Theorem of parallel axes

Options

A. applicable to body of any angle
B. needs MI about an axis passing through CM and parallel to the axis passing through CM and parallel to the axis about which we want to know the MI of the same body
C. Both (1) and (2) are correct
D. Both (1) and (2) are incorrect

Answer

To solve the question regarding the Theorem of Parallel Axes, we need to understand the theorem itself and its application. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Parallel Axis Theorem
The Parallel Axis Theorem states that if you know the moment of inertia of a body about an axis that passes through its center of mass (I_cm), you can find the moment of inertia about any parallel axis (I) that is a distance 'd' away from the center of mass axis using the formula:

\[ I = I_{cm} + m \cdot d^2 \]

Where:
- \( I \) is the moment of inertia about the new axis.
- \( I_{cm} \) is the moment of inertia about the center of mass axis.
- \( m \) is the mass of the body.
- \( d \) is the distance between the two parallel axes.

### Step 2: Identify the Conditions for Application
For the Parallel Axis Theorem to be applicable:
1. Both axes must be parallel to each other.
2. One of the axes must pass through the center of mass of the object.
3. The object can be of any shape or orientation, as long as the above conditions are satisfied.

### Step 3: Example Calculation
Let’s say we have a solid sphere of mass \( m \) and radius \( r \). The moment of inertia about an axis through its center of mass is given by:

\[ I_{cm} = \frac{2}{5} m r^2 \]

If we want to find the moment of inertia about an axis that is parallel to the center of mass axis and a distance \( d \) away, we can use the Parallel Axis Theorem:

\[ I = I_{cm} + m \cdot d^2 \]
\[ I = \frac{2}{5} m r^2 + m \cdot d^2 \]

### Step 4: Conclusion
Thus, the theorem allows us to calculate the moment of inertia about any parallel axis as long as we have the moment of inertia about the center of mass and the distance between the two axes.