Tension as a function of distance

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Homework Statement



A thin uniform metallic rod of length L rotates with an angular velocity ω in a horizontal plane about a vertical axis passing through one of its ends. Calculate the tension in the rod as function of distance from axis. The density of the rod is ρ and its area of cross-section is A.


Attempt:

If we fix our reference frame on the axis of the rod, this is the force diagram :

attachment.php?attachmentid=31548&stc=1&d=1295792549.jpg


dT = dmω2x
I am confused in calculating the tension about a point or on a small element 'dx' shown. Do I have to integrate dT from X=0 to X=x ?
 

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Answers and Replies

  • #2
tiny-tim
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Hi Abdul! :smile:
I am confused in calculating the tension about a point or on a small element 'dx' shown. Do I have to integrate dT from X=0 to X=x ?

Yes … draw a free body diagram for a piece of the rod from position x to x+dx,

then do Ftotal = ma, then divide both sides by dx, then integrate. :wink:
 
  • #3
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I attached the picture earlier but don't know why it didn't show up.

So integrating dT = dmω2x
from X=0 to X=x gives

T= Ax2ρω2/2

But this is incorrect :frown:
 
  • #4
gneill
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The tension at distance x from the center of rotation is due to the portion of the rod that is beyond x, pulling outwards.
 
  • #5
ehild
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So integrating dT = dmω2x
from X=0 to X=x gives

T= Ax2ρω2/2

Correctly: T(x)-T(0)=-Ax2ρω2/2

Do you know the tension somewhere? What it can be at x=L?

ehild
 
  • #6
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The tension at distance x from the center of rotation is due to the portion of the rod that is beyond x, pulling outwards.

Is something wrong with my figure? As we move away from the centre, does the tension increase or decrease?

Correctly: T(x)-T(0)=-Ax2ρω2/2

Do you know the tension somewhere? What it can be at x=L?

ehild

It is 0. How does that help?
 
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  • #7
gneill
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There's nothing wrong with your figure. The centrifugal force on your mass element is shown correctly. But the tension at x is due to the sum of all the mass elements beyond x (outwards) contributing their outwards pulls.

As x increases, with fewer mass elements beyond x, you would expect less tension. This is conceptually similar to a hanging chain. The tension in a given link is due to the force of gravity acting on all the links below it.
 
  • #8
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Is there any tension due to inner part of the rod (towards left of the element) ? For equilibrium, the inner part should exert more tension force than the outer part as there is already a centrifugal force acting outwards.
 
  • #9
gneill
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Unless the rod flies apart, the net force inwards must balance the net force outwards. This is called tension.
 
  • #10
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As x increases, with fewer mass elements beyond x, you would expect less tension.

Less tension in which direction?. Every point is acted upon by tension in two opposite directions in the rotational motion. For the element dm, the inward tension must be greater that the outward tension or else the rod will fly apart.
 
  • #11
gneill
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Less tension in which direction?. Every point is acted upon by tension in two opposite directions in the rotational motion. For the element dm, the inward tension must be greater that the outward tension or else the rod will fly apart.

No! Absolutely not! Unless the rod's parts are coming apart and accelerating away from each other, there is only one tension at any point along the rod, exerting equal force in both directions. If you were to divide the rod at x and insert a little spring scale to hold the sections together, it would register one amount of force -- the tension at that point.
 
  • #12
tiny-tim
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Hi Abdul! :smile:
So integrating dT = dmω2x
from X=0 to X=x gives

T= Ax2ρω2/2

But this is incorrect :frown:

No, it's correct

except you've forgotten the constant of integration. :redface:

To find that, consider the bit from L-dx to L …

that has only one tension force on it, doesn't it?

So the tension there (at x = L) is … ? :smile:
 
  • #13
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No! Absolutely not! Unless the rod's parts are coming apart and accelerating away from each other, there is only one tension at any point along the rod, exerting equal force in both directions. If you were to divide the rod at x and insert a little spring scale to hold the sections together, it would register one amount of force -- the tension at that point.

Yes there is only one tension in the sense it has only one origin. But there must be a difference between the tension forces acting at a point in opposite directions, so that the net tension force balances the centrifugal force.
 
  • #14
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except you've forgotten the constant of integration. :redface:

I performed definite integration, there is no constant.

T varies from 0 to T(x) as X varies from 0 to x.
 
  • #15
tiny-tim
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I performed definite integration

Yes, so your integral gave you T(x) - T(0)
, there is no constant.

T varies from 0 to T(x) as X varies from 0 to x.

Yes there is a constant!! … it's T(0) …

T is not 0 at x = 0.

(But, as I said, you can easily find out what T is at x = L :wink:)
 
  • #16
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T is not 0 at x = 0.

Ah yes, T(0) will be AL2ρω2/2

So that T(x) = ω2Aρ(x2 - L2)/2

But this comes out to be negative. Why?
 
  • #17
tiny-tim
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  • #18
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The centre of mass of the rod is at a distance L/2 from the axis of rotation.
So T(0) = mω2L/2 = AL2ρω2/2

T(L) will be 0. How does that help?
 
  • #19
tiny-tim
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So integrating dT = dmω2x
from X=0 to X=x gives

T= Ax2ρω2/2

But this is incorrect :frown:

As ehild :smile: pointed out, this should be minus Ax2ρω2/2 …

that explains why you were getting negative numbers.

(also you need a constant of course)
T(L) will be 0. How does that help?

Yes, T(L) is obviously zero, and you should have seen that right at the start :redface:

That gives you your constant of integration … you can now easily find T(x) - T(L). :wink:
 
  • #20
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As ehild :smile: pointed out, this should be minus Ax2ρω2/2 …

that explains why you were getting negative numbers.

Why should it be minus Ax2ρω2/2 ?

Can you explain that?
 
  • #21
tiny-tim
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Choosing the outward distance as positive:

T(x+dx) is outward, so we add it

T(x) is inward, so we subtract it

acceleration is inward, so it's negative

T(x+dx) - T(x) = minus …

so T'(x) = minus … :wink:
 
  • #22
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That means my force diagram is incorrect?
T+dT and T are in opposite directions?
 
  • #23
tiny-tim
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No, your force diagram is correct.

But it's a bad diagram because it uses T+dT and T instead of T(x) and T(x+dx) :redface:

you got confused when you converted …

T+dt becomes T(x) and T becomes T(x+dx),

so dt becomes T(x) - T(x+dx), and the derivative comes out negative. :smile:
 
  • #24
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Can we directly say that T(x) - T(x+dx) = -dT ?
 
  • #25
tiny-tim
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Can we directly say that T(x) - T(x+dx) = -dT ?

I suppose so … T(x) - T(x+dx) = -dT/dx dx = -dT.

But I wouldn't recommend taking short-cuts, simply because of the possibility of making mistakes.
 

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