Introduction

The applicability of Newton’s second legislation within the oft-quoted “common kind”  $$start{align}frac{dmathbf{P}}{dt}=mathbf{F}_{textual content{ext}}finish{align}$$ was a difficulty in a latest thread (see publish #4) in circumstances of methods with variable mass.  The next instance illustrates the sort of confusion that would come up from the (mis)software of Equation (1):

A rocket is hovering in place above floor close to the Earth’s floor. Assume that the combustion gases are expelled at fixed charge ##beta=dm/dt## with velocity ##w## relative to the rocket.  What situation should maintain for the rocket to hover in place?

A novice would possibly begin with Equation (1) and go down the backyard path solely to achieve a fast deadlock as proven under.

Tried resolution

We begin with the final type of Newton’s second legislation, Equation (1) $$frac{dP}{dt}=Mfrac{dV}{dt}+Vfrac{dM}{dt}=-Mg$$ If the rocket is simply hovering above the bottom, one would set ##V=0## and ##dfrac{dV}{dt}=0## to get ##dots## $$0+0=-Mg~~?$$

Along with the nonsensical outcome, it’s disconcerting to notice that the beginning equation doesn’t embody something associated to the expelled gases which offer the thrust. One may also ponder whether Equation (1) can be a common formulation of Newton’s second legislation and, whether it is, how it’s relevant in bodily conditions the place the system’s mass is variable.

We are going to tackle these points and, ultimately, it must be obvious to the reader that the issue just isn’t with Equation (1) however with the dearth of a transparent notion of what ##P## represents.  It’s straightforward to say “the momentum of the system” however what precisely is that when two methods are entangled within the sense that mass from one is picked up by the opposite?

We are going to derive a Newton’s second legislation equation for the variable mass system of curiosity. Then we’ll formulate a recipe for utilizing this equation.  We are going to resolve the hovering rocket paradox and clear up variable mass issues for example the tactic.  Lastly, we’ll discover whether or not it’s applicable to label Equation (1)  the final type of Newton’s second legislation.

Two entangled methods

We contemplate a system with momentum ##P_{textual content{sys}}## that’s the sum of subsystem momenta ##P_{textual content{1}}## and ##P_{textual content{2}}.## The subsystems should be chosen in order that no exterior mass crosses their mixed boundary.  All mass misplaced by one subsystem is gained by the opposite: ##d(m_1+m_2)=0.## We undertake the conference that subscript 1 labels the system of curiosity.  It consists of variable mass ##m_1## shifting with velocity ##v_1## within the lab body.  Subsystem 2 has variable mass ##m_2##.  Solely the portion of ##m_2,## ##~dm_2,## that’s exchanged with the subsystem 1 is of curiosity right here. It has velocity ##v_2## within the lab body.

We begin with Equation (1) written (in a single dimension for simplicity) within the kind $$dP_{textual content{sys}}=F_{textual content{ext}}~dt.$$ Then, $$dP_{textual content{sys}}=dP_{textual content{1}}+dP_{textual content{2}} = m_1~dv_1+v_1~dm_1+v_2~dm_2+m_2~dv_2.$$

Disentangling the system of curiosity

We notice that the exterior drive consists of two forces, ##F_{textual content{ext,1}}## appearing on system 1 completely and ##F_{textual content{ext,2}}## appearing on system 2 completely. Then $$m_1~dv_1+v_1~dm_1+v_2~dm_2+m_2~dv_2=(F_{textual content{ext,1}}+F_{textual content{ext,2}})~dt.$$ With a view to discover the acceleration of subsystem 1, we have to disentangle the 2 methods. First we substitute ##dm_2=-dm_1## and acquire $$m_1~dv_1+v_1~dm_1-v_2~dm_1+m_2~dv_2=(F_{textual content{ext,1}}+F_{textual content{ext,2}})~dt.$$ Then we summary the specified equation for subsystem 1 by deciding on phrases involving ##m_ 1## solely: $$m_1~dv_1=(v_2-v_1)~{dm_1}+F_{textual content{ext,1}}~dt.$$ From the final equation we are able to write Newton’s second legislation equation for the subsystem of curiosity when it exchanges mass on the charge ##dm/dt## with a second subsystem. We drop subscript 1 that’s now not helpful, substitute ##u=v_2## for the rate of ##dm## and divide by ##dt##: $$start{align} m~frac{dv}{dt}=(u-v)frac{dm}{dt}+F_{textual content{ext}}.finish{align}$$Right here, ##(u-v)## is the rate of the exchanged mass ##dm## relative to the system of curiosity. Equation (2) says that the instantaneous mass instances the acceleration of the system of curiosity is the same as the drive ##F_{textual content{ext}}## acts on the mass of the system plus the impulse per unit time that subsystem 2 delivers to subsystem 1.

