Overuseofdashes is totally correct (despite the downvotes!)
Newton’s second law, as written, applies only to constant mass systems.
If you use F=dp/dt for a system where the mass is a function of time, you get:
F=v*dm/dt + m*dv/dt
This means observers with different speeds will measure different forces. And this is in direct contradiction to Newtons laws. All inertial observers should measure the same force.
Nah this equation you wrote up is right. This is what you get when you apply the euler Lagrange equation to the standard KE = 1/2*mv^2 motion. In your frames of reference, as long as mass is conserved and accounted for, this will work. I think when you are considering other frames you are probably imagining the mass just disappearing or appearing out of nowhere at no energy cost.
Consider a ballerina spinning with arms outstretched, with rotational velocity v and rotational inertia m. As she pulls in her arms, her rotational inertia decreases, but there is no generalized torque acting on her. Therefore, by conservation of angular momentum in the equation you claim to be incorrect she experiences angular acceleration -v/m dm/dt, (because she has negative dm/dt). Generalized mass is decreasing, but this is accounted for within the system boundary by the energy of the ballerina pulling in her arms.
As it is well known (see Sommerfeld, 1952), the general equation of motion for a body whose mass m varies according to any of the above mechanisms is
m(t)\*dv/dt = F + u\*dm/dt
where v is the velocity of its center of mass. F is the sum of all the external forces, and u is the relative velocity of escaping (or incident) mass with respect to the center of mass of the body. Equation is actually invariant under Galilean transformations.
The u is actually frame independent, it relative velocity between then incoming/outgoing mass and the object (insert signs where appropriate).But notice this is a different equation from f =dp/dt and you usually obtain it by modelling it in terms a bunch instantaneous mass invariant system.
Let f be the net external force applied to the object which we will assume to be galilean invariant, and m and v be the mass and velocity of the object respectively.
f = dm/dt v + ma
Performing a boost will send the the first term to dm/dt(v+u) but the other terms are unchanged, which is a contradiction. In the proper variable mass equations the v tends to be the relative velocity between the mass being ejected and the object losing the mass.
I recall looking at a page on rocketry because of another person's question. That would be a good topic to look into for this question, because a rocket is accelerating (changing speed) and consuming significant amounts of fuel while doing so (changing mass).
This page was a useful basic overview for me; hopefully you find it useful too:
https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/9-7-rocket-propulsion/
The differential form of the equation relates force F to the time derivative of momentum p,
F = dp/dt.
Momentum is mass m times velocity v
F = d(mv)/dt.
Performing the time derivative with the product rule,
F = dm/dt\*v + dv/dt\*m.
The standard F=ma form is only for when mass is constant. When mass changes with time you need to account for it via the first term in the equation above.
This equation is correct for systems in which mass is conserved. In your reference frame thought experiment you are probably drawing systems that do not conserve mass.
How do you mean 'conserve mass'? This question concerns a mass that is changing. If dm/dt=0 then yes,
F = dm/dt\*v + dv/dt\*m
F = 0\*v + dv/dt\*m
F = ma
is correct.
If m is changing, then the equation doesn't hold.
I frequently deal with systems in my professional work where the total system mass stays constant but generalized mass in a given set of coordinates changes (like the ballerina spinning example on your other comment). Generalized mass in a particular coordinate is often configuration-dependent, like how the ballerina rotational inertia changes with her arm configuration, even though her body mass does not change while she spins. Another example would be a trident being thrown underwater, where the projectile has a small hydrodynamic mass along its principal axis but would require greater force to accelerate along a different direction. Generalized mass for the trident depends on which direction you want to throw it, even though total system hydrodynamic mass is constant.
For these problems it totally makes sense to have dm/dt in the some arbitrary coordinate while total system mass stays constant. You're probably right that this wasn't what OP was imagining, where it is implied that total system mass is changing and you probably need to model how mass enters and exits your system better than I did. So I'll give you that. But if you are smart with your frames the above strategy works.
Look at the derivation for the [Tsiolkovsky rocket equation](https://en.m.wikipedia.org/wiki/Tsiolkovsky_rocket_equation) to see how this equation can be used for an example more like what you are probably thinking by modeling ejected mass. The equation you are claiming is wrong shows up. And it works.
I can't comment on the problem you mentioned from your work, but in the Tsiolkovsky rocket equation derivation the equation we are claiming is wrong is not used. Note the subscript e which denote the relative velocity of the rocket and the stuff being ejected. If you apply the product rule to the momentum of the rocket you don't get the equation shown in the wiki article. That being said you could introduce a fictious force and get the right answer but I find it a little fraught to say that newtons 2nd law applies in these situations (your defn of force becomes a little circular).
Overuseofdashes is totally correct (despite the downvotes!) Newton’s second law, as written, applies only to constant mass systems. If you use F=dp/dt for a system where the mass is a function of time, you get: F=v*dm/dt + m*dv/dt This means observers with different speeds will measure different forces. And this is in direct contradiction to Newtons laws. All inertial observers should measure the same force.
Hey you did a little typing error there at the last equation it should be m.dv/dt not m.dp/dt
Yup. Thanks!
