The moment of inertia of a hollow cubical box

Asked by admin @ 20/08/2021 in Physics viewed by 277 People

The moment of inertia of a hollow cubical box of mass m and side a about an axis passing thorough the centre of the opposite faces it equal to

Answered by admin @ 20/08/2021

Final Answer : $\frac{5Ma^2}{18}$
where M is mass of Hollow cubical box.

Steps:
1) Hollow Cubical box : 6 identical Square plates of mass 'm' joined to form a cube of side 'a'.
$M = 6 m$
Moment of Inertia of square plate about COM whose axis is perpendicular to plane of square plate. :
$\frac{ma^2}{6}$

We have,
Cube whose axis passes through centre of mass of two square plate ,facing opposite to each other.
Therefore,
$I_1 = 2*\frac{ma^2}{6} = 2* \frac{M}{6}*(a/6)^2 \\ \\ = \frac{Ma^2}{18}$

2) Then,
Moment of Inertia of square plate about Com , having axis parallel to plane of square plate.
Using Parallel axis theorem,
$I_2 = \frac{ma^2}{12}$
Then, we have

3) Also, By parallel axis theorem.

Remaining Four square plates.
distance from square plate to centre of cube, d = a/2 ,
$I_{rem1} =4[ \frac{ma^2}{12}+m(a/2)^2] \\ \\ = \frac{2Ma^2}{9}$

Finally, we have
$I_{req} = I_{rem1} + I_1 = \frac{5Ma^2}{18}$

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