Question 220: Centre of Mass and Rotational Motion

Question

For a uniform rectangular sheet shown in the figure, the ratio of moments of inertia about the axes
perpendicular to the sheet and passing through O (the centre of mass) and O' (corner point) is :

Accepted Answer

$${I_O} = {M \over {12}}\left( {{a^2} + {b^2}} ight)$$ = $${M \over {12}}\left[ {{{80}^2} + {{60}^2}} ight]$$
I
O'
= I
O
+ Md
2
{parallel axis theorem}
= $${M \over {12}}\left[ {{{80}^2} + {{60}^2}} ight]$$ + M[50]
2
$$ 	herefore $$ $${{{I_O}} \over {{I_{O'}}}}$$ = $${{{M \over {12}}\left[ {{{80}^2} + {{60}^2}} ight]} \over {{M \over {12}}\left[ {{{80}^2} + {{60}^2}} ight] + M{{\left[ {50} ight]}^2}}}$$ = $${1 \over 4}$$

Answer

$${1 \over 2}$$

Answer

$${1 \over 4}$$

Answer

$${1 \over 8}$$

Answer

$${2 \over 3}$$

Centre of Mass and Rotational Motion

For a uniform rectangular sheet shown in the figure, the ratio of moments of inertia about the axes perpendicular to the sheet and passing through O (the centre of mass) and O' (corner point) is :

Choose the correct answer: