What is the moment of inertia of a disc about | vedclass

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(B) The moment of inertia of a disc about an axis passing through its centre and perpendicular to its plane is $I_z = \frac{MR^2}{2}$,where $M$ is the

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$ \frac{MR^2}{4} $

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(B) The moment of inertia of a disc about an axis passing through its centre and perpendicular to its plane is $I_z = \frac{MR^2}{2}$,where $M$ is the mass and $R$ is the radius of the disc.Since the disc is a planar body,we can apply the theorem of perpendicular axes.Let the $x$ and $y$ axes lie in the plane of the disc,intersecting at the centre $O$. These axes represent two diameters of the disc.According to the theorem of perpendicular axes,$I_z = I_x + I_y$.Due to the rotational symmetry of the disc,the moment of inertia about any diameter is the same,so $I_x = I_y = I_d$.Substituting this into the theorem,we get $I_z = I_d + I_d = 2I_d$.Therefore,$I_d = \frac{I_z}{2} = \frac{1}{2} \left( \frac{MR^2}{2} \right) = \frac{MR^2}{4}$.Thus,the moment of inertia of a disc about one of its diameters is $\frac{MR^2}{4}$.

What is the moment of inertia of a disc about one of its diameters?

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