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}
Using the theorem of perpendicular axes, we express the moment of inertia about an edge as: \[ I = I_C + m \left( \frac{a}{\sqrt{2}} \right)^2 \] Step 2: {Moment of inertia of cube about its center}
For a cube, the moment of inertia about its central axis is: \[ I_C = \frac{ma^2}{12} + \frac{ma^2}{12} = \frac{ma^2}{6} \] Step 3: {Adding the parallel axis contribution}
\[ I = \left[ \frac{ma^2}{12} + \frac{ma^2}{12} \right] + \frac{ma^2}{2} \] \[ = \frac{2}{3} ma^2 \] Thus, the correct answer is \( \frac{2}{3} ma^2 \).

Answer 2

Step 1: Moment of inertia depends on mass distribution relative to the axis of rotation.
We are asked to find the moment of inertia of a cube about one of its edges.

Step 2: Use the known result from standard geometry:
The moment of inertia of a cube of mass \( m \) and side \( a \) about an edge (say along the z-axis through one of its edges) is:
\( I = \frac{2}{3}ma^2 \)

Step 3: This result is derived using the parallel axis theorem and integrating over the cube's volume.
It accounts for rotational inertia due to mass distribution at distances from the chosen edge.

Final Answer: \( \frac{2}{3}ma^2 \)