A person drops a cylindrical steel bar (Y=6.00*1010 Pa) from a height of 4.70 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=1.100 kg, hits the floor and bounces up, maintaining its vertical orientation. Assuming that the collision with the floor is elastic and that no rotation occurs, what is the maximum compression Δ? of the bar? Use the gravitational acceleration g=9.81 m/s2.

Maximum compression; Δx = 0.003 m

Explanation:

The velocity of the bar when it hits the ground is gotten from equation of motion;

v² = u² + 2gh

u is 0 m/s since it was dropped from rest.

Thus;

v² = 2gh

Now, we know that in an elastic collision, the energy is conserved throughout. Thus, the kinetic energy just before it hits the ground is equal to the stored potential energy when it is fully compressed.

Hence, we'll say;

KE = PE.

The kinetic energy is given by the formula;

KE = ½mv²

Thus; since v² = 2gh;

KE = ½m (2gh)

KE = m•g•h

Now, the bar is essentially a spring. Thus; Spring potential energy is given by the formula; PE = ½k*Δx².

Where Δx is change in length or distance after compression.

We also know that F = k•Δx.

Young's modulus is given by the formula; Y = stress/strain

Where stress = Force/Area = F/A

And strain = change in length/original length; Δx/L

Thus, Y = (F/A) / (Δx/L) = FL/AΔx

Thus; YA/L = F/Δx

From earlier, F = kΔx or F/Δx = k

Thus k = F/Δx = YA/L.

So for potential energy:

PE = ½k•Δx²

Thus;

PE = ½YA•Δx² / L

Area = πR²

So,

PE = ½Y (πR²) (Δx²) / L

Since potential energy is equal to kinetic energy, thus;

PE = KE

½Y (πR²) (Δx²) / L = m a d

Making Δx the subject, we have;

Δx = √{2m•a•d•L/[YπR²]}

We are given;

Mass; m = 1.100 kg

Acceleration due to gravity; g = 9.81 m/s²

Height; h = 4.7m

Length of bar; L = 0.72m

Youngs modulus; Y = 6 x 10^ (10) Pa = 6 x 10^ (10) N/m²

Radius; R = 0.0065m

Δx = √{{2 (1.1) (-9.81) (-4.7) (0.72) }/[ (6 x 10^ (10)) * π (0.0065) ²]}

Δx = 0.003 m