Falling Rain Drop (Variable Mass)

  • #1

Homework Statement


Suppose a rain drop with mass ##m_0\neq 0## is falling due to gravity with initial velocity ##v_0##, assume ##\frac{dm}{dt}=k=##constant. Solve the differential equation and determine the velocity as ##t\to\infty##

Homework Equations


##F=\frac{dp}{dt}=\dot{m}v+m\dot{v}##

The Attempt at a Solution



Since ##F=mg## the D.E. is ##\dot{m}v+m\dot{v}=mg##, substituting in ##\frac{dm}{dt}=k## we find ##kv+m\dot{v}=mg##, since the mass increases with time we have ##m=m_0+kt##. One method I've attempted is essentially using the integrating factor but since the LHS is already a product rule derivative I don't need to multiply it by anything so:

$$\frac{d}{dt}(mv)=mg \\
\int \frac{d}{dt}(mv)dt=\int mg dt=\int mg \frac{dt}{dm}dm=\frac{g}{k}\int_{m_0}^{m} m\, dm\\
v(m_0+kt)+C=\frac{g}{2k}((m_0+kt)^2-m_0^2)=\frac{g}{2k}(2m_0kt+k^2t^2)
$$

at ##t=0##, ##v=v_0\Rightarrow##, ##v_0(m_0+0)+C=0\Longrightarrow C=-v_0m_0##

therefore...

$$v(m_0+kt)-v_0m_0=gm_0t+\frac{gkt^2}{2}\\
v(m_0+kt)=m_0(v_0+gt)+\frac{gkt^2}{2} \\
v=\frac{m_0(v_0+gt)}{m_0+kt}+\frac{gkt^2}{2(m_0+kt)}
$$

Now I find this solution a bit odd since my professor said that it would become asymptotic but I think he's wrong about this? The first term will approach ##\frac{m_0g}{k}## but the second term clearly diverges as ##t\to\infty## and it doesn't make sense that an object would reach a terminal velocity without drag. I suppose you could say it asymptotically goes to infinity but it's increasing linearly...
 

Answers and Replies

  • #3
haruspex
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my professor said that it would become asymptotic
By itself that means nothing. What is supposed to tend asymptotically to a constant, or to a certain function of what?
The link DrSteve posted shows the acceleration tends to a constant.
 
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  • #4
By itself that means nothing. What is supposed to tend asymptotically to a constant, or to a certain function of what?
The link DrSteve posted shows the acceleration tends to a constant.
Okay I suppose I'm just too used to thinking of functions converging to a constant when that term is mentioned.
 
  • #5
ehild
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The velocity tends to a simple function as t tends to infinity. What is that simple function?
 
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  • #6
The velocity tends to a simple function as t tends to infinity. What is that simple function?
I believe it's just ##v=\frac{m(t)g}{k}## from setting ##\dot{v}=0## in the D.E.

Simplifying my expression for ##v(t)## would give ##v(t)=\frac{m_0g}{k}+\frac{gt}{2}## though

Sorry for so many edits but just noticed that the first one becomes ##v=\frac{m_0g}{k}+gt## which is almost the same as the result from simplification.
 
  • #7
ehild
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I believe it's just ##v=\frac{m(t)g}{k}## from setting ##\dot{v}=0## in the D.E.

Simplifying my expression for ##v(t)## would give ##v(t)=\frac{m_0g}{k}+\frac{gt}{2}## though

Sorry for so many edits but just noticed that the first one becomes ##v=\frac{m_0g}{k}+gt## which is almost the same as the result from simplification.
You can not set ##\dot v = 0##. The acceleration does not tend to zero. At great t, v increases linearly with time with rate g/2 instead of g as it were for a constant-mass body.
 
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  • #8
You can not set ##\dot v = 0##. The acceleration does not tend to zero. At great t, v increases linearly with time with rate g/2 instead of g as it were for a constant-mass body.
Whoops, I realize now it doesn't make much sense to set acceleration equal to 0 :)

Interestingly at large t, it appears v always increases by ##g/2## regardless of ##\alpha## for ##\frac{dm}{dt}=k^{\alpha}##
 
  • #9
ehild
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Whoops, I realize now it doesn't make much sense to set acceleration equal to 0 :)

Interestingly at large t, it appears v always increases by ##g/2## regardless of ##\alpha## for ##\frac{dm}{dt}=k^{\alpha}##
##k^{\alpha}## is just an other constant. Why do you mix alpha in? It was said that dm/dt=k.
 
