Question 332: 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 :

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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