Moment of inertia of a disc of radius $R$ about a diametric axis is $25 \mathrm{~kg} \mathrm{~m}^2$. The moment of inertia of the disc about a parallel axis at a distance $R / 2$ from the centre is

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Aicharya · Accepted Answer

From the perpendicular axis theorem, we have $I_z=I_x+I_y$ where, $I_z$ is moment of inertia about Z-axis, $I_x$ is moment of inertia about $X$-axis and $I_y$ is moment of inertia about $Y$-axis.

Moment of inertia of a disc, $I_z=\frac{1}{2} M R^2$ Now, $I_x=I_y$, because two perpendicular diameters are equivalent. $ \begin{aligned} & \text { So, } I_{\text {diameter }}=\frac{1}{2} \times \frac{1}{2} M R^2=\frac{M R^2}{4} \\ & \Rightarrow \quad \frac{M R^2}{4}=25 (given)\\ & \Rightarrow \quad M=\frac{100}{R^2} \mathrm{~kg} \end{aligned} $ Also every diameter passes through centre of mass. So, $\quad I_{\mathrm{COM}}=\frac{M R^2}{4}=25 \mathrm{~kg}-\mathrm{m}^2$ Using parallel axis theorem, $\begin{aligned} I & =I_{\text {COM }}+M x^2 \quad\left(\text { where, } x=\frac{R}{2}\right) \\ & =25+M \frac{R^2}{4} \\ & =25+\frac{100}{R^2} \times \frac{R^2}{4} \\ & =50 \mathrm{~kg}-\mathrm{m}^2\end{aligned}$

Moment of inertia of a disc of radius $R$ about a diametric axis is $25 \mathrm{~kg} \mathrm{~m}^2$. The…

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