Question:
A thin square shaped copper plate of uniform mass distribution of side 4 m has its centre of mass at (2, 2). If at its top right corner, a square of 2 m side is cut from the plate, the centre of mass of the remaining plate is:
A thin square shaped copper plate of uniform mass distribution of side 4 m has its centre of mass at (2, 2). If at its top right corner, a square of 2 m side is cut from the plate, the centre of mass of the remaining plate is:
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Treat the cut piece as a negative mass!
Updated On: Jun 6, 2026
- 5/6, 5/6
- 5/3, 5/3
- 3/6, 3/3
- 5/3, 5/6
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The Correct Option is B
Solution and Explanation
Step 1: Set up.
Full square side 4 m, centre at $(2,2)$, area $A_1 = 16$. The removed square has side 2 m at the top right corner, centre at $(3,3)$, area $A_2 = 4$.
Step 2: Negative mass formula.
$x_{cm} = \dfrac{A_1 x_1 - A_2 x_2}{A_1 - A_2} = \dfrac{16(2) - 4(3)}{12} = \dfrac{20}{12} = \dfrac{5}{3}$.
Step 3: y by symmetry.
The shape is symmetric about $y = x$, so $y_{cm} = \dfrac{5}{3}$.
Step 4: Result.
Centre of mass $= \left(\dfrac{5}{3},\ \dfrac{5}{3}\right)$. \[ \boxed{\left(\tfrac{5}{3},\ \tfrac{5}{3}\right)} \]
Full square side 4 m, centre at $(2,2)$, area $A_1 = 16$. The removed square has side 2 m at the top right corner, centre at $(3,3)$, area $A_2 = 4$.
Step 2: Negative mass formula.
$x_{cm} = \dfrac{A_1 x_1 - A_2 x_2}{A_1 - A_2} = \dfrac{16(2) - 4(3)}{12} = \dfrac{20}{12} = \dfrac{5}{3}$.
Step 3: y by symmetry.
The shape is symmetric about $y = x$, so $y_{cm} = \dfrac{5}{3}$.
Step 4: Result.
Centre of mass $= \left(\dfrac{5}{3},\ \dfrac{5}{3}\right)$. \[ \boxed{\left(\tfrac{5}{3},\ \tfrac{5}{3}\right)} \]
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