Is this right?, conservation of energy

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.
  • #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 ##
 
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  • #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!
 
<h2>1. What is the law of conservation of energy?</h2><p>The law of conservation of energy states that energy cannot be created or destroyed, but can only be transformed from one form to another. This means that the total amount of energy in a closed system remains constant over time.</p><h2>2. How does the law of conservation of energy apply to everyday life?</h2><p>The law of conservation of energy applies to many aspects of everyday life, such as the energy we use to power our homes and vehicles. It also applies to natural processes, such as photosynthesis and the water cycle, where energy is constantly being transformed from one form to another.</p><h2>3. Is it possible to violate the law of conservation of energy?</h2><p>No, the law of conservation of energy is a fundamental law of physics and has been proven to hold true in all observed cases. If it were to be violated, it would mean that the fundamental principles of the universe are not consistent, which is highly unlikely.</p><h2>4. How does the law of conservation of energy relate to other laws of thermodynamics?</h2><p>The law of conservation of energy is the first law of thermodynamics, which states that energy cannot be created or destroyed. The second law of thermodynamics states that in any energy transformation, some energy will always be lost as heat. Together, these laws explain how energy behaves in a closed system.</p><h2>5. Can energy be converted from one form to another without any loss?</h2><p>No, according to the second law of thermodynamics, some energy will always be lost as heat in any energy transformation. This is due to the fact that no energy conversion process is 100% efficient, and some energy will always be lost as a result of friction, resistance, or other factors.</p>

1. What is the law of conservation of energy?

The law of conservation of energy states that energy cannot be created or destroyed, but can only be transformed from one form to another. This means that the total amount of energy in a closed system remains constant over time.

2. How does the law of conservation of energy apply to everyday life?

The law of conservation of energy applies to many aspects of everyday life, such as the energy we use to power our homes and vehicles. It also applies to natural processes, such as photosynthesis and the water cycle, where energy is constantly being transformed from one form to another.

3. Is it possible to violate the law of conservation of energy?

No, the law of conservation of energy is a fundamental law of physics and has been proven to hold true in all observed cases. If it were to be violated, it would mean that the fundamental principles of the universe are not consistent, which is highly unlikely.

4. How does the law of conservation of energy relate to other laws of thermodynamics?

The law of conservation of energy is the first law of thermodynamics, which states that energy cannot be created or destroyed. The second law of thermodynamics states that in any energy transformation, some energy will always be lost as heat. Together, these laws explain how energy behaves in a closed system.

5. Can energy be converted from one form to another without any loss?

No, according to the second law of thermodynamics, some energy will always be lost as heat in any energy transformation. This is due to the fact that no energy conversion process is 100% efficient, and some energy will always be lost as a result of friction, resistance, or other factors.

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