Calculate the percentage increase in the moment of inertia about the axis of symmetry of a flywheel when the diameter of the flywheel is increased by 2%.

Answer

To solve the problem of calculating the percentage increase in the moment of inertia of a flywheel when its diameter is increased by 2%, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Moment of Inertia Formula**:
The moment of inertia \( I \) of a flywheel (assuming the mass is concentrated at the periphery) is given by the formula:
\[ I = MR^2 \]
where \( M \) is the mass of the flywheel and \( R \) is the radius.

2. **Differentiate the Moment of Inertia**:
To find the change in moment of inertia, we differentiate \( I \) with respect to \( R \):
\[ dI = d(MR^2) = M \cdot d(R^2) = M \cdot (2R \cdot dR) \]
Thus, we have:
\[ dI = 2MR \cdot dR \]

3. **Express the Change in Moment of Inertia**:
We can express the relative change in moment of inertia as:
\[ \frac{dI}{I} = \frac{dI}{MR^2} = \frac{2MR \cdot dR}{MR^2} = \frac{2dR}{R} \]

4. **Relate Diameter Change to Radius Change**:
Given that the diameter of the flywheel is increased by 2%, we can relate this to the radius. Since the diameter \( D = 2R \), an increase in diameter by 2% means:
\[ \frac{dD}{D} = 0.02 \]
Therefore, since \( D = 2R \):
\[ \frac{dD}{D} = \frac{dD}{2R} = \frac{dR}{R} \]
This implies:
\[ \frac{dR}{R} = \frac{0.02}{2} = 0.01 \]

5. **Calculate the Percentage Increase in Moment of Inertia**:
Now substituting \( \frac{dR}{R} \) into the equation for the change in moment of inertia:
\[ \frac{dI}{I} = 2 \cdot \frac{dR}{R} = 2 \cdot 0.01 = 0.02 \]
To convert this into a percentage, we multiply by 100:
\[ \frac{dI}{I} \cdot 100 = 0.02 \cdot 100 = 2\% \]

6. **Final Calculation**:
Therefore, the percentage increase in the moment of inertia when the diameter of the flywheel is increased by 2% is:
\[ \text{Percentage Increase} = 4\% \]

### Final Answer:
The percentage increase in the moment of inertia of the flywheel is **4%**.