Question

A person drops a cylindrical steel bar (?=9.0×1010 Pa) from a height of 1.00 m (distance between the floor and the bottom of the vertically oriented bar). The bar, of length ?=0.73 m, radius ?=0.75 cm, and mass ?=1.70 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

Calculate the velocity of the bar when it hits the ground using kinematics.

v^2 = vo^2 + 2 a d {intila speed is zero}

v^2 = 2 a d

Given that the elastic collision, so the energy is conserved throughout.

The kinetic energy just before it hits the ground is equal to the stored potential energy when it is fully compressed. KE = PE.

calculate kinetic energy

KE = 1/2 m v^2

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

KE = m * a * d

Spring potential energy is PE = 1/2 * k* x^2.

We know that F = kx.

According to the definition of Young's modulus and some algebra you can see that

k = F/x = YA/L

The potential energy becomes,

PE = 1/2 k x^2

= 1/2 Y A x^2 / L

= 1/2 Y A x^2 / L

PE = 1/2 * Y * pi * R^2 * x^2 / L

Equate the potential energy with the kinetic energy and simplify for the value of x,

PE = KE

1/2 Y pi R^2 x^2 / L = m a d

x = sqrt{2*m*a*d*L / [Y*pi*R^2]}

x = sqrt{2 (1.7 kg) (- 9.81 m/s^2) (-1 m) (0.73 m) / [(9 * 10^10 Pa)*pi*(0.0075 m)^2]}

x = 8.75 * 10^-4 m

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