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  • #10
ehild
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It is interesting the acceleration at long time tends to g/2, regardless of k.
 
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  • #11
I was looking through my posts when I stumbled upon this one and I can't understand how they're solving the differential equation in the paper that was linked in response to this post.

The author states that when ##\frac{dm}{dt}## is independent of velocity then the accretion equation can be solved for ##m(t)## and then the Newtonian equation can be solved for ##v(t)##. I'm having trouble seeing this since for my problem ##\frac{dm}{dt} = k## which leads to ##m(t) = m_0 + kt##. I don't understand how this helps solve ##mg = \frac{dm}{dt}v + m\frac{dv}{dt}## for velocity.

Dividing through by ##m## and using ##\frac{dm}{dt}=k## I can get the result $$g = \frac{kv}{m} + \frac{dv}{dt}$$

At which point I can't see how to make progress, evidently I should apply ##m=m_0+kt## but I don't see how that helps when I can't seem to get ##\frac{dv}{v}## without a ##\frac{g}{v}## term
 
  • #12
haruspex
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I was looking through my posts when I stumbled upon this one and I can't understand how they're solving the differential equation in the paper that was linked in response to this post.

The author states that when ##\frac{dm}{dt}## is independent of velocity then the accretion equation can be solved for ##m(t)## and then the Newtonian equation can be solved for ##v(t)##. I'm having trouble seeing this since for my problem ##\frac{dm}{dt} = k## which leads to ##m(t) = m_0 + kt##. I don't understand how this helps solve ##mg = \frac{dm}{dt}v + m\frac{dv}{dt}## for velocity.

Dividing through by ##m## and using ##\frac{dm}{dt}=k## I can get the result $$g = \frac{kv}{m} + \frac{dv}{dt}$$

At which point I can't see how to make progress, evidently I should apply ##m=m_0+kt## but I don't see how that helps when I can't seem to get ##\frac{dv}{v}## without a ##\frac{g}{v}## term
I think expanding ##\frac d{dt}(mv)## was unhelpful.
Go back to ##mg= \frac d{dt}(mv)## and substitute for m using m0 etc.
Both sides are directly integrable wrt t.
 
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  • #13
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therefore...

$$v(m_0+kt)-v_0m_0=gm_0t+\frac{gkt^2}{2}\\
v(m_0+kt)=m_0(v_0+gt)+\frac{gkt^2}{2} \\
v=\frac{m_0(v_0+gt)}{m_0+kt}+\frac{gkt^2}{2(m_0+kt)}
$$

Apologies, but how has the [itex] v [/itex] appeared in the RHS?

The author states that when ##\frac{dm}{dt}## is independent of velocity then the accretion equation can be solved for ##m(t)## and then the Newtonian equation can be solved for ##v(t)##. I'm having trouble seeing this since for my problem ##\frac{dm}{dt} = k## which leads to ##m(t) = m_0 + kt##. I don't understand how this helps solve ##mg = \frac{dm}{dt}v + m\frac{dv}{dt}## for velocity.

Dividing through by ##m## and using ##\frac{dm}{dt}=k## I can get the result $$g = \frac{kv}{m} + \frac{dv}{dt}$$

At which point I can't see how to make progress, evidently I should apply ##m=m_0+kt## but I don't see how that helps when I can't seem to get ##\frac{dv}{v}## without a ##\frac{g}{v}## term

You might want to think about integrating factors. The differential equation is of the form:
[tex] \frac{dv}{dt} + f(t) v = h(t) [/tex]
where [itex] h(t) [/itex] is a constant in this case. If you use an integrating factor, you get to an expression that is almost similar to the form you derived in the original post.
 
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  • #14
I think expanding ##\frac d{dt}(mv)## was unhelpful.
Go back to ##mg= \frac d{dt}(mv)## and substitute for m using m0 etc.
Both sides are directly integrable wrt t.
$$mg = \frac{d}{dt}(mv)$$
$$d(mv) = mg\cdot dt$$
$$\int_{t=0}^{t=t} d(mv) = g\int_0^t(m_0+kt)dt$$
$$mv-m_0v_0 = g(m_0t+kt^2/2)$$
$$v = \frac{m_0v_0+g(m_0t+kt^2/2)}{m_0+kt}$$

Their expression for velocity is so simplified in the paper I can't even tell if this is correct.

