Question

A person drops a cylindrical steel bar ( Y = 1.700 × 10 11 Pa ) from a height of 3.70 m (distance between the floor and the bottom of the vertically oriented bar). The bar, of length L = 0.950 m , radius R = 0.00650 m , and mass m = 2.000 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? maximum compression: m

Answer 1

Given that, Young's modulus of steel bar is , Y = 1-7X10"Pa Height, h= 3.70m length, L=0.950m radius, R=0.00 650m mass, m=2.0kg The expression for equation of motion is wr=un+ 2as. As initially the bar is at rest , u=o. substitute in above equation, w=2as. The expression of kinetic energy is, Ek = 1/2mva = { m (295) = mas The expression botential energy is .. Ep = // kyr Here, k = ( ) = (YA) Substitute the value of k in the above equation Ep = 2 kyn -IN - = -Ź (A) yn - Ź (Y (ARM)) yn IN

As in the case of elastic collision energy remains conserved. Ex=Ep. Substitute the values, mas = Į (T (HRT)) yn or, y = /2 mash or, y = √ ZVIR Sustitute the above equation by the given values, y [2x(2.0)*(9.81)+(-3.70 )* (0.95) V (1.7*lo") x T *(0.006504h = 2.478 10-3 m - The maximum compression is 2.47 8105 m.