal increase
in radius of the band, but I can't figure out how to do that. Could
someone walk we through it?
Regards,
David
.
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| User: "John C. Polasek" |
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| Title: Re: Tension in a circular rubber band |
23 Jun 2004 06:04:37 PM |
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On 23 Jun 2004 13:18:25 -0700, (David) wrote:
I'm trying to derive the tension in a circular rubber band which is in
equilibrium and on which a constant force is being exherted on the
inside (e.g. a rubber band around a baloon).
This is what I do:
We consider half of the rubber band and put it in a coordinate system
going from -r to r. The curve describing the band would be y =
sqrt(r^2 - x^2).
At each point of the circle (band) a force f is being exherted
radially outwards. We take y-component of this force which would be
f*cos(theta), where theta is the angle between y-axis and the force
vector. cos(theta) = sqrt(r^2 - x^2)/r so the force in the y-direction
at the point (x, sqrt(r^2 - x^2)) is f*sqrt(r^2 - x^2)/r. If we sum
these up from -r to r we should get the total force in the y-direction
so we get:
T = int( -r, r, f*sqrt(r^2 - x^2)/r*dx) (T is tension)<->
T = f*int( -r, r, sqrt(r^2 - x^2)/r*dx) <->
T = f*int( -r, r, sqrt(1 - (x/r)^2)*dx) <->
set a = x/r
T = f*int( -1, 1, sqrt(1 - a^2)*da*r) <->
T = r*f*int( -1, 1, sqrt(1 - a^2)*da) <->
T = r*f*(pi/2) <->
T = f*r*pi/2
This should be the tension force in the points x = r and x = -r. But I
am pretty sure that's incorrect...
I know you could also do it by considering an infinitesimal increase
in radius of the band, but I can't figure out how to do that. Could
someone walk we through it?
Regards,
David
First you characterize the spring constant of the band:
dT = K dL/L
Now set the initial dL/L = C so initial tension is
T0 = KC
Work done in virtual displacement stretching is
dW = T0 dL
Work done radially is the given force F by dr:
dW = Fdr and equate:
T0dL = T0*2pi*dr = F*dr
T0 = F/2pi
That should be right.
Mr. Dual Space
(If you have something to say, write an equation.
If you have nothing to say, write an essay).
.
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| User: "Edward Green" |
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| Title: Re: Tension in a circular rubber band |
25 Jun 2004 05:24:25 PM |
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John C. Polasek <jpolasek@cfl.rr.com> wrote in message news:<np2kd0lk2hnulp6keh5nbblrdlk4nict9d@4ax.com>...
First you characterize the spring constant of the band:
dT = K dL/L
Now set the initial dL/L = C so initial tension is
T0 = KC
Work done in virtual displacement stretching is
dW = T0 dL
Work done radially is the given force F by dr:
dW = Fdr and equate:
T0dL = T0*2pi*dr = F*dr
T0 = F/2pi
That should be right.
Very elegant.
The only thing I blanch at is "set the initial dL/L = C". But maybe
we can pare out some steps from your already sparse argument.
Starting immediately with virtual work:
TdL = FLdr
(T is the tension, L the circumference, F the radial force/length)
L = 2*pi*r, dL = 2*pi*dr
T*2*pi*dr = F*2*pi*r*dr
F = T/r ???
One of us has made a mistake: I'm too lazy to find it. I think
instead I will give somebody the pleasure of insulting me. ;-) At
least the inverse dependence on r is reasonable: as r-> oo, why should
the cylinder care what the tension in the band is? It's just
stretched parallel to a flat surface. This also agrees with the
two-dimensional case, if I recall correctly, where the pressure
differential across a membrane is
1 1
S( --- + --- )
R r
where R,r are the principal radii of curvature, S the surface tension.
It seems to me when I have derived this I went through some ungodly
contortion, which could have been avoided, in the case of a spherical
cap, by your elegant virtual work argument.
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| User: "John C. Polasek" |
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| Title: Re: Tension in a circular rubber band |
25 Jun 2004 09:01:46 PM |
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On 25 Jun 2004 15:24:25 -0700, (Edward
Green) wrote:
John C. Polasek <jpolasek@cfl.rr.com> wrote in message news:<np2kd0lk2hnulp6keh5nbblrdlk4nict9d@4ax.com>...
First you characterize the spring constant of the band:
dT = K dL/L
Now set the initial dL/L = C so initial tension is
T0 = KC
Work done in virtual displacement stretching is
dW = T0 dL
Work done radially is the given force F by dr:
dW = Fdr and equate:
T0dL = T0*2pi*dr = F*dr
T0 = F/2pi
That should be right.
