# Solve Momentum Behavior: Get 2.72 & 2.73 Equations

• CFXMSC
In summary: V}{\partial t}=0In summary, the conversation discusses equations 2-72 and 2-73, which involve taking the divergence and curl of velocity and fluid vorticity. The conversation also mentions using vector calculus identities to solve the equations and considering steady flow problems.
CFXMSC
Can somobody help me how to get the 2.72 and 2.73 equation??

http://imageshack.com/a/img89/5594/kiw6.jpg

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2-72: have you tried: starting from 2-70 and take the divergence of both sides - note constant density and div(V)=0

2-73: this equation is the definition of "fluid vorticity" so it is not a derived thingy.
The equation just before it comes from 2-70 by taking the curl.

i tried but it not looks simple to do

Most of the terms turn out to be zero - where do you get stuck?

I really confused... There are a lot of terms in dV/dT and gravity that become 0, but in vicosity how could i do?

1 person
$\nabla p=\rho g + \mu \nabla^2 V-\rho\frac{\partial V}{\partial t}$

$\nabla^2 p=\nabla .(\rho g) + \mu\nabla . (\nabla^2 V)-\rho\nabla . \left(\frac{\partial V}{\partial t}\right)$

$\nabla^2 p=\nabla .(\rho g) + \mu\nabla^2(\nabla V)-\rho\nabla . \left(\frac{\partial V}{\partial t}\right)$

Using $\nabla . V=0$ and $\nabla .(\rho g)=0$

$\nabla^2 p=-\rho\nabla . \left(\frac{\partial V}{\partial t}\right)$

And i got stuck

Last edited:
Can i assume if $div(V)=0$ then it's a steady flow problem?

The other i got $\frac{\partial \omega}{\partial t}=\frac{\mu}{\rho}\nabla \times (\nabla^2 V)$

What now?

You are told that div.V=0.
You can take the div inside the time-partial

## 1. What is momentum behavior and why is it important?

Momentum behavior refers to the tendency of an object to maintain its current state of motion. This can be in the form of either linear momentum (when an object is moving in a straight line) or angular momentum (when an object is rotating). It is important because it helps us understand and predict the movement of objects, and is a fundamental concept in physics.

## 2. What are the equations for calculating momentum?

The equations for calculating momentum are:

• Linear Momentum: p = mv
• Angular Momentum: L = Iω

Where p is momentum, m is mass, v is velocity, L is angular momentum, I is the moment of inertia, and ω is angular velocity.

## 3. How do I solve for momentum behavior using the 2.72 & 2.73 equations?

To solve for momentum behavior using these equations, you will need to know the values of mass, velocity, and moment of inertia (for angular momentum). Plug these values into the appropriate equation and solve for the momentum. It is important to use consistent units for all values.

## 4. Can these equations be used for any object?

Yes, these equations can be used for any object as long as the values of mass, velocity, and moment of inertia are known. They are universal equations that apply to all objects, regardless of their size, shape, or composition.

## 5. How is momentum behavior related to Newton's Laws of Motion?

Momentum behavior is closely related to Newton's Laws of Motion, specifically the law of inertia. According to this law, an object will maintain its state of motion unless acted upon by an external force. This means that an object with a certain momentum will continue to move at the same velocity unless a force is applied to change its momentum. This is also known as the conservation of momentum.

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