Multiple Cholce Questlon A rotor has a weight of 100 lb with a centroidal radius of gyration of 9 in . Determine the moment of inertia about the center of gravity for the rotor. O \( 56.25 \mathrm{lb} \mathrm{ft} \cdot \mathrm{s}^{2} \) \( 1.747 \mathrm{lb} \mathrm{ft} \cdot \mathrm{s}^{2} \) O \( 5.734 \mathrm{lb} \mathrm{ft} \cdot \mathrm{s}^{2} \) O \( 2.0963 \mathrm{lb} \mathrm{ft} \cdot \mathrm{s}^{2} \)

To determine the moment of inertia \( I \) about the center of gravity for the rotor, we can use the formula: \[ I = W \cdot k^2 \] where: - \( I \) is the moment of inertia, - \( W \) is the weight of the rotor, - \( k \) is the radius of gyration. ### Given: - Weight \( W = 100 \, \text{lb} \) - Radius of gyration \( k = 9 \, \text{in} \) ### Step 1: Convert the radius of gyration to feet Since the weight is given in pounds and we want the moment of inertia in \( \text{lb} \cdot \text{ft}^2 \), we need to convert inches to feet: \[ k = \frac{9 \, \text{in}}{12 \, \text{in/ft}} = 0.75 \, \text{ft} \] ### Step 2: Calculate the moment of inertia Now we can substitute the values into the formula: \[ I = 100 \, \text{lb} \cdot (0.75 \, \text{ft})^2 \] Calculating \( (0.75 \, \text{ft})^2 \): \[ (0.75)^2 = 0.5625 \, \text{ft}^2 \] Now substituting back into the equation for \( I \): \[ I = 100 \, \text{lb} \cdot 0.5625 \, \text{ft}^2 = 56.25 \, \text{lb} \cdot \text{ft}^2 \] ### Conclusion The moment of inertia about the center of gravity for the rotor is: \[ I = 56.25 \, \text{lb} \cdot \text{ft}^2 \] Thus, the correct answer is: **\( 56.25 \, \text{lb} \cdot \text{ft}^2 \)**.