Question

The moment of inertia of a thin uniform rectangular plate of mass \(m\), having length \(a\) and width \(b\), about an axis perpendicular to the plane of the plate and passing through one of its vertices is:

• A. \(\frac{2}{3} m a b\)
• B. \(\frac{1}{3} m a b\)
• C. \(\frac{2}{3} m (a^2 + b^2)\)
• D. \(\frac{1}{3} m (a^2 + b^2)\)

Hint:

For axis passing through a vertex, always use parallel axis theorem with distance from centroid: \(I = I_{\text{centroid}} + m d^2\).

Answer 1

The Correct Option is D

Step 1: Recall parallel axis theorem.
Moment of inertia about an axis through a vertex perpendicular to the plane:
\[ I_{\text{vertex}} = I_{\text{centroid}} + m d^2 \]

Step 2: Moment of inertia about centroid.
For a rectangle, centroidal perpendicular axis:
\[ I_{\text{centroid}} = \frac{1}{12} m (a^2 + b^2) \]

Step 3: Distance from centroid to vertex.
\[ d = \sqrt{(a/2)^2 + (b/2)^2} = \frac{\sqrt{a^2 + b^2}}{2} \]

Step 4: Apply parallel axis theorem.
\[ I_{\text{vertex}} = \frac{1}{12} m (a^2 + b^2) + m \left(\frac{a^2 + b^2}{4}\right) = \frac{1}{12} m (a^2 + b^2) + \frac{1}{4} m (a^2 + b^2) \]

Step 5: Combine terms.
\[ I_{\text{vertex}} = \left(\frac{1}{12} + \frac{3}{12}\right) m (a^2 + b^2) = \frac{1}{3} m (a^2 + b^2) \]

Step 6: Final conclusion.
Hence, the moment of inertia about the vertex is:
\[ \boxed{\frac{1}{3} m (a^2 + b^2)} \]