A person drops a cylindrical steel bar $\left(Y=1.80 \times 10^{11} \mathrm{~Pa}\right)$ from a height of $1.70 \mathrm{~m}$ (distance between the floor and the bottom of the vertically oriented bar). The bar, of length $L=0.89 \mathrm{~m}$, radius $R=0.65 \mathrm{~cm}$, and mass $m=1.80 \mathrm{~kg}$, 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? \[ \Delta l= \]
Solution
Step 1 ：The potential energy of the bar before it hits the floor is given by \(mgh\), where \(m\) is the mass of the bar, \(g\) is the acceleration due to gravity, and \(h\) is the height from which the bar is dropped. For this problem, \(m = 1.8\) kg, \(g = 9.81\) m/s², and \(h = 1.7\) m. Therefore, the potential energy is \(mgh = 1.8 \times 9.81 \times 1.7 = 30.0186\) J.
Step 2 ：The elastic potential energy stored in the bar when it is compressed by a distance \(\Delta l\) is given by \(\frac{1}{2}Y\frac{\Delta l^2}{L}\), where \(Y\) is the Young's modulus of the material of the bar, \(\Delta l\) is the compression of the bar, and \(L\) is the original length of the bar. For this problem, \(Y = 1.80 \times 10^{11}\) Pa and \(L = 0.89\) m.
Step 3 ：Setting these two expressions equal to each other and solving for \(\Delta l\) will give the maximum compression of the bar. Therefore, \(\frac{1}{2}Y\frac{\Delta l^2}{L} = mgh\). Solving for \(\Delta l\), we get \(\Delta l = \sqrt{\frac{2mghL}{Y}}\). Substituting the given values, we get \(\Delta l = \sqrt{\frac{2 \times 1.8 \times 9.81 \times 1.7 \times 0.89}{1.80 \times 10^{11}}} = 1.7229352860743202 \times 10^{-5}\) m.
Step 4 ：Final Answer: The maximum compression of the bar is \(\boxed{1.72 \times 10^{-5} \mathrm{~m}}\).
