JEE MAIN - Physics (2019 - 8th April Evening Slot - No. 17)
A uniform rectangular thin sheet ABCD of
mass M has length a and breadth b, as shown
in the figure. If the shaded portion HBGO is
cut-off, the coordinates of the centre of mass
of the remaining portion will be :-
_8th_April_Evening_Slot_en_17_1.png)
$$\left( {{{5a} \over 3},{{5b} \over 3}} \right)$$
$$\left( {{{2a} \over 3},{{2b} \over 3}} \right)$$
$$\left( {{{5a} \over 12},{{5b} \over 12}} \right)$$
$$\left( {{{3a} \over 4},{{3b} \over 4}} \right)$$
Explanation
X- coordinate of CM of remaining sheet
$${X_{cm}} = {{MX - mx} \over {M - m}}$$
= $${{\left( {4m} \right) \times \left( {{a \over 2}} \right) - M\left( {{{3a} \over 4}} \right)} \over {4m - m}} = {{5a} \over {12}}$$
Similarly ycm = $${{5b} \over {12}}$$
$$ \therefore $$ $$CM\left( {{{5a} \over {12}},{{5b} \over {12}}} \right)$$
$${X_{cm}} = {{MX - mx} \over {M - m}}$$
= $${{\left( {4m} \right) \times \left( {{a \over 2}} \right) - M\left( {{{3a} \over 4}} \right)} \over {4m - m}} = {{5a} \over {12}}$$
Similarly ycm = $${{5b} \over {12}}$$
$$ \therefore $$ $$CM\left( {{{5a} \over {12}},{{5b} \over {12}}} \right)$$
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