Question

The formalism of the Young's modulus is sometimes used to calculate the reversible work involved in extending or compressing an elastic material. Assume a force \(F\) is applied to an elastic rod of cross-sectional area \(A_{0}\) and length \(L_{0}\). As a result of this force the rod changes in length by \(\Delta L\). The Young's modulus \(E\) is defined as $$E=\frac{\text { tensile stress }}{\text { tensile strain }}=\frac{F / A_{0}}{\Delta L / L_{0}}=\frac{F L_{0}}{A_{0} \Delta L}$$ a. Relate \(k\) in Hooke's Law to the Young's modulus expression just given. b. Using your result in part (a) show that the magnitude of the reversible work involved in changing the length \(L_{0}\) of an elastic cylinder of cross- sectional area \(A_{0}\) by \(\Delta L\) is \(w=\frac{1}{2}\left(\frac{\Delta L}{L_{0}}\right)^{2} E A_{0} L_{0}\)

Short answer

In this problem, we relate the spring constant 'k' in Hooke's Law to the Young's modulus expression and calculate the reversible work involved in changing the length of an elastic cylinder. We find that \(k = -\frac{E A_0}{L_0}\) and the magnitude of the reversible work is given by \(w = \frac{1}{2}(\frac{\Delta L}{L_0})^2 (-k L_0^2)\).

Step-by-step solution

Relate k in Hooke's Law to Young's modulus expression

Hooke's Law states that the force needed to extend or compress a spring by some distance 'x' is proportional to that distance. Mathematically, it can be written as: \(F = -k\Delta L\) where F is the force applied, k is the spring constant and ΔL is the change in length. We have the expression of Young's modulus as: \(E = \frac{F L_0}{A_0 \Delta L}\) We need to relate 'k' in Hooke's Law to the Young's modulus expression. First, let's isolate F in the expression for Young's modulus: \(F = \frac{E A_0\Delta L}{L_0}\) Now, compare this with Hooke's Law \(F = -k\Delta L\), to get the relationship between 'k' and 'E' as: \(k = -\frac{E A_0}{L_0}\)

Calculate the magnitude of the reversible work

Now, we need to calculate the magnitude of the reversible work involved in changing the length L₀ of an elastic cylinder of cross-sectional area A₀ by ΔL, using the expression: \(w = \frac{1}{2}\left(\frac{\Delta L}{L_0}\right)^2 E A_0 L_0 \) We already have the relationship between 'k' and 'E': \(k = -\frac{E A_0}{L_0}\) Now, solve for 'E' in terms of 'k': \(E = -\frac{k L_0}{A_0}\) Let's now substitute the expression for 'E' back into the expression for the reversible work, w: \(w = \frac{1}{2} \left(\frac{\Delta L}{L_0}\right)^2 (-\frac{k L_0}{A_0}) A_0 L_0\) Now, simplify the expression: \(w = \frac{1}{2}(\frac{\Delta L}{L_0})^2 (-k L_0^2)\) This is the magnitude of the reversible work involved in changing the length L₀ of an elastic cylinder of cross-sectional area A₀ by ΔL.

Key concepts

Hooke's Law

Understanding Hooke's Law is crucial when studying how materials respond to applied forces. Hooke's Law can be expressed as:

• \( F = -k\Delta L \)

This formula tells us that the force \( F \) needed to extend or compress a spring by a distance \( \Delta L \) is directly proportional to that distance. The symbol \( k \) stands for the spring constant, which indicates the stiffness of the spring. A high \( k \) means a stiffer spring.
If you imagine a spring being pulled, the distance it stretches is proportional to the force exerted, within the elastic limit, meaning it will return to its original shape when the force is removed.

Tensile Stress

Tensile stress is a measure of how much force is applied over a specific cross-sectional area of a material. This concept is mathematically depicted as:

• \( \text{Tensile Stress} = \frac{F}{A_0} \)

Here, \( F \) is the force applied to the material, and \( A_0 \) represents the original cross-sectional area of the material. The units of tensile stress are typically Pascals (Pa), indicating how pressure is distributed over the area.
Tensile stress is crucial in understanding how materials will behave under tension. If the stress exceeds the strength of the material, it could lead to deformation or fracture.

Tensile Strain

Tensile strain describes how much a material stretches or deforms under stress relative to its original length.

• \( \text{Tensile Strain} = \frac{\Delta L}{L_0} \)

In this formula, \( \Delta L \) is the change in length, and \( L_0 \) is the original length of the material. Tensile strain is a dimensionless quantity, meaning it has no units because it is a ratio of lengths.
The tensile strain gives insight into how a material's length changes when subjected to tensile stress. A higher strain indicates more stretching, often observed in more flexible materials.

Reversible Work

Reversible work, in the context of material deformation, is the work done in a way that can be fully recovered. When an elastic material like a spring returns to its original shape, it is said to have undergone reversible work. The expression for reversible work is provided as:

• \( w = \frac{1}{2}\left(\frac{\Delta L}{L_0}\right)^{2} E A_0 L_0 \)

Here, \( E \) is the Young's modulus, \( \Delta L \) is the change in length, \( L_0 \) is the original length, and \( A_0 \) is the original cross-sectional area.
This equation captures the essence of energy conservation in elastic deformation. The work done can be completely regained, highlighting how energy storage occurs during stretching or compressing in precisely elastic materials.