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

x_cm = ( m ₁ x ₁+ m ₂ x ₂ )/m ₁ + m ₂

y_cm = ( 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 (x_{cm ,}y_cm) = ( 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

x_cm = ( m ₁ x ₁+ m ₂ x ₂ )/m ₁ + m ₂

y_cm = ( 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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