Question 380: Centre of Mass and Rotational Motion

Question

Moment of inertia of a circular wire of mass $$M$$ and radius $$R$$ about its diameter is

Accepted Answer

Moment of Inertia of a circular wire about an axis $$nn'$$ passing through the centre of the circle and perpendicular to the plane of the circle $$ = M{R^2}$$
As shown in the figure, $$X$$-axis and $$Y$$-axis lies in the plane of the ring. Then by perpendicular axis theorem
$${I_X} + {I_Y} = {I_Z}$$
$$ \Rightarrow 2{I_X} = M{R^2}\,$$  $$\left[ \, ight.$$ as $${I_X} - {I_Y}$$ (by symmetry) and $${I_Z} = M{R^2}$$ $$\left. \, ight]$$
$$	herefore$$ $${I_X} = {1 \over 2}M{R^2}$$

Answer

$${{M{R^2}} \over 2}$$

Answer

$$M{R^2}$$

Answer

$$2M{R^2}$$

Answer

$${{M{R^2}} \over 4}$$

Centre of Mass and Rotational Motion

Moment of inertia of a circular wire of mass $M and radius R$ about its diameter is

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