Question

A person drops a cylindrical steel bar (Y = 1.000 times 10^11 Pa) from a height of 1.50 m (distance between the floor and the bottom of the vertically oriented bar). The bar, of length L = 0.720 m, radius R = 0.00650 m, and mass m = 0.600 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

Apply conservation of energy between top an bottom positions

mg(h+x) = 0.5*k*x^2

from the Young modulus relation

K = YA /L

= 1 * 10^11 * 3.14*0.0065^2 / 0.72

= 1.842 x 10^7 N / m

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0.6(9.8)(1.5+x) = 0.5*1.842 x 10^7 N / m*x^2

9210000 x^2 -5.88 x -8.82 = 0

by solving quadratic equation

x = 9.78 * 10^-4 m