Hi
I was trying to solve a problem of motion with mass changing along the way , such as in rocket : gas is going down and gaining mass while the rocket lose mass and going up.
Lets call the rocket #1 and the gas #2.
Lets say the derivative of M1 , the velocity of the gas relating to the rocket V0 , starting mass of the rocket M0 are known , and let M1,M2,V1,V2 be the mass and velocities of the rocket and gas [they change in time].
Now we know some things :
1. dM1/dt = -dM2/dt
because losing of mass to #1 is gaining mass for #2
Lets call dM1/dt = -h
so
dM2/dt = h
2. V2 = V1-V0
hence
dV1/dt = dV2/dt
because V0 don't change in time.
3. M1 = -ht+M0
M2 = ht
getting this by doing integral to the derivative of M1 and M2.
To make it easier to read i put it on artPad :
artpad.art.com/artpad/painter/?jedl9tm4yw8
The black dots above the variables means derivative in time.
As you can see , the acceleration is constant !
But in the book they say :
dV1/dt = hV0/M1
which is changing in time because M1 = -ht+M0
and then they get expression with "ln" function for the acceleration ..
I solved this thing "by the book" although the book says differently so I really don't know who is right !
Help !
Thanks for even reading this
Niv
The problem is rather subtle, because it involves the correct interpretation of the variables, rather than the math itself. The term M2v2' is where the trouble lies. V2 represents the velocity of the fuel CURRENTLY being ejected from the rocket, and so it is correct to say v2' = v1'. But M2 represents the total mass of all the fuel which has been ejected so far, and this trail of fuel is not of course all traveling at velocity v2; only the most recent increment of ejected fuel is traveling at that velocity.
There are two terms related to the rate of change of momentum of the entire fuel trail. The first, M2'v2 is certainly correct, because mass is being added to the fuel trail at a rate M2', and this additional mass enters the fuel trail at a velocity of v2, so it brings a momentum increment dM2v2 with it in a time interval dt. The value of the second term, M2v2', should be ZERO, because it represents the momentum change of the fuel trail due to the changing velocity of the fuel that is already in the trail. But the fuel in the trail is just coasting through space after it exits the rocket engine, so each individual bit of mass in it is moving at a constant velocity. All the different increments of fuel are traveling at different speeds because they were ejected at different times, but once they have been ejected, their speeds do not change.
This is a bit like a shell-game. You have to keep your eye on the mass. The bit of fuel that has velocity v2 at time t is not the same bit that has velocity v2+dv2 at time t + dt. The latter is a NEWLY EJECTED bit of fuel, and the first bit of fuel is still traveling at its old speed v2. So the short answer is, if you delete the term M2v2' from your algebra, you will get the correct result.
Another way to analyze this problem is to note that once the fuel leaves the rocket, it has no further interaction with the rocket, so it can be ignored. Therefore, you can analyze the system consisting of the rocket (m) together with a bit of fuel (dm) which it is just about to expel, and then look at momentum of the system just before and just after this bit of fuel is expelled. In that case, the initial momentum is (m+dm)v1 and the final momentum is m(v1+dv1) + dmv2. Setting Pinitial = Pfinal gives mv1 + dmv1 = mv1 + mdv1 + dmv1 - dmv0, so 0 = mdv1 - dmv0. Therefore dividing by dt gives mdv1/dt = v0dm/dt, i.e. mv' = v0m', which is the standard answer.
By the way, this rearranges to dv = v'dt = v0(dm/m) = v0d(ln(m)). Integrating this gives v(t) = v0·ln(m(t)/m(0)), so the final velocity of the rocket can be well above its exhaust velocity if the mass of the loaded rocket is mostly fuel.
Hope this helps!
--Stuart Anderson
There are two terms related to the rate of change of momentum of the entire fuel trail. The first, M2'v2 is certainly correct, because mass is being added to the fuel trail at a rate M2', and this additional mass enters the fuel trail at a velocity of v2, so it brings a momentum increment dM2v2 with it in a time interval dt. The value of the second term, M2v2', should be ZERO, because it represents the momentum change of the fuel trail due to the changing velocity of the fuel that is already in the trail. But the fuel in the trail is just coasting through space after it exits the rocket engine, so each individual bit of mass in it is moving at a constant velocity. All the different increments of fuel are traveling at different speeds because they were ejected at different times, but once they have been ejected, their speeds do not change.
This is a bit like a shell-game. You have to keep your eye on the mass. The bit of fuel that has velocity v2 at time t is not the same bit that has velocity v2+dv2 at time t + dt. The latter is a NEWLY EJECTED bit of fuel, and the first bit of fuel is still traveling at its old speed v2. So the short answer is, if you delete the term M2v2' from your algebra, you will get the correct result.
Another way to analyze this problem is to note that once the fuel leaves the rocket, it has no further interaction with the rocket, so it can be ignored. Therefore, you can analyze the system consisting of the rocket (m) together with a bit of fuel (dm) which it is just about to expel, and then look at momentum of the system just before and just after this bit of fuel is expelled. In that case, the initial momentum is (m+dm)v1 and the final momentum is m(v1+dv1) + dmv2. Setting Pinitial = Pfinal gives mv1 + dmv1 = mv1 + mdv1 + dmv1 - dmv0, so 0 = mdv1 - dmv0. Therefore dividing by dt gives mdv1/dt = v0dm/dt, i.e. mv' = v0m', which is the standard answer.
By the way, this rearranges to dv = v'dt = v0(dm/m) = v0d(ln(m)). Integrating this gives v(t) = v0·ln(m(t)/m(0)), so the final velocity of the rocket can be well above its exhaust velocity if the mass of the loaded rocket is mostly fuel.
Hope this helps!
--Stuart Anderson
Wow that was the best explanation ever !
I not only understand what was my mistake but now i also understand mechanics differently ... if you are not yet a professor , i pronounce you as one
Thanks from the moon and all the way back
Niv
I not only understand what was my mistake but now i also understand mechanics differently ... if you are not yet a professor , i pronounce you as one
Thanks from the moon and all the way back
Niv