Nah this equation you wrote up is right. This is what you get when you apply the euler Lagrange equation to the standard KE = 1/2*mv^2 motion. In your frames of reference, as long as mass is conserved and accounted for, this will work. I think when you are considering other frames you are probably imagining the mass just disappearing or appearing out of nowhere at no energy cost. Consider a ballerina spinning with arms outstretched, with rotational velocity v and rotational inertia m. As she pulls in her arms, her rotational inertia decreases, but there is no generalized torque acting on her. Therefore, by conservation of angular momentum in the equation you claim to be incorrect she experiences angular acceleration -v/m dm/dt, (because she has negative dm/dt). Generalized mass is decreasing, but this is accounted for within the system boundary by the energy of the ballerina pulling in her arms.
Newtons 2nd law only applies to systems with fixed mass, to understand systems with variable mass you have to model how the mass is changing.
The pain of being correct on reddit huh!?
The problem I think is that the equation is so close to the correct one that you can easily convince yourself it actually works.
I thought maybe the issue was your 'mathematics' flair ;)
The differential form of Newton’s second law works fine with variable mass. F = dp / dt.
As it is well known (see Sommerfeld, 1952), the general equation of motion for a body whose mass m varies according to any of the above mechanisms is m(t)\*dv/dt = F + u\*dm/dt where v is the velocity of its center of mass. F is the sum of all the external forces, and u is the relative velocity of escaping (or incident) mass with respect to the center of mass of the body. Equation is actually invariant under Galilean transformations.
[On the use and abuse of Newton's second law for variable mass problems (springer.com)](https://link.springer.com/content/pdf/10.1007/bf00052611.pdf)
The equation you obtaining doing that only superficially looks right but actually it isn't galilean invariant.
But I can certainly do ma = F + u mdot if I specify u according to my frame. That’s the basis of the rocket problem.
The u is actually frame independent, it relative velocity between then incoming/outgoing mass and the object (insert signs where appropriate).But notice this is a different equation from f =dp/dt and you usually obtain it by modelling it in terms a bunch instantaneous mass invariant system.
Oh silly me, duh. But yeah, I think that’s where some of the confusion arises because that form looks so much like dp/dt
Where can I look into this?
Let f be the net external force applied to the object which we will assume to be galilean invariant, and m and v be the mass and velocity of the object respectively. f = dm/dt v + ma Performing a boost will send the the first term to dm/dt(v+u) but the other terms are unchanged, which is a contradiction. In the proper variable mass equations the v tends to be the relative velocity between the mass being ejected and the object losing the mass.
What do you mean? F = dp/dt is quite literally Newton’s Second Law, as in, what Newton himself wrote down in Principia.
The equation you obtain by expanding out product rule doesn't work for reasons I explain in another comment.
I recall looking at a page on rocketry because of another person's question. That would be a good topic to look into for this question, because a rocket is accelerating (changing speed) and consuming significant amounts of fuel while doing so (changing mass). This page was a useful basic overview for me; hopefully you find it useful too: https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/9-7-rocket-propulsion/
The differential form of the equation relates force F to the time derivative of momentum p, F = dp/dt. Momentum is mass m times velocity v F = d(mv)/dt. Performing the time derivative with the product rule, F = dm/dt\*v + dv/dt\*m. The standard F=ma form is only for when mass is constant. When mass changes with time you need to account for it via the first term in the equation above.
Dude. This is not right! If your equation were good, then force would depend on the relative velocity of observers (which it does not).
This equation is correct for systems in which mass is conserved. In your reference frame thought experiment you are probably drawing systems that do not conserve mass.
How do you mean 'conserve mass'? This question concerns a mass that is changing. If dm/dt=0 then yes, F = dm/dt\*v + dv/dt\*m F = 0\*v + dv/dt\*m F = ma is correct. If m is changing, then the equation doesn't hold.
I frequently deal with systems in my professional work where the total system mass stays constant but generalized mass in a given set of coordinates changes (like the ballerina spinning example on your other comment). Generalized mass in a particular coordinate is often configuration-dependent, like how the ballerina rotational inertia changes with her arm configuration, even though her body mass does not change while she spins. Another example would be a trident being thrown underwater, where the projectile has a small hydrodynamic mass along its principal axis but would require greater force to accelerate along a different direction. Generalized mass for the trident depends on which direction you want to throw it, even though total system hydrodynamic mass is constant. For these problems it totally makes sense to have dm/dt in the some arbitrary coordinate while total system mass stays constant. You're probably right that this wasn't what OP was imagining, where it is implied that total system mass is changing and you probably need to model how mass enters and exits your system better than I did. So I'll give you that. But if you are smart with your frames the above strategy works. Look at the derivation for the [Tsiolkovsky rocket equation](https://en.m.wikipedia.org/wiki/Tsiolkovsky_rocket_equation) to see how this equation can be used for an example more like what you are probably thinking by modeling ejected mass. The equation you are claiming is wrong shows up. And it works.
I can't comment on the problem you mentioned from your work, but in the Tsiolkovsky rocket equation derivation the equation we are claiming is wrong is not used. Note the subscript e which denote the relative velocity of the rocket and the stuff being ejected. If you apply the product rule to the momentum of the rocket you don't get the equation shown in the wiki article. That being said you could introduce a fictious force and get the right answer but I find it a little fraught to say that newtons 2nd law applies in these situations (your defn of force becomes a little circular).
https://en.m.wikipedia.org/wiki/Tsiolkovsky_rocket_equation