A uniform rectangular thin sheet ABCD of mass | vedclass

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Question

(A) Let the total mass of the rectangular sheet be $M$. The centre of mass of the full sheet is at $\left( \frac{a}{2}, \frac{b}{2} \right)$.The shade

Vedclass · Accepted Answer

$\left( \frac{5a}{12}, \frac{5b}{12} \right)$

Vedclass · Answer

(A) Let the total mass of the rectangular sheet be $M$. The centre of mass of the full sheet is at $\left( \frac{a}{2}, \frac{b}{2} ight)$.The shaded portion $HBGO$ is a rectangle with length $\frac{a}{2}$ and breadth $\frac{b}{2}$. Its mass is $m = M 	imes \frac{(\frac{a}{2} 	imes \frac{b}{2})}{(a 	imes b)} = \frac{M}{4}$.The centre of mass of the shaded portion is at $\left( \frac{a}{2} + \frac{a}{4}, \frac{b}{2} + \frac{b}{4} ight) = \left( \frac{3a}{4}, \frac{3b}{4} ight)$.The remaining mass is $M' = M - \frac{M}{4} = \frac{3M}{4}$.The $x$-coordinate of the centre of mass of the remaining portion is:$x_{cm} = \frac{M(\frac{a}{2}) - \frac{M}{4}(\frac{3a}{4})}{M - \frac{M}{4}} = \frac{\frac{a}{2} - \frac{3a}{16}}{\frac{3}{4}} = \frac{\frac{8a-3a}{16}}{\frac{3}{4}} = \frac{5a}{16} 	imes \frac{4}{3} = \frac{5a}{12}$.Similarly,the $y$-coordinate is:$y_{cm} = \frac{M(\frac{b}{2}) - \frac{M}{4}(\frac{3b}{4})}{M - \frac{M}{4}} = \frac{5b}{12}$.Thus,the coordinates are $\left( \frac{5a}{12}, \frac{5b}{12} ight)$.

$A$ 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

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