Question

A circular disc of radius 20 cm is cut from one edge of a larger circular disc of radius 50 cm. The shift of centre of mass is

Answer 1

$\frac{40}{7}$ cm

Answer 2

Let the large disc have radius $R_L = 50$ cm and the small disc have radius $R_S = 20$ cm. Assume the center of the large disc is at the origin $(0,0)$. The center of the small disc, when cut from the edge, is at a distance $R_L - R_S$ from the origin. Let this be along the x-axis, so the center of the small disc is at $(30, 0)$.

Using the principle of superposition for the center of mass: $M_L \vec{r}_L = M_{rem} \vec{r}_{rem} + M_S \vec{r}_S$

Here, $M_L$ is the mass of the large disc, $\vec{r}_L = (0,0)$ is its center of mass. $M_S$ is the mass of the small disc, $\vec{r}_S = (30, 0)$ is its center of mass. $M_{rem} = M_L - M_S$ is the mass of the remaining part, and $\vec{r}_{rem}$ is its center of mass.

Since $\vec{r}_L = (0,0)$, we have: $0 = M_{rem} \vec{r}_{rem} + M_S \vec{r}_S$ $\vec{r}_{rem} = -\frac{M_S}{M_{rem}} \vec{r}_S$

For uniform discs, mass is proportional to area ($M \propto R^2$). $\vec{r}_{rem} = -\frac{R_S^2}{R_L^2 - R_S^2} \vec{r}_S$ $\vec{r}_{rem} = -\frac{20^2}{50^2 - 20^2} (30, 0)$ $\vec{r}_{rem} = -\frac{400}{2500 - 400} (30, 0)$ $\vec{r}_{rem} = -\frac{400}{2100} (30, 0)$ $\vec{r}_{rem} = -\frac{4}{21} (30, 0) = (-\frac{40}{7}, 0)$

The shift of the center of mass is the magnitude of $\vec{r}_{rem}$, which is $\frac{40}{7}$ cm.