QQuestionPhysics
QuestionPhysics
A person drops a cylindrical steel bar (Y = 1.70 x 1011 Pa) from a height of 1.30 m (distance between the floor and the bottom of the vertically oriented bar). The bar, or length L = 0.63 m, radius R = 0.55 cm, and mass m= 1.30 kg, hits the floow and bounces up, maintaining its vertical orientation. Assuming the collision is elastic (bounces) and no rotation occurs, what is the maximum compression of the bar?
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Answer
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Step 1:: Calculate the tension force (F) in the bar when it is at its maximum compression.
F = 12.753 \, \mathrm{N}
The tension force is equal to the weight of the bar (mg) since the bar is vertical.
Step 2:: Determine the maximum compression of the bar (ΔL) by using the formula for calculating the elastic deformation of a bar under axial load:
\Delta L = \frac{(12.753 \, \mathrm{N})(0.63 \, \mathrm{m})}{(9.621 \times 10^{-5} \, \mathrm{m}^2)(1.70 \times 10^{11} \, \mathrm{Pa})}
where F is the tension force, L is the original length of the bar, A is the cross-sectional area of the bar, and Y is Young's modulus. First, calculate the cross-sectional area (A) of the cylindrical bar: Now, substitute the values into the formula:
Step 3:: Calculate the maximum compression:
\Delta L = 5.34 \times 10^{-7} \, \mathrm{m}
Final Answer
The maximum compression of the bar is 5.34 × 10^- 7 m or 0.000534 mm.
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