Very elegant.
The only thing I blanch at is "set the initial dL/L = C". But maybe
we can pare out some steps from your already sparse argument.
I was just deriving the initial tension, which we didn't need, you're
right.
Starting immediately with virtual work:
TdL = FLdr
(T is the tension, L the circumference, F the radial force/length)
You don't have to get fancy with the force; it's just force as stated
in the problem. Alter the problem a bit:
Snip the band and unroll it flat like a solid wall. Apply force F all
along the wall, or at a point. It's just force without recourse to
2pi.
Roll it back up and just use Fdr = TdL = 2pi*Tdr.
T = F/2pi.
L = 2*pi*r, dL = 2*pi*dr
T*2*pi*dr = F*2*pi*r*dr
F = T/r ???
Ergo, delete the 2pi on the RHS.
One of us has made a mistake: I'm too lazy to find it. I think
instead I will give somebody the pleasure of insulting me. ;-) At
least the inverse dependence on r is reasonable: as r-> oo, why should
the cylinder care what the tension in the band is? It's just
stretched parallel to a flat surface. This also agrees with the
two-dimensional case, if I recall correctly, where the pressure
differential across a membrane is
1 1
S( --- + --- )
R r
where R,r are the principal radii of curvature, S the surface tension.
It seems to me when I have derived this I went through some ungodly
contortion, which could have been avoided, in the case of a spherical
cap, by your elegant virtual work argument.
Mr. Dual Space
(If you have something to say, write an equation.
If you have nothing to say, write an essay).
.
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| User: "Edward Green" |
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| Title: Re: Tension in a circular rubber band |
26 Jun 2004 06:16:44 AM |
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John C. Polasek <jpolasek@cfl.rr.com> wrote in message news:<3ulpd0p8tj2soovhh9bvrp3471spn1l9vp@4ax.com>...
On 25 Jun 2004 15:24:25 -0700, (Edward
Green) wrote:
<...>
Starting immediately with virtual work:
TdL = FLdr
(T is the tension, L the circumference, F the radial force/length)
You don't have to get fancy with the force; it's just force as stated
in the problem. Alter the problem a bit:
Snip the band and unroll it flat like a solid wall. Apply force F all
along the wall, or at a point. It's just force without recourse to
2pi.
Well, a purist might object to the idea of a "force" which was
distributed in direction -- and no (this pre-quibble is not directed
at you), it's not the net force, which is zero.
Roll it back up and just use Fdr = TdL = 2pi*Tdr.
T = F/2pi.
Ok. Or F = 2piT, and force/length f = T/r
Alright: if I change my notation to cede you use of "F", we agree.
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| User: "John C. Polasek" |
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| Title: Re: Tension in a circular rubber band |
25 Jun 2004 08:36:43 AM |
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On 23 Jun 2004 13:18:25 -0700, (David) wrote:
I'm trying to derive the tension in a circular rubber band which is in
equilibrium and on which a constant force is being exherted on the
inside (e.g. a rubber band around a baloon).
This is what I do:
We consider half of the rubber band and put it in a coordinate system
going from -r to r. The curve describing the band would be y =
sqrt(r^2 - x^2).
At each point of the circle (band) a force f is being exherted
radially outwards. We take y-component of this force which would be
f*cos(theta), where theta is the angle between y-axis and the force
vector. cos(theta) = sqrt(r^2 - x^2)/r so the force in the y-direction
at the point (x, sqrt(r^2 - x^2)) is f*sqrt(r^2 - x^2)/r. If we sum
these up from -r to r we should get the total force in the y-direction
so we get:
T = int( -r, r, f*sqrt(r^2 - x^2)/r*dx) (T is tension)<->
T = f*int( -r, r, sqrt(r^2 - x^2)/r*dx) <->
T = f*int( -r, r, sqrt(1 - (x/r)^2)*dx) <->
set a = x/r
T = f*int( -1, 1, sqrt(1 - a^2)*da*r) <->
T = r*f*int( -1, 1, sqrt(1 - a^2)*da) <->
T = r*f*(pi/2) <->
T = f*r*pi/2
This should be the tension force in the points x = r and x = -r. But I
am pretty sure that's incorrect...
I know you could also do it by considering an infinitesimal increase
in radius of the band, but I can't figure out how to do that. Could
someone walk we through it?
Regards,
David
You're welcome.
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