Find the moment of inertia of a cube of edge A about an edge.
The Correct Answer and Explanation is:
To calculate the moment of inertia of a cube about an edge, we need to use the formula for the moment of inertia of a solid body about a given axis. For a cube, we assume the mass is uniformly distributed, and we can use the perpendicular axis theorem and the parallel axis theorem to find the moment of inertia about one of its edges.
Formula for Moment of Inertia of Cube About an Edge
The moment of inertia of a cube about an edge is derived from its moments of inertia about the center of mass and its distance from the edge. The formula for the moment of inertia IedgeI_{\text{edge}}Iedge of a cube of edge length AAA about one of its edges is:Iedge=16mA2I_{\text{edge}} = \frac{1}{6} m A^2Iedge=61mA2
where:
- mmm is the mass of the cube,
- AAA is the edge length of the cube.
Step-by-Step Explanation
- Moment of Inertia About Center of Mass:
The moment of inertia of a cube about an axis through its center and perpendicular to the axis is: Icm=16mA2I_{\text{cm}} = \frac{1}{6} m A^2Icm=61mA2 This comes from the standard formula for the moment of inertia of a solid cube (for any axis through its center). - Using the Parallel Axis Theorem:
To find the moment of inertia about an edge, we use the parallel axis theorem. This theorem tells us that the moment of inertia about any axis is the moment of inertia about the center of mass plus md2md^2md2, where ddd is the distance between the two axes. The distance from the center of mass to the edge of the cube is A2\frac{A}{2}2A. So, applying the parallel axis theorem: Iedge=Icm+m(A2)2I_{\text{edge}} = I_{\text{cm}} + m \left( \frac{A}{2} \right)^2Iedge=Icm+m(2A)2 Substituting the values, we get: Iedge=16mA2+m(A2)2I_{\text{edge}} = \frac{1}{6} m A^2 + m \left( \frac{A}{2} \right)^2Iedge=61mA2+m(2A)2 - Simplifying the Expression: Iedge=16mA2+14mA2I_{\text{edge}} = \frac{1}{6} m A^2 + \frac{1}{4} m A^2Iedge=61mA2+41mA2 Combining the terms: Iedge=(16+14)mA2I_{\text{edge}} = \left( \frac{1}{6} + \frac{1}{4} \right) m A^2Iedge=(61+41)mA2 Finding the common denominator (12): Iedge=512mA2I_{\text{edge}} = \frac{5}{12} m A^2Iedge=125mA2
Therefore, the moment of inertia of a cube about one of its edges is:Iedge=512mA2I_{\text{edge}} = \frac{5}{12} m A^2Iedge=125mA2
Conclusion
The moment of inertia of a cube about an edge is 512mA2\frac{5}{12} m A^2125mA2. This result incorporates the cube’s mass and edge length, and it takes into account the rotational characteristics when the cube is rotating around an axis that is aligned with one of its edges.
