- #1
Undoubtedly0
- 98
- 0
Newton's second law gives that
[tex]\sum\mathbf{F} = \frac{d\mathbf{p}}{dt} = \frac{d(m\mathbf{v})}{dt} [/tex]
In a system where mass varies with time, m(t), one would simply think that this would lead to
[tex]\sum\mathbf{F} = m\frac{d\mathbf{v}}{dt} + \mathbf{v}\frac{dm}{dt}[/tex]
Yet everywhere online I see that this is not the case, (especially here, page 228) says that this is not the case. In fact, it is said that
[tex]\sum\mathbf F + \mathbf{u} \frac{\mathrm{d} m}{\mathrm{d}t} = m {\mathrm{d} \mathbf v \over \mathrm{d}t}[/tex]
where [itex]\mathbf{u}[/itex] is the velocity of dm, the mass entering or leaving the main body. What then is wrong with [itex]\sum\mathbf{F} = m\frac{d\mathbf{v}}{dt} + \mathbf{v}\frac{dm}{dt}[/itex]?
Happy holidays!
[tex]\sum\mathbf{F} = \frac{d\mathbf{p}}{dt} = \frac{d(m\mathbf{v})}{dt} [/tex]
In a system where mass varies with time, m(t), one would simply think that this would lead to
[tex]\sum\mathbf{F} = m\frac{d\mathbf{v}}{dt} + \mathbf{v}\frac{dm}{dt}[/tex]
Yet everywhere online I see that this is not the case, (especially here, page 228) says that this is not the case. In fact, it is said that
[tex]\sum\mathbf F + \mathbf{u} \frac{\mathrm{d} m}{\mathrm{d}t} = m {\mathrm{d} \mathbf v \over \mathrm{d}t}[/tex]
where [itex]\mathbf{u}[/itex] is the velocity of dm, the mass entering or leaving the main body. What then is wrong with [itex]\sum\mathbf{F} = m\frac{d\mathbf{v}}{dt} + \mathbf{v}\frac{dm}{dt}[/itex]?
Happy holidays!