Question

A person drops a vertically-oriented, cylindrical steel bar from a height of 1.80 m (measured from the floor to the bottom of the bar). The bar has length L=0.67 m, radius R=0.70 cm, mass m=0.60 kg, and Young's modulus Y=1.80×10¹¹ Pa. The bar hits the floor and bounces back up, maintaining its vertical orientation. Use the gravitational acceleration g=9.81 m/s² . Assume that the collision with the floor is elastic, and that no rotation occurs. What is the maximum compression Δl of the bar?

Δl=

Answer 1

The maximum compression Δl of the bar is approximately 2.97x10^-13 m.

Step-by-step explanation:

To find the maximum compression Δl of the bar, we can use the principle of conservation of energy. When the bar hits the floor, it loses its initial gravitational potential energy and gains an equal amount of elastic potential energy due to compression.

Using the equation for elastic potential energy Ue = 0.5kΔl2, where k is the spring constant and Δl is the compression distance, we can solve for Δl.

First, we need to find the spring constant k. The spring constant is given by k = (YπR2)/L, where Y is the Young's modulus, R is the radius of the bar, and L is the length of the bar.

Substituting the given values, we have k = (1.80x1011 Pa)(π(0.70x10-2 m)2)/(0.67 m).

Next, we can calculate the change in potential energy ΔU = mgΔl, where m is the mass of the bar and g is the gravitational acceleration. Setting this equal to the elastic potential energy, we have mgΔl = 0.5kΔl2.

Simplifying, we get Δl = 2mg/k.

Substituting the known values, we have Δl = 2(0.60 kg)(9.81 m/s2)/((1.80x1011 Pa)(π(0.70x10-2 m)2)/(0.67 m)).

Calculating this expression gives Δl ≈ 2.97x10-13 m.