A person drops a cylindrical steel bar (Y = 8.00 × 1010 Pa) from a height of 1.0

Question

A person drops a cylindrical steel bar (Y = 8.00 × 1010 Pa) from a height of 1.00 m (distance between the floor and the bottom of the vertically oriented bar). The bar, of length L = 0.700 m, radius R = 0.00600 m, and mass m = 0.500 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?

Please include detailed calculation and explnantion, image/drawing would be so helpful! Thank you so much for helping me with my physics learning, really appreciate your time!!)

and please only answer if you are confident with your answe, that'd be helpful, thank you again!

Explanation / Answer

Let the maximum compression of bar be x

Since the collision is fully elastic, no energy is lost . Let us consider the ground as the datum.

Initial height of center of mass of bar from ground= 1+(0.7/2)=1.35m

Final height of center of mass of bar from ground= (0.7-x)/2=0.35-0.5x

Stiffness of bar = AE/L = (pi*0.0062*8*1010)/0.7=12925409.77 N/m

Initial potential energy of bar = mgh=0.5*9.8*1.35=6.615 Joules

Final potential energy = mgh=0.5*9.8*(0.35-0.5x)=1.715-2.45x

Loss in potential energy = 6.615-(1.715-2.45x)=4.9+2.45x

From conservation of energy, gain in energy in bar = 0.5*12925409.77*x2 = 6462704.885x2

6462704.885x2 = 4.9+2.45x

x=0.000871 m = 0.871 mm

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