The moment of inertia of a uniform rod of mass m = 0.50 kg and length `I=1m` is `I=0.10 kg-m^(2)` about a line perpendicular to the rod. The distance of this line from the middle point of the rod is `sqrt(a/b)m` rod where a and b are coprime. Find a + b

Explanation

To solve the problem, we will use the concept of the moment of inertia and the parallel axis theorem. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify Given Values
We have:
- Mass of the rod, \( m = 0.50 \, \text{kg} \)
- Length of the rod, \( L = 1 \, \text{m} \)
- Moment of inertia about a line perpendicular to the rod, \( I_0 = 0.10 \, \text{kg m}^2 \)

### Step 2: Calculate the Moment of Inertia about the Center of Mass
The moment of inertia of a uniform rod about an axis passing through its center of mass is given by the formula:
\[ I_{cm} = \frac{1}{12} m L^2 \]
Substituting the values:
\[ I_{cm} = \frac{1}{12} \times 0.50 \, \text{kg} \times (1 \, \text{m})^2 = \frac{0.50}{12} = \frac{1}{24} \, \text{kg m}^2 \]

### Step 3: Apply the Parallel Axis Theorem
The parallel axis theorem states:
\[ I_0 = I_{cm} + m x^2 \]
Where \( x \) is the distance from the center of mass to the new axis. Rearranging gives:
\[ x^2 = \frac{I_0 - I_{cm}}{m} \]

### Step 4: Substitute Known Values
Substituting \( I_0 \), \( I_{cm} \), and \( m \) into the equation:
\[ x^2 = \frac{0.10 \, \text{kg m}^2 - \frac{1}{24} \, \text{kg m}^2}{0.50 \, \text{kg}} \]
Calculating \( I_0 - I_{cm} \):
\[ I_0 - I_{cm} = 0.10 - \frac{1}{24} = \frac{2.4}{24} - \frac{1}{24} = \frac{1.4}{24} = \frac{7}{120} \, \text{kg m}^2 \]
Now substituting back:
\[ x^2 = \frac{\frac{7}{120}}{0.50} = \frac{7}{120} \times 2 = \frac{7}{60} \]

### Step 5: Solve for \( x \)
Taking the square root:
\[ x = \sqrt{\frac{7}{60}} \, \text{m} \]

### Step 6: Identify \( a \) and \( b \)
From the expression \( x = \sqrt{\frac{a}{b}} \), we have \( a = 7 \) and \( b = 60 \). Since \( a \) and \( b \) are coprime, we can now find \( a + b \).

### Step 7: Calculate \( a + b \)
\[ a + b = 7 + 60 = 67 \]

Thus, the final answer is:
\[ \boxed{67} \]