Question

The moment of inertia of a cube of mass \( m \) and side \( a \) about one of its edges is equal to:

• A. \( \frac{2}{3} ma^2 \)
• B. \( \frac{4}{3} ma^2 \)
• C. \( 3 ma^2 \)
• D. \( \frac{8}{3} ma^2 \)

Hint:

For any solid body, the moment of inertia about an edge can be found using the parallel axis theorem: \[ I = I_C + m d^2 \] where \( d \) is the perpendicular distance from the center to the edge.

Answer 1

The Correct Option is A

Step 1: {Apply the perpendicular axis theorem}
By the theorem of perpendicular axes, the moment of inertia about an edge is calculated as:\[I = I_C + m \left( \frac{a}{\sqrt{2}} \right)^2\]Step 2: {Moment of inertia of cube about its center}
The moment of inertia of a cube about its central axis is given by:\[I_C = \frac{ma^2}{12} + \frac{ma^2}{12} = \frac{ma^2}{6}\]Step 3: {Adding the parallel axis contribution}
\[I = \left[ \frac{ma^2}{12} + \frac{ma^2}{12} \right] + \frac{ma^2}{2}\]\[= \frac{2}{3} ma^2\]The final result is \( \frac{2}{3} ma^2 \).