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I am reading text on motion when mass varies. The equation of motion comes out to be:

$$\frac{d}{dt}(mv) = F + V\frac{dm}{dt}$$

where $V$ is the velocity with which the incremented mass was moving.

Now I have two exercise questions:

  1. Trailer full of sand is pulled by constant force F and sand is leaking at constant rate.

  2. Snow slides off a (inclined) roof clearing away a part of uniform breadth. All snow slides at once.

What will be the value of $V$ in both cases.

In texts it gives:

  1. V = v (velocity of trailer)
  2. V = 0

I understand the first part that sand is initially moving with velocity of trailer. But I dont understand second part. The snow that is sliding off must have the velocity of snow at that instant(??). Please explain.

Edit: Here is the snippet:

enter image description here

And I will reframe the second question if I said it wrong. Quoting from the tests, "Snow slides off a roof clearing away a part of uniform breadth. If it all slide at once, find the time in which the roof will be cleared."

Here is snippet for that too: question solution

Golden_Hawk
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MWD
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  • Hint: Suppose instead of "snow" you had just one snowflake on the roof. What force equation would you write? What changes if you now have many snowflakes? – Amit Feb 23 '23 at 16:31
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    The correct version of Newton's second law is always $F = \frac{d}{dt} ( m v )$ which simplifies to $F=ma$ when mass is constant. The equation in the question is incorrect. – Prahar Feb 23 '23 at 16:46
  • @Prahar - It seems to me they just used the Leibniz rule on what you wrote: $\frac{d}{dt}(mv)=\frac{dm}{dt}v+m\frac{dv}{dt}=\frac{dm}{dt}v+ma$ So I don't think it's a mistake if they define $F$ differently (not as the total force, but only force due to acceleration). I agree it is not the same $F$ from Newton's second law. – Amit Feb 23 '23 at 17:11
  • @Amit - OP should then clarify that they are using non-standard notation. Further, in the rest of the post, they use the term force and that force is not the $F$ that they have earlier in the equation. This can cause significant confusion. – Prahar Feb 23 '23 at 18:00
  • @Prahar ..I agree. I hope that the text clarifies it, otherwise indeed it is a very confusing text to learn from. – Amit Feb 23 '23 at 18:03
  • @MWD - the notation in this book seems atrocious. What book is this? – Amit Feb 23 '23 at 19:21
  • This looks like a bad quality book. 4) and 5) contradict 1). 4) and 5) are correct. – Ján Lalinský Feb 23 '23 at 19:39
  • @Prahar that is not true. $\frac{d(mv)}{dt}$ is Galilei-variant, but force is not. 1D motion of variable mass systems obeys 5). – Ján Lalinský Feb 23 '23 at 19:40
  • @JánLalinský he wrote his comment before the OP shared the images.... – Amit Feb 23 '23 at 19:46
  • @JánLalinský can you please recommend better texts that are little easier to understand too. – MWD Feb 23 '23 at 19:49
  • @JánLalinský This page on wikipedia(https://en.m.wikipedia.org/wiki/Variable-mass_system) gives same equation exception being V is relative velocity. Is that they(my texts) are missing? – MWD Feb 23 '23 at 19:54
  • @Amit It's a book from a local publication(prakashan kendra) that my college teachers often recommend. – MWD Feb 23 '23 at 19:57
  • $V$ is magnitude of velocity (thus positive number) of the expelled mass in the lab frame where the body has velocity $v$; if $v$ is positive, the difference $V-v$ is velocity of the expelled mass in the frame of the body. It's better to write this equation as $ma = F_{ext} + \frac{dm}{dt}c$ where $c$ is velocity of the expelled mass in body's frame. – Ján Lalinský Feb 23 '23 at 20:04
  • @MWD -- see if you can find something here: https://physics.stackexchange.com/questions/12175/resource-recommendations – Amit Feb 23 '23 at 20:06
  • @JánLalinský If I take c as relative velocity, then in first case of sand trail the value of c will be 0. And in second case it will be v. Please Correct me if I am wrong. – MWD Feb 23 '23 at 20:16
  • @MWD I don't have full text of those problems. $c$ is relative velocity always, depending on the problem, it can be zero, positive or negative. – Ján Lalinský Feb 23 '23 at 22:31
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    @MWD In the example of snow sliding down and acquiring more and more mass, $V=0$ (since the newly added snow is at rest), and relative velocity of this newly added snow before it gets added is $-v$. – Ján Lalinský Feb 23 '23 at 22:49
  • @MWD for mechanics in general, e.g. Berkeley Physics Course, Mechanics by Kittel et al. For variable mass systems I don't know a good reference, perhaps search for good explanations of the rocket equation on the internet. – Ján Lalinský Feb 23 '23 at 22:52

1 Answers1

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The below answer was written before significant changes were made to the OP and may no longer be appropriate to its contents.

What a mess. If this is independent study, you might want to consider obtaining a new textbook by a different author.

Temporarily forget all physics. Our variables are just variables. Go back to the product rule for derivatives:

$$(mv)' = mv'+ vm'$$

for any variables $m, v$.

In physics the function $(mv)'$ is called Force and represented with the variable $F$. We obtain the elementary physics form $F = mv'$ only if $m' = 0$. The equation in your textbook appears to define $F := mv'$ in the context of changing mass, which is wrong for physics (although strictly speaking you can define any variable however you like). Then it does the math wrong by introducing a different variable $V$.

$(mv)' = mv' + Vm'$ only if $V=v$, which makes introducing the new variable pointless and any answers other than $V=v$ false.

Your first scenario (sand leaving a truck accelerating under constant net force) should read:

$F = (mv') = mv' + vm'$

We then examine the scenario to determine what our terms actually represent, noting that the changing mass that is relevant to this equation must be the mass that the force $F$ is pushing on and is traveling at $v$. We find that it is the mass of the sand in the truck (which is decreasing at a constant rate) which satisfies these conditions.

Your second scenario is unphysical. Snow doesn't leave roofs all at once. If it did (perhaps a wizard zaps it out of existence) we'd have $m$ go from $m=m_0$ to $m=0$ in $\Delta t = 0$ hence $vm' \to -\infty$ with $F$ and $mv'$ undefined, which is unphysical. If we replace the scenario with a fixed quantity of snow sliding down a long section of otherwise bare roof in a solid block, then we just have a block sliding down an inclined plane.

g s
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