Question 377: Centre of Mass and Rotational Motion

Question

A circular disc $$X$$ of radius $$R$$ is made from an  iron plate of thickness $$t,$$ and another disc $$Y$$ of radius $$4$$ $$R$$ is made from an iron plate of thickness $${t \over 4}.$$ Then the relation between the moment of inertia $${I_X}$$  and  $${I_Y}$$ is

Accepted Answer

We know that density $$\left( d ight) = {{mass\left( M ight)} \over {volume\left( V ight)}}$$
$$	herefore$$ Mass of disc $$M = d 	imes V = d 	imes \left( {\pi {R^2} 	imes t} ight).$$
The moment of inertia of any disc is $$I = {1 \over 2}M{R^2}$$
$$	herefore$$ $$I = {1 \over 2}\left( {d 	imes \pi {R^2} 	imes t} ight){R^2} = {{\pi d} \over 2}t 	imes {R^4}$$
$$	herefore$$ $${{{I_X}} \over {{I_Y}}} = {{{t_X}R_X^4} \over {{t_Y}R_Y^4}}$$
$$ = {{t 	imes {R^4}} \over {{t \over 4} 	imes {{\left( {4R} ight)}^4}}}$$
$$ = {1 \over {64}}$$

Answer

$${I_Y} = 32{I_X}$$

Answer

$${I_Y} = 16{I_X}$$

Answer

$${I_Y} = {I_X}$$

Answer

$${I_Y} = 64{I_X}$$

Centre of Mass and Rotational Motion

A circular disc $X of radius R is made from an iron plate of thickness t, and another disc Y of radius 4 R is made from an iron plate of thickness {t \over 4}. Then the relation between the moment of inertia {I_X} and {I_Y}$ is

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