Sunday

April 26, 2015

April 26, 2015

Posted by **lin** on Wednesday, June 26, 2013 at 9:54am.

- heeeeeeeelp physics -
**Count Iblis**, Wednesday, June 26, 2013 at 1:39pmThis problem is fundamentally flawed, because you cannot treat the collision as elastic (in the sense that all the energy stays in the center of mass motion). You can't on the one hand consider the effect of the finite elasticity coefficient making the collision last a finite time and the same time pretend that the rods won't oscillate.

This is how you have to solve the problem. The relation between the deformation tensor e_{i,j} and the stress tensor t_{i,j} is

e_{i,j} = (1+nu)/E t_{i,j} - nu/E t_{k,k} delta_{i,j}

where

e_{i,j} = 1/2 [du_i/dx_j + du_j/dx_i]

with u the displacement vector and

t_{i,j} is the ith component of the force per unit area with normal in the jth direction that the part the normal is pointing in exerts on the other part.

Taking the 1-direction to be the direction the rods move in, we can assume that the only nonzero component of t is t_{1,1}. At the point where the two rods meet this will be positive at first. If you then pretend that the two rods form one big rod, the collision time is the time during which t_{1,1} at that points is larger than zero.

The equation of motion is (Newton's second law in local form):

rho d^u_i/dt^2 = d/dx_j t_{i,j}

Express t_{i,j} in terms of u_{i,j} from:

e_{i,j} = (1+nu)/E t_{i,j} - nu/E t_{k,k} delta_{i,j}

Take the trace:

e_{i,i} = (1-2nu)/E t_{k,k} ----->

t_{k,k} = E/(1-2nu) e_{k,k}

so, you can write:

t_{i,j} =

E/(1+nu) e_{i,j} + E nu/[(1-2nu)(1+nu)] e_{k,k} delta_{i,j}

The equation of motion

rho d^2u_i/dt^2 = d/dx_j t_{i,j}

can then be written in terms of the displacement u alone as:

rho d^2u_i/dt^2 =

1/2 E/(1+nu) d^2u_i/dx_jdx_j +

E/[2(1+nu)(1-2nu)] d^2u_j/dx_jdx_i

You need to solve these partial differential equation with the appropriate boundary and initial conditions to solve the problem.

- heeeeeeeelp physics -
**4823**, Thursday, June 27, 2013 at 11:39am0.00012