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2001, W. E. Haisler Chapter 3: Conservation of Linear Momentum 1

Conservation of Linear MomentumChapter 3

The linear momentum of a mass (m) with a velocity is defined to be . Important: linear

momentum is a vector.


or, for a time period t:

For a differential volume element with density , the linear momentum is given by .


Consider mass flow in the x direction with density and velocity flowing into the volume xyz through the surface yz during a time increment of t. Total mass entering and leaving through the “x face” in time t is:


For the “left x-face”, the total mass input is:

and its momentum is: .

For the “right x-face”, the total mass output is:

and its momentum is:

momentum flux in x direction= momentum of mass flow in x direction (per unit area

and per unit time)

2001, W. E. Haisler Chapter 3: Conservation of Linear Momentum 5The momentum of the mass flow will tend to try to move the control volume (body) and therefore external surface tractions (forces/unit area) may be required to keep the body in equilibrium. For example, consider mass flowing through a pipe:

Think of the fluid within the pipe as the "system." For such a system, the pipe boundary must exert forces on the fluid boundary. These normal boundary forces on the fluid are what make the fluid turn as it flows through the pipe. The freebody of the fluid in the pipe is shown by:


Note that the boundary forces are distributed along the pipe in some fashion that has yet to be determined. Because of viscosity effects, we will find that the pipe will also exert shear (or drag type) forces on the fluid which tend to slow the fluid near the pipe boundary.

Traction forces will exist in both solid and fluid systems. Consider a frame problem from ENGR 211:


At some point x in the beam, there is a distribution of normal (axial and shear forces) that can be represented by a "traction" vector.

2001, W. E. Haisler Chapter 3: Conservation of Linear Momentum 8Now consider a general shaped body (system) with forces applied to its external surface as shown below:

With a cutting plane "A" which has an outward unit normal on freebody 2, make two freebody diagrams. At the cut section we must place internal reactions (forces) as shown. In general, these forces will be forces distributed over the entire area of the cut section. For freebody 2, consider the force applied to an area A that has unit normal . Now define the traction on this surface as


The notation means the traction vector (vector force per unit area) acting on the surface with unit normal . This vector is not necessarily normal to the surface. is shown below:

For freebody 1, we note that the force at a given point must be equal and opposite to the force on freebody 2 and hence the tractions must

2001, W. E. Haisler Chapter 3: Conservation of Linear Momentum 10also be equal and opposite. The normal on FB 1also points in the opposite direction to FB 2. The traction on the surface with unit normal is denoted as and we have that

Now lets consider a differential volume system in the fluid (or solid). Just as for the pipe example, the differential system (xyz) can also have tractions on its boundary. Define the tractions on the x-face as shown below:


The traction vector on the left x face is denoted by: and on the

right face by . Note: Traction is defined as positive when it

acts in the positive coordinate direction (just like fluid velocity ).


The traction vector on the +i face may be written in terms of its components as: . The subscripts on the vector components mean the following:

1st subscript: face the force acts on (i means x)2nd subscript: direction of the force.

For example, is the traction on the i face acting in the y direction.

For the tractions acting on the -i and +i faces, the sum of the external forces (on x-faces only) is given by

.


We should also consider body forces that act on the volume. Assume that the body force per unit mass (usually due to gravity) is given . The total body force (vector) on the volume during a time increment t is


Now, add the momentum terms in the y and z directions:In the above, we have looked at momentum associated with mass flow in the x direction only. We can write similar momentum terms for mass flow in the y and z directions. Similarly, we can define the forces on the volume due to tractions on the y and z faces.

Now, the change in momentum with respect to time:

The conservation of linear momentum (including linear momentum terms, tractions and boundary forces for each of the 3 coordinate directions) becomes


=

+

+

+ +

+

+

Divide by xyzt and take the limit of each term; x, y, z, t 0, to obtain


Use the calculus product rule on the 1st and 2nd term in above equation. The second term (double underlined) becomes


and the first term (single underline) becomes

Group these two expansions as follows

Conservation of mass is:

so that the two triple-underlined terms (multiplied by ) are conservation of mass and sum to zero.


Thus, after incorporating conservation of mass (continuity) into the linear momentum equation, and rearranging the result, we obtain:

Note that the above is a vector equation. Note that since conservation of mass was incorporated into the above, this linear momentum equation automatically satisfies conservation of mass. Also, the term in brackets can be written in vector notation: so that

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