BITSAT 2024 Shift 1 Paper

To calculate the moment of inertia of a cube about one of its edges, we can use the theorem of perpendicular axes. Here's how it is derived:First, we consider the moment of inertia about the center of the cube. According to the theorem of perpendicular axes, the total moment of inertia $I$ can be found by adding the contribution from each axis. For a cube, with side length $a$ and mass $m$, the moment of inertia about one of its center axes is:$I_{\;\text{center }\;}=\;\frac{m a^2}{12}$Since the cube is symmetric, this value is the same for any of the three perpendicular axes intersecting at the center. The moment of inertia through the axis parallel to a face through the center can similarly be calculated.Now, when calculating the moment of inertia about an edge, we can add the parallel axis contribution:$I = I_{\;\text{center }\;} + m \left( \;\frac{a}{\sqrt{2}} \right)^2$This accounts for the shift in the axis position. With simplification:$I\; = \left[ \;\frac{m a^2}{12} + \;\frac{m a^2}{12} \right] + \;\frac{m a^2}{2}$$\; = \frac{2}{12} m a^2 + \;\frac{m a^2}{2}$$\; = \frac{m a^2}{6} + \;\frac{m a^2}{2}$$\; = \frac{2}{3} m a^2$Thus, the moment of inertia of a cube of mass $m$ and side $a$ about one of its edges is $\;\frac{2}{3} m a^2$.