Question

A person drops a cylindrical steel bar (Y = 8.00 × 1010 Pa) from a height of 3.20 m (distance between the floor and the bottom of the vertically oriented bar). The bar, of length L = 0.820 m, radius R = 0.00500 m, and mass m = 1.300 kg, hits the floor and bounces up, maintaining its vertical orientation. Assuming the collision with the floor is elastic, and that no rotation occurs, what is the maximum compression of the bar?

Answer 1

You can the velocity of the bar when it hits the ground using kinematics:
v^2= vo² + 2 a d
v^2 = 2*a*d

It's an elastic collision, Kinetic Energy is conserved !!

KE = 1/2*m*v²
KE = 1/2*m*2 a d
KE = m*a*d

Spring potential energy, PE = 1/2 k x²
We know that F = kx.
k = F/x = YA/L

PE = 1/2 k x²
PE = 1/2 Y A x² / L
PE = 1/2 Y A x² / L
PE = 1/2 Y ? R² x² / L

Using Energy Conservation, Potential Energy = Kinetic Energy
PE = KE
1/2 Y ? R² x² / L = m a d
x = sqrt( (2 m a d L) / [Y ? R²] )
x = sqrt( (2*1.3*9.81*3.20*0.82) / (8.0 * 10^10 * ? * 0.005^2) )
x = 0.00326 m
Maximum compression of the bar in meters, x = 3.26 * 10^-3 m