Given the moment of inertia of a thin uniform disc about its diameter to be $\frac{1}{4} M R^2$, where $M$ and $R$ are respectively the mass and radius of the disc, find its moment of inertia about an axis normal to the disc and passing through a point on its edge.
Q 81
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Consider a thin uniform disc of mass $\mathrm{M}$ and radius $\mathrm{R}$ in the $\mathrm{xy}$ plane. Let $\mathrm{I} \mathrm{x}$, ly and $\mathrm{Iz}$ be the moments of inertia of the disc about the $x, y$ and $z$ axes respectively.
Now, $I_x=I_y$
since each represents the moment of inertia (MI) of the disc about its diameter and, by symmetry, the $\mathrm{Ml}$ of the disc about any diameter is the same.
$\therefore \mathrm{I}_{\mathrm{x}}=\mathrm{I}_{\mathrm{y}}=\frac{1}{4} \mathrm{MR}^2$ (Given)
According to the theorem of perpendicular axes,
$\mathrm{I}_z=\mathrm{I}_{\mathrm{x}}+\mathrm{I}_{\mathrm{y}}=2\left(\frac{1}{4} M R^2\right)=\frac{1}{2} \mathrm{MR}^2$
Let I be the $\mathrm{Ml}$ of the disc about a tangent normal to the disc and passing through a point on its edge (i.e., a tangent perpendicular to its plane). According to the theorem of parallel axis,
$\mathrm{I}=\mathrm{I}_{\mathrm{CM}}+\mathrm{Mh}^2$
Here, $I_{C M}=I_z=\frac{1}{2} M R^2$ and $h=R$.
$\therefore \mathrm{I}=\frac{1}{2} \mathrm{MR}^2+\mathrm{MR}^2=\frac{3}{2} \mathrm{MR}^2$
which is the required expression.
Now, $I_x=I_y$
since each represents the moment of inertia (MI) of the disc about its diameter and, by symmetry, the $\mathrm{Ml}$ of the disc about any diameter is the same.
$\therefore \mathrm{I}_{\mathrm{x}}=\mathrm{I}_{\mathrm{y}}=\frac{1}{4} \mathrm{MR}^2$ (Given)
According to the theorem of perpendicular axes,
$\mathrm{I}_z=\mathrm{I}_{\mathrm{x}}+\mathrm{I}_{\mathrm{y}}=2\left(\frac{1}{4} M R^2\right)=\frac{1}{2} \mathrm{MR}^2$
Let I be the $\mathrm{Ml}$ of the disc about a tangent normal to the disc and passing through a point on its edge (i.e., a tangent perpendicular to its plane). According to the theorem of parallel axis,
$\mathrm{I}=\mathrm{I}_{\mathrm{CM}}+\mathrm{Mh}^2$
Here, $I_{C M}=I_z=\frac{1}{2} M R^2$ and $h=R$.
$\therefore \mathrm{I}=\frac{1}{2} \mathrm{MR}^2+\mathrm{MR}^2=\frac{3}{2} \mathrm{MR}^2$
which is the required expression.
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