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. 5/6, 5/6
• B. 5/3, 5/3
• C. 3/6, 3/3
• D. 5/3, 5/6

Hint:

Treat the cut piece as a negative mass!

Answer 1

The Correct Option is B

Step 1: Picture the plate.
The copper plate is a square of side 4 m. Put its corners at $(0,0)$ and $(4,4)$. Since it is uniform, its centre of mass is at the middle, $(2,2)$.
Step 2: The piece that is cut.
A square of side 2 m is removed from the top right corner. It covers $x$ from 2 to 4 and $y$ from 2 to 4, so its own centre is at $(3,3)$.
Step 3: Treat the cut as negative mass.
Removing a part is the same as adding negative mass. For a uniform sheet mass goes with area, so use \[ x_{cm} = \frac{A_1 x_1 - A_2 x_2}{A_1 - A_2}, \] with big area $A_1 = 16$ and cut area $A_2 = 4$.
Step 4: Put in the numbers.
$x_{cm} = \dfrac{16(2) - 4(3)}{16 - 4} = \dfrac{32 - 12}{12} = \dfrac{20}{12} = \dfrac{5}{3}$.
Step 5: Use symmetry for y.
The plate and the cut are mirror images about the line $y = x$, so $y_{cm}$ has to equal $x_{cm}$. That gives $y_{cm} = \dfrac{5}{3}$. This is why any choice with x and y different from each other cannot be right.
Step 6: Conclusion.
The new centre of mass is $\left(\dfrac{5}{3},\ \dfrac{5}{3}\right)$. \[ \boxed{\left(\tfrac{5}{3},\ \tfrac{5}{3}\right)} \]