(B) The moment of inertia $(I)$ of a thin circular disc about a tangent in the plane of the disc is given by the parallel axis theorem: $I = I_{cm} +

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(B) The moment of inertia $(I)$ of a thin circular disc about a tangent in the plane of the disc is given by the parallel axis theorem: $I = I_{cm} + MR^2 = \frac{1}{4}MR^2 + MR^2 = \frac{5}{4}MR^2$. From this,we get $MR^2 = \frac{4}{5}I$. The moment of inertia of the disc about an axis perpendicular to the plane and passing through its centre is $I_z = \frac{1}{2}MR^2$. Substituting the value of $MR^2$,we get $I_z = \frac{1}{2} \times (\frac{4}{5}I) = \frac{2}{5}I$.

Class 11 Physics System of Particles and Rotational Motion Moment of Inertia of Compound Bodies and Theorem of Moment of Inertia

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If $I$ is the moment of inertia of a thin circular disc about an axis passing through the tangent of the disc and in the plane of the disc,what is the moment of inertia of the same circular disc about an axis perpendicular to the plane and passing through its centre?

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A
$\frac{4 I}{5}$
B
$\frac{2 I}{5}$
C
$\frac{4 I}{3}$
D
$\frac{2 I}{3}$

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System of Particles and Rotational Motion

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Moment of Inertia of Compound Bodies and Theorem of Moment of Inertia

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If $I$ is the moment of inertia of a thin circular disc about an axis passing through the tangent of the disc and in the plane of the disc,what is the moment of inertia of the same circular disc about an axis perpendicular to the plane and passing through its centre?

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