# Is this right?, conservation of energy

• sg001
In summary, the problem involves a water tank on a frictionless surface and the goal is to determine the final velocity of the tank once the water has run out. Different approaches are suggested, including using conservation of energy and momentum. The idea of the horizontal center of mass is discussed and it is determined that the center of mass cannot change.
sg001

## Homework Statement

The problem relates to a water tank see link
http://www.phys.unsw.edu.au/~jw/1131/IntroLS.pdf
and scoll down a little!

## The Attempt at a Solution

so I think to go about this problem I need to use a conservation of energy approach.
ie ke(i) + gpe(i)= ke(f) + gpe(f)

so...
initially the tap is not realesed thererore there is no kinetic energy at this time
if I add the element (h) height, I can say the system at this a t=0 has gpe = (M + m)g(h)

Hence, (M+m)g(h) = 1/2mv^2
therefore, v^2 = (2(M+m)g(h))/m
v= √ (2(M+m)g(h))/m (m/s)

Is this approach correct.
And I am assuming that the tank would keep on going in the -x direction as the surface is frictionless...correct?

And can I introduce the element (h)?
Or would I have to use calculus to find a solution?

Any help is appreciated!

what do you mean keep going in the -x direction? why do you think it was going in the -x direction in the first place?

beardo34 said:
what do you mean keep going in the -x direction? why do you think it was going in the -x direction in the first place?

I would assume that since when the tap is released a force would be created in the x direction..ie, the water flowing to the end of the pipe before it is released... so there would be an equal but opposite force in the -x direction..Newtons third law.
Is this wrong?

EDIT: I have thought about this problem a bit and it seems to me that the tank should move to -x direction with a constant velocity(conservation of momentum) until the water hits the vertical part of the tap after which the tank would stop moving. Therefore the tank would be displaced a little to the left.

Last edited:
Ok I think I agree

When the tank is full the tank + water have a horizontal center of mass. Since the surface is frictionless and all horizontal forces are internal to the system, the center of mass should not change. Where's the center of mass after the water has run out?

This is a duplicate thread! I guess sg001 wanted more advice from other people. Which is fair enough I guess.

can some explain why the final velocity is not zero?

I have the equation vf = √((2mgh)/M) m/s
I thought this eqn was for the water... perhaps it is for the tank?

sg001 said:
can some explain why the final velocity is not zero?
Do you know for a fact that it is nonzero?

Yes
according to my physics teacher, when I saaid
vf=0 and distance traveled =0.
he said no.

sg001 said:
Yes
according to my physics teacher, when I saaid
vf=0 and distance traveled =0.
he said no.

Did he say "No" to both?

no, he just said no in general.
So are you saying that it did have a final vf ≠0

sg001 said:
no, he just said no in general.
So are you saying that it did have a final vf ≠0

Not necessarily It could have a nonzero displacement and vf = 0, for example.

gneill said:
Not necessarily It could have a nonzero displacement and vf = 0, for example.

But the tank is on a frictionless surface so I would assume that once it had a velocity it would remain constant.

And I don't see where the force would come from to make the object have a vf =0, once the object was in motion.

sg001 said:
But the tank is on a frictionless surface so I would assume that once it had a velocity it would remain constant.

And I don't see where the force would come from to make the object have a vf =0, once the object was in motion.

Suppose that you stand on a frictionless surface and you have a heavy ball in your arms. Now you extend your arms out in front of you holding the ball. After your arms are extended you drop the ball straight down. Describe the motions and positions.

gneill said:
When the tank is full the tank + water have a horizontal center of mass. Since the surface is frictionless and all horizontal forces are internal to the system, the center of mass should not change. Where's the center of mass after the water has run out?
what exactly do you mean by HORIZONTAL centre of mass??
does this mean the mass is evenly distributed along the horizonatl axis of the tank(not including the pipe).

If the centre of mass changes could this provide the fore need to make vf=0??

gneill said:
Suppose that you stand on a frictionless surface and you have a heavy ball in your arms. Now you extend your arms out in front of you holding the ball. After your arms are extended you drop the ball straight down. Describe the motions and positions.

This seems weird but I guess you would fall on your back, after the ball is released?

