Helium balloon energy conservation

[member 731016]

Homework Statement

Please see below

Relevant Equations

Please see

For this problem,

How can energy be conserved if the bit highlighted in orange is true?Many thanks!

 

[hmmm27]

What do you think the "bit highlighted in orange" refers to ? the balloon, or the air.

 

[member 731016]

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!

 

[Steve4Physics]

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

 

[member 731016]

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!

 

[jbriggs444]

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?

 

[member 731016]

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!

 

[jbriggs444]

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.

 

[Steve4Physics]

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.

 

[jbriggs444]

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.

 

[Steve4Physics]

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.

 

[kuruman]

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?

 

[member 731016]

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.

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.

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.

Thank you for your replies @jbriggs444 and @Steve4Physics !

I think I understand now :)

 

[member 731016]

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!

 

[kuruman]

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.

 

[member 731016]

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.