Parallel Axis Theorem

Question

The moment of inertia of a disc of mass M and radius R about its center is MR²/2. What is its moment of inertia about a tangent in its plane?

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1. **Step 1: Disc Diameter**: By the perpendicular axis theorem ($I_z = I_x+I_y$), for a symmetric disc, $I_z = 2 I_{diameter}$.
   - $I_{diameter} = I_z / 2 = (MR^2/2) / 2 = MR^2/4$.
2. **Step 2: Axis Selection**: A tangent **in the plane** is parallel to the diameter.
3. **Step 3: Parallel Axis Theorem**: $I_{tangent} = I_{diameter} + M h^2$, where $h = R$.
4. **Calculation**:
   - $I = \frac{MR^2}{4} + MR^2 = \frac{5}{4} MR^2$.
5. **Note**: If the tangent was perpendicular to the plane, the starting axis would be the center axis ($MR^2/2$), giving $3MR^2/2$.

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