Is this right?, conservation of energy

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Homework Help Overview

The problem involves a water tank and the principles of conservation of energy and momentum. Participants are exploring the dynamics of the tank as water is released, questioning the implications of forces and motion on a frictionless surface.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to apply conservation of energy to relate gravitational potential energy and kinetic energy, while also questioning the motion of the tank in relation to the water flow. Some participants explore the implications of Newton's third law and the conservation of momentum in this context.

Discussion Status

Participants are actively questioning assumptions about the direction of motion and the final velocity of the tank. There is a mix of interpretations regarding the conservation of momentum and the behavior of the system as water is released, with some guidance offered on the implications of center of mass and internal forces.

Contextual Notes

There are discussions about the assumptions made regarding the frictionless nature of the surface and the initial conditions of the system, as well as the definitions of forces involved in the motion of the tank and water.

  • #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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