Question

A person drops a cylindrical steel bar (Y = 1.300 x 1011 Pa) from a height of 4.90 m (distance between the floor and the bottom of the vertically oriented bar). The bar, of length L = 0.990 m, radius R = 0.00500 m, and mass m = 1.600 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:

Answer 1

Given: Y= 1.3x10"Pa h=4.90m L=0.990 m R=0.005 m and, m=1.6kg Here the spring constant of the bar is: K= Y TUR? ik= (1.3 x 10"Pa) (II) (0.005 m) - 1.03 x107 Nm (0.990 m) Nows According to consewation of energy: PES = PEG $k(AXC) = mgh Ax= 2mgh 211.6kg) (9.8 m433) (4.90 m) ( 63 x * Nm) 1.OX= 3.86 X10-3 m

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Answer 2

NOTE: There are different values in the following method but the method is correct. Put your question values and solve, the answer will be surely correct.

The spring constant of the bar is given by

$$k=\frac{YA}{L}$$

where $k$ is the elastic or spring constant of the bar, $Y$ is Young's modulus of the bar, $L$ is the length of the bar, and $A$ is the cross-sectional area of the bar.

$$k=\frac{YA}{L}=\frac{Y\pi R^2}{L}=\frac{1.500\times {10}^{11}\text{ Pa}\times \pi \times {\left(0.00750\text{ m}\right)}^2}{0.940\text{ m}}=2.82\times {10}^7\text{ N}/\text{m}$$

The total energy is conserved, so the gravitational potential energy of the bar-earth system is equal to the elastic potential energy of the bar as it hits the earth and compresses.

$$mgh=\frac{1}{2}k{\left(\mathrm{\Delta }l\right)}^2,$$

where Δ?$\mathrm{\Delta }l$ is the length by which the bar compresses. Thus,

Δ?=2??ℎ?⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√=2×2.000 kg×9.81 m/s2×3.90 m2.82×107 N/m⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√=0.00233 m