Question 534: Centre of Mass and Rotational Motion

Question

A disc and a ring both have same mass and radius. The ratio of moment of inertia of the disc about its diameter to that of a ring about a tangent in its plane is

Accepted Answer

For ring, by using perpendicular and parallel axis theorem,
$$\begin{array}{ll} 
& \mathrm{I}_{\mathrm{R}}=\mathrm{I}_{\mathrm{y}}+\mathrm{Mx}^2=\frac{\mathrm{MR}^2}{2}+\mathrm{MR}^2=\frac{3}{2} \mathrm{MR}^2 \
	herefore \quad & \frac{\mathrm{I}_{\mathrm{D}}}{\mathrm{I}_{\mathrm{R}}}=\frac{\mathrm{MR}^2}{4} 	imes \frac{2}{3 \mathrm{MR}^2} \
	herefore \quad & \frac{\mathrm{I}_{\mathrm{D}}}{\mathrm{I}_{\mathrm{R}}}=\frac{1}{6}
\end{array}$$

Answer

$1: 2$

Answer

$1: 4$

Answer

$1: 6$

Answer

$1: 8$

Centre of Mass and Rotational Motion

A disc and a ring both have same mass and radius. The ratio of moment of inertia of the disc about its diameter to that of a ring about a tangent in its plane is

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