Question 12: Centre of Mass and Rotational Motion

Question

The moment of inertia of a cube of mass $ m $ and side $ a $ about one of its edges is equal to

Accepted Answer

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_{	ext{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_{	ext{center}} + m \left(\frac{a}{\sqrt{2}}ight)^{2} $
This accounts for the shift in the axis position. With simplification:
$ \begin{aligned} I & = \left[\frac{m a^{2}}{12} + \frac{m a^{2}}{12}ight] + \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}  \end{aligned} $
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} $.

Answer

$ \frac{2}{3} m a^{2} $

Answer

$ \frac{4}{3} m a^{2} $

Answer

$ 3 m a^{2} $

Answer

$ \frac{8}{3} m a^{2} $

Centre of Mass and Rotational Motion

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

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