Question

A person drops a cylindrical steel bar (Y = 8.00 × 1010 Pa) from a height of 3.20 m (distance between the floor and the bottom of the vertically oriented bar). The bar, of length L = 0.820 m, radius R = 0.00500 m, and mass m = 1.300 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

By using third equation of motion we can find the velocity of the bar when it hits the ground

Initial velocity is zero hence

eq-1

It's an elastic collision, so we know that the energy is conserved throughout. This means that the kinetic energy just before it hits the ground is equal to the stored potential energy when it is fully compressed. KE = PE. First let's calculate kinetic energy:

by plugging the value from eq-1 we get

the bar is essentially a spring. Spring potential energy is PE = 1/2 k x². We know that F = kx.

Using the definition of Young's modulus and some algebra you can see that k = F/x = YA/L.

So the potential energy is

PE = 1/2 k x²
PE = 1/2 Y A x² / L
PE = 1/2 Y A x² / L
PE = 1/2 Y ? R² x² / L

Set potential energy equal to kinetic energy and solve for x:
PE = KE
1/2 Y ? R² x² / L = m a s

which gives

by plugging all the values we get

which gives

or