Apologies, but how has the [itex] v [/itex] appeared in the RHS?
I have basically no idea how I derived anything in the original post. I don't even understand how I got this
$$

\frac{d}{dt}(mv)=mg \\

\int \frac{d}{dt}(mv)dt=\int mg dt
$$

Does this part make sense to you?

You might want to think about integrating factors. The differential equation is of the form:
[tex] \frac{dv}{dt} + f(t) v = h(t) [/tex]
where [itex] h(t) [/itex] is a constant in this case. If you use an integrating factor, you get to an expression that is almost similar to the form you derived in the original post.

I thought I couldn't use integrating factor here because the ##m## is actually ##m(t)## and it's in the denominator for the ##f(t)## function. I might be wrong I haven't done integrating factor in years.
 
  • #15
haruspex
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$$mg = \frac{d}{dt}(mv)$$
$$d(mv) = mg\cdot dt$$
$$\int_{t=0}^{t=t} d(mv) = g\int_0^t(m_0+kt)dt$$
$$mv-m_0v_0 = g(m_0t+kt^2/2)$$
$$v = \frac{m_0v_0+g(m_0t+kt^2/2)}{m_0+kt}$$

Their expression for velocity is so simplified in the paper I can't even tell if this is correct.
That all looks right, and you can quickly see the asymptotic behaviour is v=gt/2.
Does this answer your question in post #11?
 
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  • #16
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I thought I couldn't use integrating factor here because the ##m## is actually ##m(t)## and it's in the denominator for the ##f(t)## function. I might be wrong I haven't done integrating factor in years.

Just thought I would respond to this part for now. I believe you can use an integrating factor here.

Background on integrating factor (might be useful if you haven't seen it for a while; feel free to skip this section):
Without just jumping to the formula, the integrating factor can be thought of as the answer to the following question: "what term do I need to multiply the equation by such that I can write it as a derivative of a product?" I.e. if we have:
[tex] \frac{dv}{dt} + f(t){v} = h(t) [/tex] it would be much more convenient to write the LHS in a form that is more easily integrated (wrt. [itex] t [/itex]). If we have integrating factor [itex] P(t) [/itex] and we multiply it through to get:
[tex] P(t) \frac{dv}{dt} + P(t)f(t){v} = h(t)P(t) [/tex]
we now want the LHS to be expressed as [itex] \frac{d}{dt}(P(t)v(t)) [/itex]. By using the product rule and comparing coefficients, we can see that [itex] \frac{dP(t)}{dt} = P(t)f(t) [/itex]. This can be solved to give the form of the integrating factor [itex] P(t) = e^{ \int f(t) dt } [/itex]

Therefore, if we integrate the equation with respect to [itex] t [/itex]:
[tex] P(t)v(t) = \int P(t)h(t) dt[/tex]

Specific response to your question:
So we have [itex] f(t) = \frac{k}{m_0 + kt} [/itex]. Can that be integrated? (Hint: think about natural logarithms). Don't forget that the integrating factor is [itex] P(t) = e^{ \int f(t) dt } [/itex]

One final note:
I hasten to add this as you have already been helped, but I just thought to let you know of another way of getting to the differential equation (it is my preferred approach when doing these types of questions). It is very similar to what you have written, but you can perhaps use it to double-check whether your ODE is correct in future problems. The whole principle here (as you know) is that Impulse = change in momentum. So we can draw a diagram of the mass at time [itex] t [/itex]: it has mass [itex] m [/itex] and velocity [itex] v [/itex] (downwards). It has a weight [itex] mg [/itex] acting on it downwards. We can then draw an 'after' picture at time [itex] \delta t [/itex] later (so at time [itex] t + \delta t [/itex]). It now mass mass [itex] m + \delta m [/itex] (as it is accumulating mass) and has velocity [itex] v + \delta v [/itex] (downwards). During the time [itex] \delta t [/itex], an impulse equal to [itex] mg \delta t [/itex] acted on the raindrop. If we resolve vertically downwards and write out impulse = change in momentum:
[tex] mg \delta t = (m + \delta m)(v + \delta v) - mv [/tex]
We can then simplify, divide by [itex] \delta t [/itex] and let [itex] \delta t [/itex] tend to 0 to form the same ODE that you derived.

