- #1

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I see ##F = m dv/dt## so frequently that I found myself asking this question.

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- Thread starter kent davidge
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- #1

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I see ##F = m dv/dt## so frequently that I found myself asking this question.

- #2

Chestermiller

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Can you give an example of how the mass is changing with time?

I see ##F = m dv/dt## so frequently that I found myself asking this question.

- #3

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Perhaps a rocket expelling fuel?Can you give an example of how the mass is changing with time?

- #4

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$$

\vec F = \frac{d\vec p}{dt}.

$$

For a variable mass system where ##m## is a function of ##t##, you need to be careful though. Just applying the definition

$$

\vec F = \frac{d(m\vec v)}{dt} = \dot m \vec v + m \vec a

$$

without considering what goes into ##\vec F## can lead you astray and the equation is not Galilei invariant (##\vec v## depends on the inertial frame). In order to get it right, you must consider both external forces that act upon the system as well as the loss of momentum due to the mass leaving. If the mass leaving does so with velocity ##\vec u##, the force equation becomes

$$

\vec F_{\rm ext} + \dot m \vec u = \dot m \vec v + m \vec a,

$$

where ##\vec F_{\rm ext}## is the external force acting on the system, which can be rewritten as

$$

m\vec a = \vec F_{\rm ext} + \dot m (\vec u - \vec v) = \vec F_{\rm ext} + \dot m \Delta \vec v,

$$

where ##\Delta\vec v## is the difference between the velocity of the exhaust and that of your object, i.e., the exhaust velocity relative to the object, which is Galilei invariant as it is the difference of two velocities.

- #5

Nugatory

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That's the apppropriate simplification to ##F=\frac{dp}{dt}## when the mass is a constant. You see it so often because it because so many interesting and important problems involve a constant mass. The reader is expected to recognize it as the constant-mass special case without further reminders.I see ##F = m dv/dt## so frequently that I found myself asking this question.

- #6

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- #7

Chestermiller

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$$m(t)\mathbf{v}(t)-\int_0^t{\dot{m}(t')\mathbf{u}(t')dt'}=\mathbf{C}$$

where m(t) is the mass of rocket+fuel at time t, ##\mathbf{v}## is the velocity of the rocket+fuel, ##\dot{m}=dm/dt##, and ##\mathbf{u}##is the velocity of the discharge gas. If I differentiate this equation with respect to t, I obtain:

$$m\mathbf{a}+\dot{m}\mathbf{v}-\dot{m}\mathbf{u}=0$$

or $$m\mathbf{a}=\dot{m}(\mathbf{u}-\mathbf{v})$$

Is there another alternate derivation that is easy to make sense of.

- #8

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That is exactly the correct way to get the rocket equation, with the exception that there is nothing temporary about the external force being zero.

- #9

Chestermiller

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I was thinking that the external force might be gravity, for example.As ChesterMiller said,The only way I was able to derive the equation for a rocket was to first (temporarily) assume that the external force is zero, and then to apply conservation of momentum.

That is exactly the correct way to get the rocket equation, with the exception that there is nothing temporary about the external force being zero.

- #10

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Newton's Second Law says that "For a particle, F = m*a."

No, it says F=dp/dt. There is no limitation to particles in the second law.

http://cudl.lib.cam.ac.uk/view/PR-ADV-B-00039-00001/46

The only way I was able to derive the equation for a rocket was to first (temporarily) assume that the external force is zero, and then to apply conservation of momentum.

And you can do that using forces because they imply conservation of momentum (due to the 3rd law).

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