A flywheel can rotate in order to store kinetic energy. The flywheel is a uniform disc made of a material with a density `rho` and tensile strength `sigma` (measured in Pascals), a radiys `r`, and a thickness `h`. The flywheel is rotating at the maximum possible angular velocity so that it does not break. Which of the following expression correctly gives the maximum kinetic energy kilogram that can be stored in the flywheel? Assume that `alpha` is a dimensionless constant.

Options

A. `alphasqrt(rhosigma//r)`
B. `alphahsqrt(rhosigma//r)`
C. `sigmaalphasqrt((h//r)rho)`
D. `alphasigma//rho`

Explanation

To find the maximum kinetic energy per kilogram that can be stored in the flywheel, we need to analyze the relationship between the tensile strength of the material, its density, and the geometry of the flywheel.

### Step-by-Step Solution:

1. **Understanding Tensile Strength and Density**:
- The tensile strength (σ) is defined as the maximum stress that a material can withstand while being stretched or pulled before failing or breaking. It is measured in Pascals (N/m²).
- The density (ρ) of the material is defined as mass per unit volume (kg/m³).

2. **Volume and Mass of the Flywheel**:
- The flywheel is a uniform disc with radius \( r \) and thickness \( h \).
- The volume \( V \) of the flywheel can be calculated using the formula for the volume of a cylinder:
\[ V = \pi r^2 h \]
- The mass \( m \) of the flywheel can be calculated using the density:
\[ m = \rho V = \rho (\pi r^2 h) = \pi \rho r^2 h \]

3. **Maximum Tensile Stress**:
- The maximum tensile stress (σ) can be expressed as the force (F) divided by the cross-sectional area (A). For a disc, the area is:
\[ A = \pi r^2 \]
- Therefore, the tensile stress can be expressed as:
\[ \sigma = \frac{F}{A} = \frac{F}{\pi r^2} \]
- Rearranging gives:
\[ F = \sigma \pi r^2 \]

4. **Work Done by the Tensile Force**:
- The work done (W) by the tensile force when the flywheel rotates can be expressed as:
\[ W = F \cdot L \]
- Here, \( L \) is the effective distance over which the force acts. In this case, we can consider \( L \) to be the thickness \( h \) of the flywheel.

5. **Kinetic Energy per Unit Mass**:
- The kinetic energy (KE) per unit mass can be expressed as:
\[ KE = \frac{W}{m} \]
- Substituting the expressions for \( W \) and \( m \):
\[ KE = \frac{F \cdot h}{m} = \frac{(\sigma \pi r^2) h}{\pi \rho r^2 h} \]
- Simplifying this gives:
\[ KE = \frac{\sigma}{\rho} \]

6. **Including the Dimensionless Constant**:
- Since the problem states that there is a dimensionless constant \( \alpha \), we include it in the final expression:
\[ KE = \alpha \frac{\sigma}{\rho} \]

### Final Expression:
Thus, the maximum kinetic energy per kilogram that can be stored in the flywheel is given by:
\[ KE = \alpha \frac{\sigma}{\rho} \]