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A square of side 4 units is divided into four equal squares. the square portion ABCH is removed and placed over HCDE. What is the new centre of mass of the system
Dear Sai,
First we can consider the square slab is placed such that its centre of mass is at C(0,0)
Now we can consider this centre of mass to be contributed by the square slab ABCH and the remaining part HCBGIFDE
and let the position of centre of mass of this part is (x1, y1) and the removed square part will be (1,1)
From these we can find the position of th Centre of mass of the part remaining after removing the square ABCH
xcm = ( m 1 x 1+ m 2 x 2 )/m 1 + m 2
ycm = ( m 1 y 1 + m 2 y2 )/ m 1+ m 2
Where m1 = mass of HCBGIFDE = (Area of it) * thicknesss * density of the slab = A1 * t * d
where A1 = 3* area of each square = 12
And similarly m2 = 4 * t * d
And ( x2, y2 ) = ( 1, 1 )
and (xcm ,ycm) = ( 0,0 )
thus we would get ( x1, y1 ) = ( -1/3, -1/3 )
Now if the removed plate is kept on the Plate HCDE
then its centre of mass becomes ( x2, y2 ) = ( -1, 1 )
Now we again calculate the centre of mass of the system bye using the formula
ycm = ( m 1 y 1 + m 2 y2 )/ m 1 + m 2
we get the centre of mass of the system to be ( -1/2, 0 )
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Regards,
Askiitians Experts
Adapa Bharath
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