We notice that the variable mass ##m_1## will need to have a relentless nonzero part ##M_0##, i.e. be within the kind $$m_1(t)=M_0+int_0^tf(t’)dt’$$in any other case the acceleration of subsystem 1 at ##t=0## just isn’t outlined. The transferred mass ##dm## can’t be the primary nor the final little bit of ##m_1##.  Such a requirement just isn’t crucial for ##m_2## until it too is a system of curiosity for, say, half (b) of a query.

Conservation of momentum

When there isn’t any exterior drive appearing on the system of curiosity, we count on momentum conservation. To discover what kind it takes, we set ##F_{textual content{ext}}=0## in Equation (2), separate variables, after which combine
$$start{align} & mdv=(u-v)dm implies frac{dv}{u-v}=frac{dm}{m} nonumber
& nonumber
& int_{v_0}^vfrac{dv}{u-v}=int_{m_0}^m frac{dm}{m} implies -lnleft(frac{u-v}{u-v_0} proper)=lnleft(frac{m}{m_0}proper) implies frac{u-v_0}{u-v}=frac{m}{m_0}. nonumber
finish{align}$$The final equation may be rearranged to $$m_0v_0=(m_0-m)u+mv.$$It’s the momentum conservation equation for a wonderfully inelastic collision when lots with speeds ##v## and ##u## stick collectively and transfer as one with velocity ##v_0##.  For a system ranging from relaxation, e.g. a rocket,
##v_0=0##; ##~m=## the mass of the rocket plus unspent gasoline; ##~m_0-m=## the mass of the spent gasoline.
This ends in $$mv+(m_0-m)u=0$$ which is the momentum conservation equation for an explosion.

Thus, the conservation of momentum holds all through the method of mass switch. Equation (2) offers the technique of acquiring the rate of the system of curiosity as a perform of time whereas the mass switch takes place. The system whose momentum is conserved has a mass that features all of the mass that has been or will probably be a part of the system of curiosity.

A particular case

A particular case of Equation (2) is when ##u=0##, i.e. the system of curiosity sweeps up mass at relaxation. Then $$start{align} & m~frac{dv}{dt}=-vfrac{dm}{dt}+F_{textual content{ext}} nonumber
& m~frac{dv}{dt}+vfrac{dm}{dt}=F_{textual content{ext}}implies frac{dP}{dt}=F_{textual content{ext}}. nonumber finish{align}$$ the place ##P## is the momentum of the system of curiosity.

A working recipe

We define a step-by-step process for utilizing Equation (2) to deal with variable mass issues.

1. Declare the subsystem of curiosity and the subsystem exchanging mass with it.
2. Write expressions for the variables showing in Equation (2) and substitute.
3. Use the tailored equation to reply the query posed.

Its software to particular examples follows under.

Examples

Hovering rocket

A rocket is hovering in place above floor close to the Earth’s floor. Assume that the combustion gases are expelled at a relentless charge ##beta=dm/dt## with velocity ##w## relative to the rocket.  What situation should maintain for the rocket to hover in place?

Step 1:
The subsystem of curiosity is the rocket of mass ##M_0## plus the unspent gasoline. The subsystem interacting with it’s the expelled gases.  The preliminary mass of all of the gasoline is ##m_0.##

Step 2:
##m=M_0+m_0-beta~t~;~~v=V~;~~u=V-w.## $$start{align}
&(M_0+m_0-beta~t)~frac{dV}{dt}=(-w)(-beta)- (M_0+m_0-beta~t)g=w~beta- (M_0+m_0-beta~t)g nonumber
nonumber
&(M_0+m_0-beta~t)~frac{dV}{dt}=w~beta- (M_0+m_0-beta~t)g. nonumber
finish{align}$$

Step 3:
When the rocket hovers and its acceleration ##dfrac{dV}{dt}## is zero, $$w~beta-(M_0+m_0-beta~t)~g=0.$$ It says that as time will increase and the load of the unspent gasoline decreases, the product ##w~beta## (a.ok.a. thrust) should lower to keep up zero acceleration.  When all of the gasoline runs out at time ##T=m_0/beta##, the thrust goes to zero as a result of ##dm/dt=0## and the rocket with closing mass ##M_0## is in free fall.

Who would have thought that

##dfrac{dP}{dt}=Mdfrac{dV}{dt}-wdfrac{dM}{dt}~## as an alternative of ##~dfrac{dP}{dt}=dfrac{d(MV)}{dt}= Mdfrac{dV}{dt}+Vdfrac{dM}{dt}~?##

Cart within the rain

An empty cart of mass ##M_0## begins shifting when fixed drive ##F## pulls on it on the identical time that it begins raining.  Raindrops fall vertically and water accumulates contained in the cart at a relentless charge ##beta =dm/dt.## The cart strikes on a horizontal floor.  Discover an expression for the rate of the cart as a perform of time.

