Calculate the moment of inertia of a ring of mass $500 \mathrm{~g}$ and radius $0.5 \mathrm{~m}$ about an axis of rotation passing through
(i) its diameter
(ii) a tangent perpendicular to its plane.
(i) its diameter
(ii) a tangent perpendicular to its plane.
Q 96.1
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Data : $M=500 \mathrm{~g} 0.5 \mathrm{~kg}, \mathrm{R}=0.5 \mathrm{~m}$
(i) The moment of inertia of the ring about its
$
\begin{aligned}
& \text { diameter }=\frac{M R^2}{2}=\frac{0.5 \times(0.5)^2}{2} \\
& =0.0625 \mathrm{~kg} \mathrm{~m}{ }^2=6.25 \times 10^{-2} \mathrm{~kg} \cdot \mathrm{m}^2
\end{aligned}
$
(ii) The moment of inertia of the ring about a tangent perpendicular to its plane $=2 \mathrm{MR}^2=2 \times 0.5 \times(0.5)^2=0.25 \mathrm{~kg} \cdot \mathrm{m}^2$
(i) The moment of inertia of the ring about its
$
\begin{aligned}
& \text { diameter }=\frac{M R^2}{2}=\frac{0.5 \times(0.5)^2}{2} \\
& =0.0625 \mathrm{~kg} \mathrm{~m}{ }^2=6.25 \times 10^{-2} \mathrm{~kg} \cdot \mathrm{m}^2
\end{aligned}
$
(ii) The moment of inertia of the ring about a tangent perpendicular to its plane $=2 \mathrm{MR}^2=2 \times 0.5 \times(0.5)^2=0.25 \mathrm{~kg} \cdot \mathrm{m}^2$
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