JEE MAIN - Physics (2019 - 11th January Evening Slot - No. 6)
A circular disc D1 of mass M and radius R has two identical discs D2 and D3 of the same mass M and radius R attached rigidly at its opposite ends (see figure). The moment of inertia of the system about the axis OO' ,passing through the centre of D1 as shown in the figure, will be :
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3MR2
MR2
$${2 \over 3}$$ MR2
$${4 \over 5}$$ MR2
Explanation
I $$=$$ $${{M{R^2}} \over 2} + 2\left( {{{M{R^2}} \over 4} + M{R^2}} \right)$$
$$ = {{M{R^2}} \over 2} + {{M{R^2}} \over 2} + 2M{R^2}$$
$$=$$ $$3M{R^2}$$
$$ = {{M{R^2}} \over 2} + {{M{R^2}} \over 2} + 2M{R^2}$$
$$=$$ $$3M{R^2}$$
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