Please note that the 'before' and 'after' diagrams may look different for different types of problems (e.g. rocket burning fuel). While diagrams may differ, the underlying concepts are the same.

Hope that was useful. If not, let me know which parts could be more clear. Apologies if you already knew all of that, but perhaps it might be helpful to another reader.
 
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  • #17
That all looks right, and you can quickly see the asymptotic behaviour is v=gt/2.
Does this answer your question in post #11?
Yes it does thanks.

Just thought I would respond to this part for now. I believe you can use an integrating factor here.
Background on integrating factor (might be useful if you haven't seen it for a while; feel free to skip this section):
Without just jumping to the formula, the integrating factor can be thought of as the answer to the following question: "what term do I need to multiply the equation by such that I can write it as a derivative of a product?" I.e. if we have:
[tex] \frac{dv}{dt} + f(t){v} = h(t) [/tex] it would be much more convenient to write the LHS in a form that is more easily integrated (wrt. [itex] t [/itex]). If we have integrating factor [itex] P(t) [/itex] and we multiply it through to get:
[tex] P(t) \frac{dv}{dt} + P(t)f(t){v} = h(t)P(t) [/tex]
we now want the LHS to be expressed as [itex] \frac{d}{dt}(P(t)v(t)) [/itex]. By using the product rule and comparing coefficients, we can see that [itex] \frac{dP(t)}{dt} = P(t)f(t) [/itex]. This can be solved to give the form of the integrating factor [itex] P(t) = e^{ \int f(t) dt } [/itex]

Therefore, if we integrate the equation with respect to [itex] t [/itex]:
[tex] P(t)v(t) = \int P(t)h(t) dt[/tex]

Specific response to your question:
So we have [itex] f(t) = \frac{k}{m_0 + kt} [/itex]. Can that be integrated? (Hint: think about natural logarithms). Don't forget that the integrating factor is [itex] P(t) = e^{ \int f(t) dt } [/itex]

Interesting for some reason I had it in my mind that they had to be polynomials.
$$\frac{dv}{dt}+\frac{k}{m}v=g $$ $$\ln P(t)=\int \frac{k}{m_0+kt}dt=\ln{\left (m_0+kt\right )}+C $$ $$P(t)=e^C(m_0+kt)$$
$$\frac{d}{dt} ( v(m_0+kt)e^C ) = (m_0+kt)e^Cg$$
$$v(m_0+kt)-v_0m_0 = g(m_0t+kt^2/2)$$
$$v = \frac{v_0m_0+g(m_0t+kt^2/2)}{m_0+kt}$$
resulting in the same answer as from the previous method, nice.
One final note:
I hasten to add this as you have already been helped, but I just thought to let you know of another way of getting to the differential equation (it is my preferred approach when doing these types of questions). It is very similar to what you have written, but you can perhaps use it to double-check whether your ODE is correct in future problems. The whole principle here (as you know) is that Impulse = change in momentum. So we can draw a diagram of the mass at time [itex] t [/itex]: it has mass [itex] m [/itex] and velocity [itex] v [/itex] (downwards). It has a weight [itex] mg [/itex] acting on it downwards. We can then draw an 'after' picture at time [itex] \delta t [/itex] later (so at time [itex] t + \delta t [/itex]). It now mass mass [itex] m + \delta m [/itex] (as it is accumulating mass) and has velocity [itex] v + \delta v [/itex] (downwards). During the time [itex] \delta t [/itex], an impulse equal to [itex] mg \delta t [/itex] acted on the raindrop. If we resolve vertically downwards and write out impulse = change in momentum:
[tex] mg \delta t = (m + \delta m)(v + \delta v) - mv [/tex]
We can then simplify, divide by [itex] \delta t [/itex] and let [itex] \delta t [/itex] tend to 0 to form the same ODE that you derived.

Please note that the 'before' and 'after' diagrams may look different for different types of problems (e.g. rocket burning fuel). While diagrams may differ, the underlying concepts are the same.

Hope that was useful. If not, let me know which parts could be more clear. Apologies if you already knew all of that, but perhaps it might be helpful to another reader.

I have seen this before but mostly forgotten about it, I will try to use this method in the future when possible.

I also just realized the equations in the original post used to be separated in latex using "\ \" which no longer creates new lines, the ##v## in post #13 is actually at the start of a newline.
 

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