TryingToThink said:
Thanks for your response Chet. I would greatly appreciate it if you would show how to derive the analytical solution for the situation you described. Thanks!
Incidentally, Welcome to Physics Forums.
I'm not going to go through all the details since that would be too lengthy for this venue. But, I will give you an abridged analysis. The compression wave and recovery waves travel along the cylinder at the speed of sound in the metal during the time that the cylinder is in contact with the wall. From the analysis of the dynamics in conjunction with hooke's law, the speed of sound in the metal is [itex]\sqrt{\frac{E}{ρ}}[/itex], where E is the young's modulus and ρ is the density. If the cylinder is of length L, the amount of time that it is in contact with the wall is [itex]\frac{2L}{\sqrt{\frac{E}{ρ}}}[/itex]. During the contact, the analysis shows that the force is constant. The impulse of the force is equal to the change in momentum. Therefore,
[tex]F\frac{2L}{\sqrt{\frac{E}{ρ}}}=2mv[/tex]
where m is the mass of the cylinder, and v is its approach and rebound speed. The mass m is equal to the cross sectional area A times the length times the density:
m=ALρ
If we substitute this into the previous momentum balance equation, we get:
[tex]\frac{F}{A}=\sqrt{Eρ}v=E\frac{v}{\sqrt{\frac{E}{ρ}}}[/tex]
The left hand side of this equation is the compressive stress at the contact surface (as well as throughout the compression region). The compressive strain in this region is, according to the above equation, given by [itex]\frac{v}{\sqrt{\frac{E}{ρ}}}[/itex]. This is the ratio of the approach speed to the speed of sound in the metal.
I hope this helps.
Chet