Or even possibly land flat on your stomach if you were to quickly extend your arms out.?

sg001 said:
what exactly do you mean by HORIZONTAL centre of mass??
does this mean the mass is evenly distributed along the horizonatl axis of the tank(not including the pipe).
The center of mass will have some location (x,y). The tank cannot move in the y direction, but it can in the x direction.
If the centre of mass changes could this provide the fore need to make vf=0??
The center of mass (at least for those parts of the system that we consider to be free to move) cannot change. That's the whole point of center of mass and conservation of momentum.

sg001 said:
This seems weird but I guess you would fall on your back, after the ball is released?

Or even possibly land flat on your stomach if you were to quickly extend your arms out.?

Okay, so you're sitting stably on a sled and can't fall over.

gneill said:
Okay, so you're sitting stably on a sled and can't fall over.

To me it would seem that you would go towards the left because if you can't fall over you should at least have some "kick" to move you a little.

sg001 said:
To me it would seem that you would go towards the left because if you can't fall over you should at least have some "kick" to move you a little.

Apply conservation of momentum. You thrust the ball out and hold it steady. Compared to before, where are you, where is the ball, and what are the velocities?

gneill said:
Apply conservation of momentum. You thrust the ball out and hold it steady. Compared to before, where are you, where is the ball, and what are the velocities?

initially
pball= 0 & pyou=0 therefore ptot (i) = 0

final
pball = mv & pyou = -(mv) therefore ptot (f) = 0

therefore you move in the opposite direction of the ball.
Assuming your mass is greater than that of the ball, you have a slower velocity than that of the ball and in the opposite direction.

But then once your in motion there has to be another force to make your speed = 0.

sg001 said:
initially
pball= 0 & pyou=0 therefore ptot (i) = 0

final
pball = mv & pyou = -(mv) therefore ptot (f) = 0

therefore you move in the opposite direction of the ball.
Assuming your mass is greater than that of the ball, you have a slower velocity than that of the ball and in the opposite direction.

But then once your in motion there has to be another force to make your speed = 0.

I said that you hold the ball out, not throw it. You retain the ball. The only motion occurs as you move the ball from your lap to the extended position where you hold it firm.

gneill said:
I said that you hold the ball out, not throw it. You retain the ball. The only motion occurs as you move the ball from your lap to the extended position where you hold it firm.

oh ok...

So the initial momentum is zero, so the final momentum must also be zero. (conservation)

But when you hold the ball out, then I would assume that it takes momentum to do this from yourself,
so the ball must have the minus momentum that it takes for you to hold the ball out?

sg001 said:
oh ok...

So the initial momentum is zero, so the final momentum must also be zero. (conservation)

But when you hold the ball out, then I would assume that it takes momentum to do this from yourself,
so the ball must have the minus momentum that it takes for you to hold the ball out?

Momentum is only nonzero when things are moving. While the ball is being moved you are moving in the opposite direction in such a way that momentum is conserved and the center of mass stays fixed. When the relative velocity between you and the ball stops, all motion stops. Total momentum remains zero, your velocity is zero, the ball's velocity is zero, but both have new locations. Surely you've seen the concept of center of mass before?

gneill said:
Momentum is only nonzero when things are moving. While the ball is being moved you are moving in the opposite direction in such a way that momentum is conserved and the center of mass stays fixed. When the relative velocity between you and the ball stops, all motion stops. Total momentum remains zero, your velocity is zero, the ball's velocity is zero, but both have new locations. Surely you've seen the concept of center of mass before?

It's seems very simple and logical at that, but no I have not had the concept "explained" to me, I just took it as a general term.

So in terms of the question now...
I would assume that when the water is released and is traveling in the x direction, consequently the tank moves to the left.
Now does the tank keep moving until the last drop of water has been expelled in the x direction?

and I need to come up with an expression for vf and the distance travelled.
So I would assume under the "concept" of cenntre of mass the vf of tank would be zero.
Would I need an expression to "claim" this or is a simple answer justified?

As far as the distance travelled...
the water does not come out horizontally therefore it can't have traveled further than it initially has when the tap is released.
So the tank has moved in the - x direction, now any ideas how to approach an "expression" for this??

Thanks!

sg001 said:
It's seems very simple and logical at that, but no I have not had the concept "explained" to me, I just took it as a general term.

