Helium balloon energy conservation

In summary: OP's logic when they said "But the finial and initial KE are zero, so cancel in the conservation of energy equation. ". So X (correctly) does not include losses during the ascent.
  • #1
ChiralSuperfields
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Homework Statement
Please see below
Relevant Equations
Please see
For this problem,
1678736286318.png

How can energy be conserved if the bit highlighted in orange is true?Many thanks!
 
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  • #2
What do you think the "bit highlighted in orange" refers to ? the balloon, or the air.
 
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  • #3
hmmm27 said:
What do you think the "bit highlighted in orange" refers to ? the balloon, or the air.
Thank you for your reply @hmmm27!

Air.

Many thanks!
 
  • #4
Callumnc1 said:
View attachment 323567
How can energy be conserved if the bit highlighted in orange is true?
Remember that the balloon starts moving (gains kinetic energy). Can you take it from there?

Edit - typo's
 
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  • #5
Steve4Physics said:
Remember that the balloon starts moving (gains kinetic energy). Can you take it from there?

Edit - typo's
Thank you for your reply @Steve4Physics !

But the finial and initial KE are zero, so cancel in the conservation of energy equation.

Many thanks!
 
  • #6
Callumnc1 said:
But the finial and initial KE are zero, so cancel in the conservation of energy equation.
But there was kinetic energy in between. Where did it go?
 
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  • #7
jbriggs444 said:
But there was kinetic energy in between. Where did it go?
Thank you for your reply @jbriggs444!

Maybe internal energy since the balloon would collide with air molecules? Or are we assuming no non-conservative forces are acting?

Many thanks!
 
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  • #8
Callumnc1 said:
Maybe internal energy since the balloon would collide with air molecules?
Right. That is the place where kinetic energy almost always goes.

The scenario with the helium baloon is not very different in principle from a stationary block that is held aloft by a string in a closed room. If the string breaks, the block falls and winds up stationary on the floor. The center of mass of the block has descended. The center of mass of the air has ascended. The total potential energy for the things in the room has decreased. The reduction in total mechanical energy is balanced by an increase in internal energy.

The only difference between that situation and the case at hand is that it is the air that has fallen and the helium baloon which has been displaced thereby.
 
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  • #9
Callumnc1 said:
Thank you for your reply @Steve4Physics !

But the finial and initial KE are zero, so cancel in the conservation of energy equation.
GPE lost by air = GPE gained by balloon + X

X is the kinetic energy gained by the balloon up to just before impact of balloon with ceiling (while the balloon is still moving).

After impact, X represents the heat generated due to the inelastic collision between the balloon and the ceiling.
 
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  • #10
Steve4Physics said:
After impact, X represents the heat generated due to the inelastic collision between the balloon and the ceiling.
And, of course, the heat generated due to turbulence and viscous drag during the ascent.
 
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  • #11
jbriggs444 said:
And, of course, the heat generated due to turbulence and viscous drag during the ascent.
Yes, thanks. The other losses must be accounted.

I defined 'X' a the balloon's kinetic energy just before impact with the ceiling. I was trying to point out the flaw in the OP's logic when they said "But the finial and initial KE are zero, so cancel in the conservation of energy equation. ". So X (correctly) does not include losses during the ascent.

I should have written:
GPE lost by air = GPE gained by balloon + X +Y
where Y covers the other losses. That was careless of me.
 
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  • #12
This problem can be modeled as exchanging the positions of two spheres of equal volume but different densities and hence masses (see figure below.) The masses are connected with a rigid rod and pivoted at the midpoint O. Initially the smaller mass is at the bottom (##m_1<m_2##). A disembodied hand exerts non-conservative forces and rotates the rod by 180° at constant speed. What is the change in potential energy?

HeliumBalloonRising.png
 
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  • #13
jbriggs444 said:
Right. That is the place where kinetic energy almost always goes.

The scenario with the helium baloon is not very different in principle from a stationary block that is held aloft by a string in a closed room. If the string breaks, the block falls and winds up stationary on the floor. The center of mass of the block has descended. The center of mass of the air has ascended. The total potential energy for the things in the room has decreased. The reduction in total mechanical energy is balanced by an increase in internal energy.

