Question

From a circular disc of radius \(2R\), a smaller circular disc is cut with radius of the larger disc as its diameter. The centre of the hole is at a distance of \(R\) from the centre of the original disc. The distance of the centre of mass of the remaining portion from the centre is:

• A. \( \frac{R}{3} \)
• B. \( \frac{R}{4} \)
• C. \( \frac{R}{2} \)
• D. \( R \)

Hint:

For removed parts, treat them as negative mass and apply centre of mass formula.

Answer 1

The Correct Option is A

Step 1: Understand geometry.
Original disc radius = \(2R\). Small disc radius = \(R\) (since diameter = \(2R\)).

Step 2: Mass proportional to area.
Mass of disc ∝ area:
\[ M \propto (2R)^2 = 4R^2 \] \[ m \propto R^2 \]
So ratio:
\[ M : m = 4 : 1 \]

Step 3: Treat removed disc as negative mass.
Centre of big disc at origin. Removed disc at distance \(R\).

Step 4: Centre of mass formula.
\[ x_{cm} = \frac{0 \cdot M - R \cdot m}{M - m} \]

Step 5: Substitute values.
\[ x_{cm} = \frac{-R \cdot 1}{4 - 1} \]
\[ x_{cm} = -\frac{R}{3} \]

Step 6: Magnitude of shift.
Distance from centre is:
\[ \frac{R}{3} \]

Step 7: Final conclusion.
\[ \boxed{\frac{R}{3}} \] Hence, correct answer is option (A).