Question 215: 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: