대학원 공부노트
Momentum 본문
운동량(Momentum)
$$p=m\vec{v}$$
운동량의 시간에 대란 변화량
$$\frac{\partial p}{\partial t}=\frac{\partial m}{\partial t}\vec{v}+m\frac{\partial \vec{v}}{\partial t}$$
Let's assume \(\vec{v}=const\)
$$\frac{\partial p}{\partial t}=\frac{dp}{dt}=\frac{dm}{dt}\vec{v}=\dot{m}\vec{v}\cdots(1)$$
Bring the concept 'mass flowrate' \(\dot{m}\)
$$m=\rho V$$
$$\dot{m}=\frac{dm}{dt}=\frac{d\rho}{dt}V+\rho\frac{dV}{dt}$$
Let's assume \(\rho=const\)
$$\dot{m}=\rho\frac{dV}{dt}$$
And volume\(V\) can be express with 'area times height'.
$$V=Ah$$
So, when we develope the equation under the assumption: '\(A=const\)', we can derive
$$\frac{dV}{dt}=\frac{dA}{dt}h+A\frac{dh}{dt}=A\frac{dh}{dt}=A\vec{v}$$
Then, bring mass flowrate into the equation \((1)\)
$$\dot{m}\vec{v}=\rho\dot{V}\vec{v}$$
$$=\rho A \vec{v} \vec{v}$$
$$\Rightarrow \frac{dP}{dt}=\rho A \vec{v}\vec{v}$$
Now, let's check the unit
$$\frac{\rm{kg}}{\rm{m}^{3}}\times \rm{m}^{2} \times \frac{\rm{m}}{\rm{s}} \times \frac{\rm{m}}{\rm{s}}=\rm{kg}\times\frac{\rm{m}}{\rm{s}^{2}}$$
And, unit of 'Force' is...
$$F=ma$$
$$\rm{kg}\times\frac{\rm{m}}{\rm{s}^{2}}$$
That's the reason why \(\rho A \vec{v} \vec{v}\) is mentioned with the 'force'.
And when we talk about the 'flux' we use \(\rho\vec{v}\vec{v}\).
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