Question

A steel cylindrical rod of length $l$ and radius $r$ is suspended by its end from the ceiling. (a) Find the elastic deformation energy U of the rod. (b) Define U in terms of tensile strain $\Delta l/l$ of the rod.

• A. (a) $U = \frac{1}{2} Y (\pi r^2 l) \epsilon^2$, (b) $U = \frac{1}{2} Y (\pi r^2 l) \left(\frac{\Delta l}{l}\right)^2$
• B. (a) $U = Y (\pi r^2 l) \epsilon^2$, (b) $U = Y (\pi r^2 l) \left(\frac{\Delta l}{l}\right)^2$
• C. (a) $U = \frac{1}{2} Y (\pi r^2) \epsilon^2$, (b) $U = \frac{1}{2} Y (\pi r^2) \left(\frac{\Delta l}{l}\right)^2$
• D. (a) $U = \frac{1}{2} Y (\pi r^2 l) \epsilon$, (b) $U = \frac{1}{2} Y (\pi r^2 l) \left(\frac{\Delta l}{l}\right)$

Answer 1

Answer: (A)

(a) The elastic deformation energy $U$ of the rod is $U = \frac{1}{2} Y (\pi r^2 l) \epsilon^2$. (b) Defined in terms of tensile strain $\Delta l/l$: $U = \frac{1}{2} Y (\pi r^2 l) \left(\frac{\Delta l}{l}\right)^2$.

Answer 2

• Part (a): The elastic deformation energy $U$ of a rod is given by $U = \frac{1}{2} Y V \epsilon^2$, where $Y$ is Young's modulus, $V$ is the volume of the rod, and $\epsilon$ is the tensile strain. The volume of the cylindrical rod is $V = \pi r^2 l$. Thus, $U = \frac{1}{2} Y (\pi r^2 l) \epsilon^2$.
• Part (b): Tensile strain is defined as $\epsilon = \Delta l/l$. Substituting this into the formula from part (a) gives $U = \frac{1}{2} Y (\pi r^2 l) \left(\frac{\Delta l}{l}\right)^2$.