The moment of inertia of a thin disc about axes $a, b, c, d$ are $I_{1}, I_{2}, I_{3}$ and $I_{4}$ respectively, as shown in figure. If the moment of inertia about an axis passing through the centre and perpendicular to the plane of the disc is $I$ then,
The problem asks us to determine how the moment of inertia \( I \) of a thin disc, about an axis perpendicular to its plane passing through its center, relates to the moments of inertia \( I_1, I_2, I_3, \) and \( I_4 \) about other specific axes.
The moment of inertia of a thin disc about an axis perpendicular to its plane and passing through its center \( O \) is given by \( I = \frac{1}{2} M R^2 \), where \( M \) is the mass and \( R \) is the radius of the disc.
For axes \( a \) and \( b \), and \( c \) and \( d \), which are in the plane and perpendicular to one another, the perpendicular axis theorem states:
\(I = I_{1} + I_{2}\)
and
\(I = I_{3} + I_{4}\)
This theorem applies because the axes are within the plane of the disc, and according to the theorem:
\(I_{\perp} = I_x + I_y\)
where \( I_x \) and \( I_y \) are the moments of inertia about two perpendicular axes in the plane of the object.
Therefore, the valid answers based on this conclusion are:
- \(I = I_{1} + I_{2}\)
- \(I = I_{3} + I_{4}\)
These are the correct options based on the given figure and problem conditions.
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