Is this right?, conservation of energy

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SUMMARY

The discussion centers on the application of the conservation of energy principle to analyze the motion of a water tank on a frictionless surface when water is released. The equation used is ke(i) + gpe(i) = ke(f) + gpe(f), leading to the derived velocity equation v = √((2(M+m)g(h))/m). Participants debate the implications of Newton's third law and conservation of momentum, concluding that the tank will move in the negative x-direction as water exits, maintaining a constant velocity until the water is fully expelled. The center of mass remains unchanged due to internal forces, which is crucial for understanding the system's dynamics.

PREREQUISITES
  • Understanding of conservation of energy principles
  • Familiarity with Newton's laws of motion
  • Basic knowledge of momentum conservation
  • Concept of center of mass in physics
NEXT STEPS
  • Study the implications of Newton's third law in fluid dynamics
  • Learn about conservation of momentum in closed systems
  • Explore kinetic and potential energy equations in various scenarios
  • Investigate the concept of center of mass and its applications in mechanics
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, fluid dynamics, and energy conservation principles. This discussion is beneficial for anyone looking to deepen their understanding of motion on frictionless surfaces and the behavior of systems under internal forces.

  • #31
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.
 
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  • #32
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?
 
  • #33
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.
 
  • #34
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?
 
  • #35
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.
 
  • #36
Rethinking it a bit, a simpler approach would be to let M be the total original mass of the tank plus water. You want to move a quantity m of that total mass (representing the water) through the pipe. Then

##(M - dm)dR = dm\;dr ~~~~## and as before, ##~~dr = L - dR##

##dR = L \frac{dm}{M}##

Integration is trivial, with limits 0 → m

##R = \frac{m}{M}L ##
 
  • #37
gneill said:
Rethinking it a bit, a simpler approach would be to let M be the total original mass of the tank plus water. You want to move a quantity m of that total mass (representing the water) through the pipe. Then

##(M - dm)dR = dm\;dr ~~~~## and as before, ##~~dr = L - dR##

##dR = L \frac{dm}{M}##

Integration is trivial, with limits 0 → m

##R = \frac{m}{M}L ##

Ok so now the eqn for the distance traveled x = R=∫m/M *L with limits 0 &m

and final velocity =0.

Thanks!
 
  • #38
sg001 said:
Ok so now the eqn for the distance traveled x = R=∫m/M *L with limits 0 &m
That's R=∫(L/M)*dm
and final velocity =0.

Thanks!
 
  • #39
Thanks for the continuing help
Greatly appreciated.
 
  • #40
sg001 said:
Thanks for the continuing help
Greatly appreciated.

We'll see... :smile:
I've decided to add some confusion back into the mix. Upon rethinking the rethinking, I've decided that the previous form of the integration makes more sense :frown:

At some time partway through the emptying of the tank, let's say that an amount x of the water has already gone down the pipe. That leaves the tank plus remaining water with mass M-x. It's this remaining mass for which we need to find the dR associated with the next dx that's moved.

##((M - x) - dx) dR = dx\;dr##

##((M - x) - dx) dR - dx (L - dR) = 0~~~~~## because ##dr = L - dR##

##dR = L \frac{dx}{M - x} ##

Integrate that for x going from 0 to m, the mass of water drained. M is still the mass of the tank plus the initial amount of water.

EDIT: Changed variable dm to dx in order to have it reflect that it's affecting x, the water that's being shifted.
 
Last edited:
  • #41
gneill said:
We'll see... :smile:
I've decided to add some confusion back into the mix. Upon rethinking the rethinking, I've decided that the previous form of the integration makes more sense :frown:

At some time partway through the emptying of the tank, let's say that an amount x of the water has already gone down the pipe. That leaves the tank plus remaining water with mass M-x. It's this remaining mass for which we need to find the dR associated with the next dx that's moved.

##((M - x) - dx) dR = dx\;dr##

##((M - x) - dx) dR - dx (L - dR) = 0~~~~~## because ##dr = L - dR##

##dR = L \frac{dx}{M - x} ##


Integrate that for x going from 0 to m, the mass of water drained. M is still the mass of the tank plus the initial amount of water.

EDIT: Changed variable dm to dx in order to have it reflect that it's affecting x, the water that's being shifted.


ok so I have R = ∫L(dx/M-x) = ∫L (M-x)^-1 dx.

But then I have R= -L

Or could it be,,

R = -L(ln(M-x) + c

??
 
  • #42
sg001 said:
ok so I have R = ∫L(dx/M-x) = ∫L (M-x)^-1 dx.

But then I have R= -L

Or could it be,,

R = -L(ln(M-x) + c

??

You need to plug in both limits (0 and m, the total mass of water drained).
 
  • #43
ok...
so x = R = (-L(ln(M-m)) - (-L(ln(M-0))
 
  • #44
sg001 said:
ok...
so x = R = (-L(ln(M-m)) - (-L(ln(M-0))

Pull out the L and combine the ln's.
 
  • #45
Finally...
x = R= -L(ln(M-m)-(M-0))
 
  • #46
sg001 said:
Finally...
x = R= -L(ln(M-m)-(M-0))

You need to check your log manipulation math. ln(x) - ln(y) ≠ ln(x - y).
 
  • #47
gneill said:
You need to check your log manipulation math. ln(x) - ln(y) ≠ ln(x - y).

Ohh yeah My bad simple yet stupid mistake.

x= R = -L(ln(M-m)-ln(M-0))
 
  • #48
sg001 said:
Ohh yeah My bad simple yet stupid mistake.

x= R = -L(ln(M-m)-ln(M-0))

Still review your log math operations; Those logs can be combined into a single log.
 
  • #49
woops

R = -L(ln((M-m)/(M-0)) = x
 
  • #50
sg001 said:
woops

R = -L(ln((M-m)/(M-0)) = x

It can still be cleaned up some. Drop the zero and use the fact that -ln(b/a) = +ln(a/b).
 
  • #51
gneill said:
It can still be cleaned up some. Drop the zero and use the fact that -ln(b/a) = +ln(a/b).

Ok so I have now

R = L (ln(M/M-m)) = x
 
  • #52
sg001 said:
Ok so I have now

R = L (ln(M/M-m)) = x

That looks much neater.
 
  • #53
gneill said:
That looks much neater.

Thanks again gneil!
 
  • #54
sorry gneill!
 

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