Calculating Centripetal Force in a Bent Tube

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In summary, the conversation discusses two problems related to fluid dynamics. The first problem involves finding the external force required to hold a bent horizontal tube stationary, while the second problem involves finding the force with which a rubber tube is stretched due to the flow of water through it. The conversation includes a discussion of the equations and concepts involved in solving these problems, as well as a free body diagram to illustrate the forces at play. It is determined that the force on a circular cross section of the tube is equal to half of the force on the left half of the tube.
  • #1
Tanya Sharma
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Homework Statement



A horizontal tube of uniform cross sectional area 'A' is bent in the form of U. If the liquid of density 'ρ' enters and leaves the tube with velocity 'v' then how much external force is required to hold the bend stationary ?

Homework Equations


The Attempt at a Solution



I think the bend in the tube provides centripetal force to the fluid flowing through it .As a result the fluid also exerts force on the bend. But then radius of the curve is not given.

In 1 sec liquid of mass ρAv is flowing through any cross section of the tube.

I would be grateful if somebody could help me with the problem.
 
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  • #2
You can work the problem using centripetal force, but there's an easier way.
momentum
 
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  • #3
The change in momentum/sec 2ρAv2 of the liquid is the force by the tube on the liquid.Same force is applied by liquid on the tube .So external force F = 2ρAv2 needs to be applied , in opposite direction to the force by liquid to keep the tube at rest .

Is it correct ?
 
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  • #4
Looks good!
 
  • #5
Thanks TSny !

There is a similar problem in which I am having difficulty .

Q . Water flows with a velocity v along a rubber tube having the form of a ring. The radius of the ring is R and diameter of the tube is d (d<<R). Find the force with which the rubber tube is stretched.(ρ is the density of water).

My reasoning : The mass/sec of water flowing through any cross section of the tube is given by ρπd2v/4 .

Not sure what to do next .
 
  • #6
Can you relate a ring to a U-tube?
 
  • #7
Three similarities I can think

1) Liquid goes in liquid comes out.
2) In U tube a part is curved whereas a ring is completely circular.
3) Change of momentum/sec of liquid is same in both the cases i.e 2ρAv2
 
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  • #8
Tanya Sharma said:
3) Change of momentum/sec of liquid is same in both the cases i.e 2ρAv2

For what part of the circumference of the ring does this expression apply?
 
  • #9
Approx. 100% of the circumference .Almost a full circle .
 
  • #10
No. Think about where the factor of 2 comes from in 2ρAv2 in the result for the U-tube.
 
  • #11
Factor of 2 comes from a reversal of momentum of the fluid which had entered the curved part.
 
  • #12
Yes. How much of the circumference corresponds to reversal of momentum?
 
  • #13
Can't assign a numeric value ,but a very small fraction of the circumference .
 
  • #14
Consider a cross section of the ring and the direction of momentum flowing through it. How far around the ring do you need to go before you reach a second cross section where the direction of momentum is reversed relative to the first cross section?
 
  • #15
TSny said:
Consider a cross section of the ring and the direction of momentum flowing through it. How far around the ring do you need to go before you reach a second cross section where the direction of momentum is reversed relative to the first cross section?

The distance around the ring is ∏R i.e half the circumference.
 
  • #16
OK. Draw a free body diagram for the portion of the rubber tube representing half a circumference. Include the total force due to the motion of the fluid in this section of the tube as well as any other forces acting on this section of the tube.
 
  • #17
In the attachment I have shown the left half of the tube.

The net force due to the fluid is towards left having magnitude 2ρAv2. Since it is in equilibrium an equal amount of force is exerted by right half of the rubber tube.
 

Attachments

  • ring.GIF
    ring.GIF
    2.1 KB · Views: 453
  • #18
OK. Can you be more specific about the force exerted by the right half? At what point(s) of the left half does the force of the right half act?
 
  • #19
TSny said:
OK. Can you be more specific about the force exerted by the right half? At what point(s) of the left half does the force of the right half act?

The force on the left half is exerted on the circumference of two circular end points of the tube (in red). It's exaggerated view is shown towards right . The blue lines indicate the force on the circumference .The right half exerts force through the two end points .
 

Attachments

  • ring.GIF
    ring.GIF
    4.1 KB · Views: 452
  • #20
OK. How would you relate the question "Find the force with which the rubber tube is stretched" to the force on a circular cross section (i.e., one of your blue forces in your figure)?
 
  • #21
I am not entirely convinced but I will give a try

The force with which left half of the rubber tube is stretched = Sum of the forces on the two circular cross sections = 2(force on a circular cross section)
 
  • #22
So, it comes down to interpreting the question. I tend to think that the question is asking for the "tension" force in the rubber tube created by the moving fluid. If you take a cross section of the tube, then the tension is the total force on the circumference of the tube at the cross section, as shown in your figure. If that's what the problem is asking for, then you should be able to get the answer easily from your free-body diagram.
 