Step 1:
The subsystem of curiosity is the cart of mass ##M_0## plus the rain it accommodates. The subsystem interacting with it’s the vertically falling rain that collects within the cart.

Step 2:
##m=M_0+beta~t~;~~v=V~;~~u=0.##
We acknowledge this because the particular case ##u=0## talked about earlier and instantly write $$frac{dP}{dt}=F.$$Step 3 $$start{align} & frac{dP}{dt}=F implies P=F~t nonumber
& V=frac{P}{m}=frac{F~t}{M_0+beta~t}.
nonumber finish{align}$$

Grappling hook

A grappling hook of mass ##M_0## is hooked up to a supple rope of linear mass density ##lambda## and size ##L.##  The hook is projected straight up with preliminary velocity ##v_0.##  Discover the utmost peak ##y_{textual content{max}}## to which the hook rises.  Assume that the rope has enough size, i.e. ##y_{textual content{max}}<L##, and that no lateral forces act on it.

Step 1:
The subsystem of curiosity is the grappling hook plus the rope hanging within the air. The subsystem that interacts with it’s the portion of the rope that’s at relaxation on the bottom.

Step 2:
##m=(M_0+lambda~y)~;~~v=V~;~~u=0.##
The exterior drive appearing on the whole system is the drive of gravity ##-(M_0 +lambda~L)g## that acts on all all of the mass plus the conventional drive ##N=lambda(L-y)## that acts solely on system 2. Then ##F_{textual content{ext}}=-(M_0+lambda~y)g=-mg.## As soon as once more we acknowledge the particular case ##u=0## and write $$frac{dP}{dt}=-mg.$$

Step 3:
First, we alter variables. $$start{align} & frac{dP}{dt}= frac{dP}{dm}frac{dm}{dy}frac{dy}{dt}=lambda~V frac{dP}{dm}=lambda~frac{P}{m} frac{dP}{dm}nonumber
implies & lambda~frac{P}{m} frac{dP}{d m}=-m~g.nonumber
finish{align}$$Then we separate variables. $$P~dP=-frac{g}{lambda}m^2~dm.$$ Lastly we combine noting that

• initially when ##y=0##, ##P=M_0v_0## and ##m=M_0##;
• lastly when ##y=y_{max}##, ##P=0## and ##m=M_0+lambda~y_{max}##

$$start{align} & int_{M_0v_0}^0P~dP=-frac{g}{lambda}int_{M_0}^{M_0+lambda~y_{max}}m^2~dm nonumber
nonumber
implies &-frac{1}{2}(M_0v_0)^2=-frac{g}{3lambda}left[ (M_0+lambda~y_{max})^3-M_0^3)right]. nonumber
finish{align}$$ After some easy algebra, we discover that $$y_{max}=frac{M_0}{lambda}left[left(frac{3lambda v_0^2}{M_0 g}right)^{1/3}+1right].$$

Afterthoughts

A normal formulation of Newton’s first legislation is “A physique stays at relaxation, or in movement at a relentless velocity in a straight line until acted upon by an unbalanced drive.” A fast internet survey confirmed that every one formulations start with “A physique ##~dots~##” which suggests fixed mass. The formulations of the second and third legal guidelines additionally assume our bodies of fixed mass and there’s no point out of variable mass methods. Equation (1) that entails including forces is relevant. We draw free-body diagrams, add all of the forces and discover the acceleration of any mounted mass or system of mounted lots. Newton’s second legislation is ##Mdfrac{dV}{dt}=F_{textual content{ext}}## and that’s a longtime reality.

Nonetheless, it’s no coincidence that derivations of the rocket equation don’t contain free-body diagrams. With variable mass methods, one should substitute Equation (1) with $$dP=F_{textual content{ext}}~dt.$$Then, as an alternative of contemplating forces, one considers impulses to be able to guarantee momentum conservation through the mass switch course of. Thus, within the case of variable mass methods, one should first receive an impulse equation relating ##dV## and ##dm## as a crucial intermediate step for acquiring an expression for ##Mdfrac{dV}{dt}.## Calling Equation (1) “the final type of Newton’s second legislation”, hides this step and leads one to consider that merely taking the time by-product of the momentum will produce an expression for ##Mdfrac{dV}{dt}.## This mistaken perception is all too apparent within the hovering rocket instance.

Identical to Equation (1), Equation (2) reduces to the usual type of Newton’s second legislation when the system’s mass is fixed. Nonetheless, along with that, and similar to its constant-mass counterpart, Equation (2)

• has ##Mdfrac{dV}{dt}## on the left-hand aspect
• may be utilized on to variable mass issues.

I feel that these extra attributes make Equation (2) a extra appropriate candidate than Equation (1) to bear the title “generalized type of Newton’s second legislation.”  Moreover, I feel that Equation (2), not Equation (1), must be the start line for fixing variable mass issues.