Question 331: Centre of Mass and Rotational Motion

Question

Moment of inertia of an equilateral triangular lamina ABC, about the axis
passing through its centre O and perpendicular to its plane is I
o
as shown in the figure. A cavity DEF is cut out from the lamina, where D, E, F are the mid points of the sides. Moment of inertia of the remaining part of lamina about the same axis is :
![](https://app-content.cdn.examgoal.net/fly/@width/image/1l83nwl5t/d66d76f8-ddeb-44da-9693-d6b03b3a41df/2f25a910-354b-11ed-8c5c-f3f0850e6b94/file-1l83nwl5u.png?format=png)

Accepted Answer

Let, side of triangle ABC = $\ell $
According to perpendicular axes theorem, moment of inertia of triangle about it center and perpendicular to its plane,
I
O
= ${1 \over {12}}$ m$\ell $
2
In, triangle DEF,
DE = DF = EF = ${1 \over 2}$ AB = ${1 \over 2}$ $\ell $
$	herefore\,\,\,$ moment of inertia of triangle DEF,
I
DEF
= ${1 \over {12}} 	imes {m \over 4} 	imes {\left( {{\ell  \over 2}} ight)^2}$
= ${1 \over {12}} 	imes {{m{\ell ^2}} \over {16}}$
= ${{{I_O}} \over {16}}$
$	herefore\,\,\,$ Moment of inertia of the remaining part,
I
remain
= I
O
$-$ ${{{{m I}_O}} \over {16}}$ = ${{15\,{{m I}_O}} \over {16}}$

Answer

${7 \over 8}$ I
o

Answer

${15 \over 16}$ I
o

Answer

${{3\,{{\rm I}_o}} \over 4}$

Answer

${{31\,{{\rm I}_o}} \over 32}$

Centre of Mass and Rotational Motion

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