Velocity and acceleration of a particle of a fluid

In summary: Thanks again, using Gaussian elimination as you suggested I found X1,X2 and X3 all to be equal to d(1-αt)/(1-(α^3)(t^3)).
  • #1
Woolyabyss
143
1

Homework Statement


Fluid mech.png

Homework Equations


x = (x1,x2,x3) are the coordinates of the current configuration and X = (X1,X2,X3) are the coordinates of the reference configuration where x = y(X,t) where y is a deformation of X.

The Attempt at a Solution


(a)
taking the partial derivatives of xi with respect to t:
∂(x1)/∂t = αX2 + 2(α^2)X3*t
∂(x2)/∂t = 2(α^2)*X1*t + αX3
∂(x3)/∂t = αX1 +2(α^2)X2t

so v(X,t) = (αX2 + 2(α^2)X3*t , 2(α^2)*X1*t + αX3 , αX1 +2(α^2)X2t )
(1) v((d,d,d),0) = ( αd , αd , αd )

a(X,t) = ∂v/∂t = ( 2(α^2)X3, 2(α^2)*X1 , 2(α^2)X2 )

(2) a((d,d,d),0) = ( 2(α^2)d , 2(α^2)*d, 2(α^2)d )(b)
for this part we know that x = (d,d,d) , so we wan't to find the X that corresponds to this point so we can plug it into the functions v(X,t) and a(X,t).

rearranging the three equations of motion:
X1 = x1 - αX2t - (α^2)X3t^2 (a)
X2 = x1 - αX3t - (α^2)X1t^2 (b)
X3 = x1 - αX1t - (α^2)X2t^2 (c)

now, subbing (b) and (c) into (a) we have:

X1 = x1 - α( x1 - αX3t - (α^2)X1t^2 )t - (α^2)( x1 - αX1t - (α^2)X2t^2 )t^2

from the above it seems like like I won't be able to express X1 solely in terms of spatial coordinates because if I expand the second and third term I won't be able to get rid of X3 and X2.
Any help would be appreciated.
 
Last edited:
Physics news on Phys.org
  • #2
You have 3 linear algebraic equations in three unknowns, X1, X2, and X3 corresponding to (d,d,d) @ t.
 
  • Like
Likes Woolyabyss
  • #3
Chestermiller said:
You have 3 linear algebraic equations in three unknowns, X1, X2, and X3 corresponding to (d,d,d) @ t.

Thanks for the reply,
looking at the system of equations:
d = X1 + αX2t + (α^2)X3t^2
d = X2 + αX3t + (α^2)X1t^2
d = X3 + αX1t + (α^2)X2t^2

if I begin subtracting the equations from one another to remove the d terms and then manipulating the resultant equations I will get:
Xi(terms involving α and t ) = 0 , for i =1,2,3
I'm not sure what I can say about this other than that either the Xi or () must be zero.
 
  • #4
Woolyabyss said:
Thanks for the reply,
looking at the system of equations:
d = X1 + αX2t + (α^2)X3t^2
d = X2 + αX3t + (α^2)X1t^2
d = X3 + αX1t + (α^2)X2t^2

if I begin subtracting the equations from one another to remove the d terms and then manipulating the resultant equations I will get:
Xi(terms involving α and t ) = 0 , for i =1,2,3
I'm not sure what I can say about this other than that either the Xi or () must be zero.
Just use Gaussian elimination.
 
  • Like
Likes Woolyabyss
  • #5
Chestermiller said:
Just use Gaussian elimination.
Thanks again, using Gaussian elimination as you suggested I found X1,X2 and X3 all to be equal to d(1-αt)/(1-(α^3)(t^3)).
I suspect this to be the correct answer as its easy to see why when t approaches α^-1 this motion is unrealistic.
 
Last edited:

1. What is the difference between velocity and acceleration of a particle of a fluid?

Velocity is the rate of change of an object's position, while acceleration is the rate of change of an object's velocity. In the context of a fluid, velocity refers to the speed and direction at which a particle of the fluid is moving, while acceleration refers to how quickly the velocity of the particle is changing.

2. How is the velocity of a particle of a fluid measured?

The velocity of a particle of a fluid can be measured using a variety of techniques, such as using a flow meter or tracking the movement of particles in the fluid using imaging techniques. The velocity can also be calculated by measuring the distance the particle travels over a period of time.

3. What factors affect the velocity of a particle of a fluid?

The velocity of a particle of a fluid can be affected by various factors, including the properties of the fluid (such as viscosity and density), the shape and size of the particle, and the forces acting on the particle (such as gravity or external forces).

4. How is the acceleration of a particle of a fluid related to the forces acting on it?

The acceleration of a particle of a fluid is directly related to the net force acting on the particle. If the net force is zero, the particle will have no acceleration and will move at a constant velocity. However, if there is a net force acting on the particle, it will experience an acceleration in the direction of the force.

5. Can the velocity and acceleration of a particle of a fluid change over time?

Yes, the velocity and acceleration of a particle of a fluid can change over time. This can be due to various factors, such as changes in the properties of the fluid, changes in the forces acting on the particle, or interactions with other particles in the fluid. Understanding these changes is important in studying the behavior of fluids and their impact on various systems.

Similar threads

Replies
5
Views
3K
  • Special and General Relativity
Replies
27
Views
2K
  • Advanced Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
861
  • Differential Geometry
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
4
Views
3K
  • Advanced Physics Homework Help
Replies
8
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
Replies
7
Views
2K
Back
Top