So in terms of the question now...
I would assume that when the water is released and is traveling in the x direction, consequently the tank moves to the left.
Now does the tank keep moving until the last drop of water has been expelled in the x direction?
Isn't that one of the things that the problem is asking you to determine? What are your thoughts? Have you considered extending the little thought experiment that we just did with the heavy ball so that it becomes a better analogy to the water tank?
and I need to come up with an expression for vf and the distance travelled.
So I would assume under the "concept" of cenntre of mass the vf of tank would be zero.
Would I need an expression to "claim" this or is a simple answer justified?
The original problem statement only asked "After, where is the tank and which way is it moving?" You might only need to state whether or not there has been a shift of the tank either left or right and whether or not the tank is still in motion (and in what direction) when the tank is finally empty. Having a well reasoned argument handy would help.
As far as the distance travelled...
the water does not come out horizontally therefore it can't have traveled further than it initially has when the tap is released.
So the tank has moved in the - x direction, now any ideas how to approach an "expression" for this??
I'd suggest thinking about the problem by successively improving an analogous model.

Take the discrete approach first (heavy ball analogy) to see what the math looks like. Then see what happens if there are several balls. Eventually see what happens as the balls are made smaller but more numerous, until you approach a continuous stream (fluid).

Note that I haven't done a close analysis myself; I'm just offering the thoughts that occurred to me when I saw the problem statement.

Ok so I had a look at centre of mass concept on hyperphysics and from what I gathered the eqn of distance traveled should look something like this...

x= ∫(with limits (M+m) & 0)

x=∫(x.dm)/(M+m)

Is this corect or am I way off??

sg001 said:
Ok so I had a look at centre of mass concept on hyperphysics and from what I gathered the eqn of distance traveled should look something like this...

x= ∫(with limits (M+m) & 0)

x=∫(x.dm)/(M+m)

Is this corect or am I way off??

I think you're getting warm, but I think that the upper limit should reflect the amount of water moved, not the total mass of the tank+water.

When a small bit of mass dm is moved to a position L away from the center of mass of the tank, the remaining tank+water must move so that the overall center of mass occupies the same place as before (the center of mass of the system must remain fixed). So in the diagram above, the tank has moved distance dR from the old center of mass, while the bit of water dm has moved a distance dr from the center of mass. Then, if M is the tank's mass, m is water's mass, then

##(M + m - dm)\;dR = dr\;dm ~~~~~~~~~ dR + dr = L##

Solve for dR in terms of dm, then integrate to sum up all the dR's to find the total shift.

#### Attachments

• Fig1.gif
1.6 KB · Views: 466
ok so dR =(dr*dm)/(M+m-dm)

But then how would I integrate that??
doesnt it have to many terms in it?

sg001 said:
ok so dR =(dr*dm)/(M+m-dm)

But then how would I integrate that??
doesnt it have to many terms in it?

You didn't take advantage of the fact that L = dr + dR in order to eliminate dr. Take the initial expression and replace dr right away.

gneill said:
You didn't take advantage of the fact that L = dr + dR in order to eliminate dr. Take the initial expression and replace dr right away.

Ok so I have dR= ((L-dR)dm)/(M+m-dm)

So integrating

R=∫((L-dR)dm)/(M+m-dm)

But how do I integrate that?
I still have to many terms?

sg001 said:
Ok so I have dR= ((L-dR)dm)/(M+m-dm)
You haven't isolated dR. It still appears on both sides of the expression.

gneill said:
You haven't isolated dR. It still appears on both sides of the expression.

I am aware but from the eqn's

dR +dr = L & (M+m-dm)dR=dr(dm)

You can't isolate dR without introducing dr...
am I missing something?

sg001 said:
I am aware but from the eqn's

dR +dr = L & (M+m-dm)dR=dr(dm)

You can't isolate dR without introducing dr...
am I missing something?

It's just algebra. Replace dr using dr = L - dR. Move everything to the L.H.S. Expand and collect all the dR terms together. Rearrange with dR alone on the L.H.S.

• Introductory Physics Homework Help
Replies
3
Views
574
• Introductory Physics Homework Help
Replies
5
Views
975
• Introductory Physics Homework Help
Replies
44
Views
4K
• Introductory Physics Homework Help
Replies
2
Views
353
• Introductory Physics Homework Help
Replies
6
Views
540
• Introductory Physics Homework Help
Replies
7
Views
251
• Introductory Physics Homework Help
Replies
9
Views
538
• Introductory Physics Homework Help
Replies
11
Views
1K
• Introductory Physics Homework Help
Replies
6
Views
1K
• Introductory Physics Homework Help
Replies
10
Views
802