An engineer has odd-shaped 10 kg object needs to find its rotational inertia about an axis through its center of mass. The object is supported on a wire stretched along the desired axis. The wire has a torsion constant `k = 0.50 N.m`. If this torsion pendulum oscillates through 20 cycles in 50 s, what is the rotational inertia `("in" 10^(-2) kg m^(2))` of the object?

Answer

To find the rotational inertia of the object, we will follow these steps:

### Step 1: Determine the Time Period
The problem states that the torsion pendulum oscillates 20 cycles in 50 seconds. We can find the time period \( T \) of one cycle using the formula:

\[ T = \frac{\text{Total Time}}{\text{Number of Cycles}} = \frac{50 \text{ s}}{20} = 2.5 \text{ s} \]

### Step 2: Use the Formula for the Time Period of a Torsion Pendulum
The time period \( T \) of a torsion pendulum is given by the formula:

\[ T = 2\pi \sqrt{\frac{I}{k}} \]

Where:
- \( I \) is the rotational inertia,
- \( k \) is the torsion constant.

### Step 3: Rearrange the Formula to Solve for Rotational Inertia
We can rearrange the formula to solve for \( I \):

\[ I = \frac{T^2 \cdot k}{4\pi^2} \]

### Step 4: Substitute the Known Values
We know:
- \( T = 2.5 \text{ s} \)
- \( k = 0.50 \text{ N.m} \)

Now substitute these values into the formula:

\[ I = \frac{(2.5)^2 \cdot 0.50}{4\pi^2} \]

### Step 5: Calculate the Value
First, calculate \( (2.5)^2 \):

\[ (2.5)^2 = 6.25 \]

Now substitute this back into the equation:

\[ I = \frac{6.25 \cdot 0.50}{4\pi^2} \]

Calculating \( 4\pi^2 \):

\[ 4\pi^2 \approx 39.478 \]

Now substitute this value:

\[ I = \frac{6.25 \cdot 0.50}{39.478} = \frac{3.125}{39.478} \approx 0.0792 \text{ kg m}^2 \]

### Step 6: Convert to the Required Units
The problem asks for the answer in \( 10^{-2} \text{ kg m}^2 \):

\[ I \approx 0.0792 \text{ kg m}^2 = 7.92 \times 10^{-2} \text{ kg m}^2 \]

### Final Answer
Rounding this to one decimal place, we get:

\[ I \approx 8 \times 10^{-2} \text{ kg m}^2 \]