The only difference between that situation and the case at hand is that it is the air that has fallen and the helium baloon which has been displaced thereby.
Steve4Physics said:
GPE lost by air = GPE gained by balloon + X

X is the kinetic energy gained by the balloon up to just before impact of balloon with ceiling (while the balloon is still moving).

After impact, X represents the heat generated due to the inelastic collision between the balloon and the ceiling.
jbriggs444 said:
And, of course, the heat generated due to turbulence and viscous drag during the ascent.
Steve4Physics said:
Yes, thanks. The other losses must be accounted.

I defined 'X' a the balloon's kinetic energy just before impact with the ceiling. I was trying to point out the flaw in the OP's logic when they said "But the finial and initial KE are zero, so cancel in the conservation of energy equation. ". So X (correctly) does not include losses during the ascent.

I should have written:
GPE lost by air = GPE gained by balloon + X +Y
where Y covers the other losses. That was careless of me.
Thank you for your replies @jbriggs444 and @Steve4Physics !

I think I understand now :)
 
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  • #14
kuruman said:
This problem can be modeled as exchanging the positions of two spheres of equal volume but different densities and hence masses (see figure below.) The masses are connected with a rigid rod and pivoted at the midpoint O. Initially the smaller mass is at the bottom (##m_1<m_2##). A disembodied hand exerts non-conservative forces and rotates the rod by 180° at constant speed. What is the change in potential energy?

View attachment 323613
Thank you for your reply @kuruman!

##\Delta U_g = U_{gf} - U_{gi}##
##\Delta U_g = m_1gh + m_2gh - m_2g2h = m_1gh - m_2gh## Taking the origin to be at the COM of ##m_1## and the distance to the midpoint from COM of m_1 to be h.

Many thanks!
 
  • #15
Callumnc1 said:
Thank you for your reply @kuruman!

##\Delta U_g = U_{gf} - U_{gi}##
##\Delta U_g = m_1gh + m_2gh - m_2g2h = m_1gh - m_2gh## Taking the origin to be at the COM of ##m_1## and the distance to the midpoint from COM of m_1 to be h.

Many thanks!
This doesn't look right. To begin with, the masses are not equal therefore they are not equidistant from point O. You will be better off taking the zero of potential energy at the starting point of the balloon.

Also, the equation for ##\Delta U_g## that wrote down is incorrect. ##U_{gf}## and ##U_{gi}## should each have two terms because you have chosen the zero at the CM where neither mass can be found.
 
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  • #16
kuruman said:
This doesn't look right. To begin with, the masses are not equal therefore they are not equidistant from point O. You will be better off taking the zero of potential energy at the starting point of the balloon.

Also, the equation for ##\Delta U_g## that wrote down is incorrect. ##U_{gf}## and ##U_{gi}## should each have two terms because you have chosen the zero at the CM where neither mass can be found.
Thank you for your reply @kuruman!

I will come back to this problem once I have some more time from uni work.
 

1. What is helium balloon energy conservation?

Helium balloon energy conservation is the practice of using helium balloons as a source of renewable energy. This involves harnessing the energy produced by the movement of helium balloons to power various devices and reduce the use of non-renewable energy sources.

2. How do helium balloons generate energy?

Helium balloons generate energy through a process called piezoelectricity. When the balloons move, they create an electric charge that can be captured and used to power devices such as sensors, lights, or even generators.

3. What are the benefits of using helium balloons for energy conservation?

There are several benefits to using helium balloons for energy conservation. Firstly, helium is a renewable resource, so it does not contribute to the depletion of non-renewable energy sources. Additionally, helium balloons are lightweight and can be easily transported, making them a versatile source of energy. Furthermore, using helium balloons for energy conservation can reduce carbon emissions and help combat climate change.

4. Are there any limitations to using helium balloons for energy conservation?

While helium balloons have many benefits, there are also some limitations to consider. One limitation is that the amount of energy produced by a single balloon is relatively small, so a large number of balloons would be needed to power larger devices. Additionally, helium is a finite resource, so its use for energy conservation must be managed carefully to avoid depletion.

5. How can I incorporate helium balloon energy conservation in my daily life?

There are several ways to incorporate helium balloon energy conservation in your daily life. You can start by using helium balloons to power small devices like sensors or lights. You can also support companies and organizations that are researching and implementing helium balloon energy conservation. Additionally, you can reduce your overall energy consumption by using more energy-efficient devices and practices, making the use of helium balloons for energy conservation even more impactful.

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