Question

A circular plate of radius \( r \) is removed from a uniform circular plate P of radius \( 4r \) to form a hole. If the distance between the centre of the hole formed and the centre of the plate P is \( 2r \), then the distance of the centre of mass of the remaining portion from the centre of the plate P is?

• A. \( \frac{r}{3} \)
• B. \( \frac{r}{15} \)
• C. \( \frac{2r}{15} \)
• D. \( 2r \)

Hint:

To find the center of mass of a system with a removed section, treat the missing part as negative mass and use the weighted average formula.

Answer 1

The Correct Option is C

We apply the concept of the center of mass for a system with a removed section. 

Step 1: Define the mass distribution 
Let the mass per unit area of the uniform plate be \( \sigma \). Then, the mass of the original plate of radius \( 4r \) is: \[ M = \sigma \pi (4r)^2 = 16 \pi \sigma r^2 \] The mass of the removed portion (hole) of radius \( r \) is: \[ m = \sigma \pi r^2 \] 

Step 2: Use the concept of the center of mass shift 
The center of mass of the remaining portion is found using the formula: \[ X_{\text{cm}} = \frac{M X_M - m X_m}{M - m} \] Since the center of mass of the original plate is at the origin (\(X_M = 0\)), and the center of the removed portion is at \(X_m = 2r\): \[ X_{\text{cm}} = \frac{0 \times 16 \pi \sigma r^2 - (\sigma \pi r^2 \times 2r)}{16 \pi \sigma r^2 - \sigma \pi r^2} \] \[ = \frac{-2 \sigma \pi r^3}{15 \sigma \pi r^2} \] \[ = -\frac{2r}{15} \] Thus, the magnitude of the center of mass shift is: \[ \frac{2r}{15} \]