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Step-by-Step Solution
Step 1: Understand the Problem
We have a larger disc of radius $2R$. From this disc, a smaller disc of radius $R$ is removed. The smaller disc is removed in such a way that its circumference lies along the circumference of the bigger disc. We need to find the distance of the center of mass of the remaining portion from the center of the bigger disc, expressed as $\alpha R$.
Step 2: Assign Coordinate System and Notation
Place the center of the bigger disc at the origin (O). Since the two discs are tangential at the outer edge, the smaller disc’s center (call it C) will lie at a distance $R$ from O along, say, the positive x-axis. Hence, if O is at $(0,0)$, center C of the smaller disc is at $(R, 0)$.
Step 3: Calculate Masses of the Discs
Let the surface mass density of both discs be $\sigma$ (mass per unit area).
Mass of bigger disc ($M_1$):
$$
M_1 = \sigma \times \text{Area} = \sigma \times \pi\,(2R)^2 = 4\pi\,R^2\,\sigma.
$$
Mass of smaller disc ($M_2$):
$$
M_2 = \sigma \times \pi\,R^2 = \pi\,R^2\,\sigma.
$$
Step 4: Use the Concept of Negative Mass for Removal
The remaining shape can be viewed as the bigger disc of mass $M_1$ plus a “negative mass” of $-M_2$ at the position $(R,0)$. This is a common method to find the center of mass of a body from which a portion has been removed.
Step 5: Write Down the Formula for the Net Center of Mass
Let $\vec{R}_{\text{CM}}$ be the position vector of the center of mass of the remaining portion. Using the principle of superposition (treating the removed part as negative mass):
$$
\vec{R}_{\text{CM}}
= \frac{M_1 \,\vec{r}_1 + (-M_2)\,\vec{r}_2}{M_1 + (-M_2)}.
$$
Here:
$\vec{r}_1 = \vec{0}$ (the center of the bigger disc is at the origin),
$\vec{r}_2 = (R,\,0)$ (the center of the smaller disc),
$M_1 = 4\pi R^2 \sigma$ and $M_2 = \pi R^2 \sigma$.
Step 6: Substitute the Values
$$
\vec{R}_{\text{CM}}
= \frac{(4\pi R^2 \sigma)\vec{0} \;+\; (-\pi R^2 \sigma)(R,\,0)}{4\pi R^2 \sigma - \pi R^2 \sigma}
= \frac{-\pi R^2 \sigma \,(R,0)}{3\pi R^2 \sigma}.
$$
Canceling the common factors:
$$
\vec{R}_{\text{CM}}
= \left(\frac{-\pi R^2 \sigma R}{3 \pi R^2 \sigma},\,0\right)
= \left(\frac{-R}{3},\,0\right).
$$
Step 7: Determine the Magnitude and Direction
The negative sign indicates that the center of mass is located along the negative x-axis relative to the center of the larger disc if we assumed the smaller disc’s center at $(+R, 0)$. However, physically, we usually measure the distance from O in the direction from O to the center of the void. The magnitude of this distance is $R/3$.
Therefore, $\alpha = \frac{1}{3}$.
Final Answer
The distance of the center of mass of the remaining disc from the center of the bigger disc is $\frac{1}{3}R$. Hence, $\alpha = \frac{1}{3}$.
Diagram
