Question:
A circular portion of radius \(R_2\) has been removed from one edge of a circular disc of radius \(R_1\). The correct expression for the centre of mass for the remaining portion of the disc is
A circular portion of radius \(R_2\) has been removed from one edge of a circular disc of radius \(R_1\). The correct expression for the centre of mass for the remaining portion of the disc is
Show Hint
For bodies with a portion removed, use the negative mass method. Treat the removed part as having negative mass and apply the centre of mass formula:
\[
x_{\text{cm}}=\frac{\sum m_ix_i}{\sum m_i}.
\]
Updated On: Jun 18, 2026
- \(-\dfrac{R_2^2}{R_1+R_2}\)
- \(-\dfrac{R_2^2}{R_1-R_2}\)
- \(\dfrac{R_2^2}{R_1+R_2}\)
- \(-\dfrac{R_1^2}{R_1+R_2}\)
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The Correct Option is A
Solution and Explanation
Step 1: Choose the origin and axis.
Let the centre of the original circular disc of radius \(R_1\) be the origin.
The smaller circular portion of radius \(R_2\) is removed from one edge of the disc.
Since the removed circle touches the outer edge internally, the distance between the centres is \[ R_1-R_2 \] Assume the removed part is on the positive \(x\)-axis.
So, the centre of the removed portion is at \[ x=R_1-R_2 \]
Step 2: Use the method of negative mass.
The remaining body can be treated as:
Original full disc \(+\) removed circular portion with negative mass.
For a uniform disc, mass is proportional to area. Hence, \[ M_1\propto \pi R_1^2 \] and \[ M_2\propto \pi R_2^2 \] So, we may take \[ M_1=R_1^2,\qquad M_2=R_2^2 \]
Step 3: Write centre of mass formula.
The centre of mass of the remaining portion is \[ x_{\text{cm}} = \frac{M_1(0)-M_2(R_1-R_2)}{M_1-M_2} \] Substituting \[ M_1=R_1^2,\qquad M_2=R_2^2, \] we get \[ x_{\text{cm}} = \frac{0-R_2^2(R_1-R_2)}{R_1^2-R_2^2} \] \[ x_{\text{cm}} = -\frac{R_2^2(R_1-R_2)}{(R_1-R_2)(R_1+R_2)} \] Cancelling \((R_1-R_2)\), \[ x_{\text{cm}} = -\frac{R_2^2}{R_1+R_2} \]
Step 4: Final conclusion.
Therefore, the centre of mass of the remaining portion is \[ \boxed{-\frac{R_2^2}{R_1+R_2}} \]
Let the centre of the original circular disc of radius \(R_1\) be the origin.
The smaller circular portion of radius \(R_2\) is removed from one edge of the disc.
Since the removed circle touches the outer edge internally, the distance between the centres is \[ R_1-R_2 \] Assume the removed part is on the positive \(x\)-axis.
So, the centre of the removed portion is at \[ x=R_1-R_2 \]
Step 2: Use the method of negative mass.
The remaining body can be treated as:
Original full disc \(+\) removed circular portion with negative mass.
For a uniform disc, mass is proportional to area. Hence, \[ M_1\propto \pi R_1^2 \] and \[ M_2\propto \pi R_2^2 \] So, we may take \[ M_1=R_1^2,\qquad M_2=R_2^2 \]
Step 3: Write centre of mass formula.
The centre of mass of the remaining portion is \[ x_{\text{cm}} = \frac{M_1(0)-M_2(R_1-R_2)}{M_1-M_2} \] Substituting \[ M_1=R_1^2,\qquad M_2=R_2^2, \] we get \[ x_{\text{cm}} = \frac{0-R_2^2(R_1-R_2)}{R_1^2-R_2^2} \] \[ x_{\text{cm}} = -\frac{R_2^2(R_1-R_2)}{(R_1-R_2)(R_1+R_2)} \] Cancelling \((R_1-R_2)\), \[ x_{\text{cm}} = -\frac{R_2^2}{R_1+R_2} \]
Step 4: Final conclusion.
Therefore, the centre of mass of the remaining portion is \[ \boxed{-\frac{R_2^2}{R_1+R_2}} \]
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