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 $$ 2.0963 \mathrm{lb} \mathrm{ft} \cdot \mathrm{s}^{2} $$

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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:

**$$ 56.25 \, \text{lb} \cdot \text{ft}^2 $$**.
The moment of inertia about the center of gravity for the rotor is $$ 56.25 \, \text{lb} \cdot \text{ft}^2 $$.

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

O
O

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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 O O