A person drops a cylindrical steel bar (Y = 1.00 × 1011 Pa) from a height of 1.30 m (distance between the floo?

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A person drops a cylindrical steel bar (Y = 1.00 × 1011 Pa) from a height of 1.30 m (distance between the floor and the bottom of the vertically oriented bar). The bar, of length L = 0.95 m, radius R = 0.0075 m, and mass m = 0.50 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?Please answer in meters! Thank you! 🙂

Hermina Ferry · Accepted Answer

You can the velocity of the bar when it hits the ground using kinematics: v² = vo² + 2 a dv² = 2 a dIt’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:KE = 1/2 m v² KE = 1/2 m 2 a dKE = m a dNext… I’m not sure how YOU have been taught to go about this, but 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 for potential energy:PE = 1/2 k x²PE = 1/2 Y A x² / L PE = 1/2 Y A x² / L PE = 1/2 Y π R² x² / LSet potential energy equal to kinetic energy and solve for x:PE = KE1/2 Y π R² x² / L = m a dx = √{2 m a d L / [Y π R²]}x = √{2 (0.50 kg) (-9.81 m/s²) (-1.30 m) (0.95 m) / [(1.00 * 10^11 Pa) π (0.0075 m)²]}x = 8.28 * 10^-4 mx = 8.28 * 10^-4 m = 0.000828 m

Jakob Littel

A person drops a cylindrical steel bar (Y = 1.00 × 1011 Pa) from a height of 1.30 m (distance between the floo?

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