  • #23
Is post#21 right ?

Force on a circular cross section = 1/2(Force on the left half of the tube) = ρAv2 .

Is it correct ?

TSny said:
So, it comes down to interpreting the question. I tend to think that the question is asking for the "tension" force in the rubber tube created by the moving fluid.

Okay

TSny said:
If you take a cross section of the tube, then the tension is the total force on the circumference of the tube at the cross section, as shown in your figure.

I think this is the crux . But ,sorry I didn't understand .Could you please elaborate
 
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  • #24
Tanya Sharma said:
Is post#21 right ?

I would say that's right. But I think the force they want is the force on a circular cross section.

Force on a circular cross section = 1/2(Force on the left half of the tube) = ρAv2 .

Is it correct ?

Yes. I think that's the answer to the question.

But ,sorry I didn't understand .Could you please elaborate .

I was just trying to say that my interpretation of the question is to calculate the force on a circular cross section, which is what you have done. Hope I haven't misinterpreted the question.
 
  • #25
Okay...so our goal was to find the force at any cross section of the tube and for that we chose half the circumference as it is easy to find the force on it with the help of change of momentum of the liquid . Right ?

Now after doing the problem ,I am not able to relate this to the OP i.e the case of a U tube.In the OP we applied a similar approach but we were not calculating the force at any cross section ,rather the net force on the U tube .

How does the same concept fit in both the cases ?
 
  • #26
Tanya Sharma said:
Okay...so our goal was to find the force at any cross section of the tube and for that we chose half the circumference as it is easy to find the force on it with the help of change of momentum of the liquid . Right ?

Yes.

Now after doing the problem ,I am not able to relate this to the OP i.e the case of a U tube.In the OP we applied a similar approach but we were not calculating the force at any cross section ,rather the net force on the U tube .

How does the same concept fit in both the cases ?

The curved portion of the U-tube may be considered a semicircle. In solving the U-tube question, you found the force that the fluid exerts on the curved, semicircular portion of the U-tube. So, this is also the force that the fluid exerts on a semicircle of the rubber tube. That was the purpose of relating the circular rubber tube question to the U-tube question.

Once you have the net force on a semicircular portion of the rubber tube, you can deduce the tension force at a cross section.
 
  • #27
Excellent ! Thank you very much TSny.

OK...enough of physics :smile:

By the way , "Thanks ∞ " looks very impressive :wink:

In between I didn't realize I crossed 1000 posts :biggrin:
 
  • #28
Tanya Sharma said:
In between I didn't realize I crossed 1000 posts :biggrin:

Congrats! you're now an old timer.
 
  • #29
Hello TSny

I have come up with another method to solve the problem in post#5 ,so I would like to discuss it with you .

I feel this problem is similar to finding the tension T in a ring of mass M and length L rotating with a constant angular velocity ω.If we consider a differential element dθ of the ring and do a force balance ,then we arrive at the result i.e ## T= \frac{MLω^2}{4\pi^2} ## .We will use this result in our problem .

In this problem ## M = \frac{ρ(\pi d^2)(2\pi R)}{4} = \frac{ρ\pi^2d^2R}{2} ,L = 2\pi R ,ω = \frac{v}{R} ##

$$ T = \frac{ρ\pi d^2v^2}{4} = ρAv^2 $$ .This is same result as we obtained in post#23 :smile:

What is your opinion ?
 
  • #30
That looks good. Nice work.
 

1. How do you calculate centripetal force in a bent tube?

To calculate centripetal force in a bent tube, you need to know the mass of the object moving in the tube, the radius of the curve, and the speed of the object. The formula for centripetal force is Fc = (mv^2)/r, where Fc is the centripetal force, m is the mass, v is the velocity, and r is the radius of the curve.

2. What is the difference between centripetal force and centrifugal force?

Centripetal force is the force that keeps an object moving in a circular path, while centrifugal force is the apparent outward force that seems to push an object away from the center of rotation. Centrifugal force is a fictitious force and does not actually exist, but it is used to explain the motion of objects in a rotating frame of reference.

3. Can centripetal force change the direction of an object's motion?

Yes, centripetal force is responsible for changing the direction of an object's motion and keeping it moving in a circular path. Without centripetal force, an object would move in a straight line tangent to the curve.

4. How does the radius of the curve affect the centripetal force?

The radius of the curve is directly proportional to the centripetal force. This means that as the radius decreases, the centripetal force increases, and vice versa. This is because a smaller radius requires a greater force to keep an object moving in a circular path.

5. What happens to the centripetal force if the speed of the object increases?

If the speed of the object increases, the centripetal force must also increase to keep the object moving in a circular path. This is because a greater speed means a greater velocity, which in turn requires a greater force to maintain the circular motion.

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