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: Set up axes along the two edges meeting at the vertex.
Place the vertex at the origin with one edge of length $a$ along the x-axis and the other edge of length $b$ along the y-axis. Since the plate lies flat in this plane, the perpendicular axis theorem lets us build the perpendicular moment of inertia $I_z$ directly from the two in-plane moments, $I_z = I_x + I_y$.

Step 2: Find $I_x$, the moment about the edge lying along the x-axis.
Treating the plate as strips parallel to the x-axis at height $y$, with mass per unit area $\sigma = \dfrac{m}{ab}$:
\[ I_x = \int_0^b y^2(\sigma a)\,dy = \sigma a\cdot\frac{b^3}{3} = \frac{m}{ab}\cdot a\cdot\frac{b^3}{3} = \frac{mb^2}{3} \]

Step 3: Find $I_y$ the same way, about the edge along the y-axis.
\[ I_y = \int_0^a x^2(\sigma b)\,dx = \sigma b\cdot\frac{a^3}{3} = \frac{ma^2}{3} \]

Step 4: Add them using the perpendicular axis theorem.
\[ I_z = I_x + I_y = \frac{mb^2}{3} + \frac{ma^2}{3} = \frac{1}{3}m(a^2+b^2) \]

Step 5: Conclusion.
\[ \boxed{\frac{1}{3}m(a^2+b^2)} \]