Momentum

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 \(\rho\vec{v}\vec{v}\).