Question

From a uniform disk of radius $\text{R}$, a circular hole of radius ${\displaystyle \frac{\text{R}}{2}}$ is cut out. The centre of the hole is at ${\displaystyle \frac{\text{R}}{2}}$ from the centre of the original disc. Locate the centre of gravity of the resulting flat body.

Answer 1

Hint: Assume that the mass per unit area of the uniform disc is $\sigma$ therefore the mass of the uniform disc is $\text{M = }\sigma \pi {\text{R}}^2$ and the mass of the small disc is $\sigma \pi {\left({\displaystyle \frac{\text{R}}{2}}\right)}^2={\displaystyle \frac{\sigma \pi {\text{R}}^2}{4}}={\displaystyle \frac{\text{M}}{4}}$
Formula used:
$\text{x = }{\displaystyle \frac{\left({\text{m}}_1{\text{r}}_1\text{ + }{\text{m}}_2{\text{r}}_2\right)}{\left({\text{m}}_1\text{ + }{\text{m}}_2\right)}}$
where $\text{x}$ is the distance through which the centre of gravity of the remaining portion shifts
and ${\text{m}}_1,{\text{r}}_1$ are the mass & radius of uniform disc and ${\text{m}}_2,{\text{r}}_2$ are the mass & radius of the small disc that has been cut.

Complete step-by-step solution -

Given that,
Radius of uniform disc $=\text{R}$
Radius of the smaller disc $={\displaystyle \frac{\text{R}}{2}}$
Let the mass per unit area of the original disc $=\sigma$
Therefore mass of the uniform disc $=\text{M = }\sigma \pi {\text{R}}^2$
And the mass of the small disc $\text{ = }\sigma \pi {\left({\displaystyle \frac{\text{R}}{2}}\right)}^2={\displaystyle \frac{\sigma \pi {\text{R}}^2}{4}}={\displaystyle \frac{\text{M}}{4}}$
Now as the small disc has been cut from the uniform disc, the remaining portion is considered to be a system of two masses.
The two masses are: $\text{M}$ (concentrated at O) & $-\text{M}$(concentrated at O')
(negative sign indicating above that the portion is removed from the uniform disc)
Let $\text{x}$ be the distance through which the centre of mass of the remaining portion shifts from point O.
The relation between the centre of masses of two masses is give as:
$$\begin{gathered} \text{x = }{\displaystyle \frac{\left({\text{m}}_1{\text{r}}_1\text{ + }{\text{m}}_2{\text{r}}_2\right)}{\left({\text{m}}_1\text{ + }{\text{m}}_2\right)}} \\ \text{x = }{\displaystyle \frac{\left[\left(\text{M}\times \text{0 - }{\displaystyle \frac{\text{M}}{4}}\right)\times \left({\displaystyle \frac{\text{R}}{2}}\right)\right]}{\left(\text{M - }{\displaystyle \frac{\text{M}}{4}}\right)}} \\ ={\displaystyle \frac{\left({\displaystyle \frac{-\text{MR}}{8}}\right)}{\left({\displaystyle \frac{3\text{M}}{4}}\right)}} \\ ={\displaystyle \frac{-4\text{R}}{24}} \\ \text{x = }{\displaystyle \frac{-\text{R}}{6}} \end{gathered}$$.

Note: The relation between the centre of masses of two masses is calculated by the formula $\text{x = }{\displaystyle \frac{\left({\text{m}}_1{\text{r}}_1\text{ + }{\text{m}}_2{\text{r}}_2\right)}{\left({\text{m}}_1\text{ + }{\text{m}}_2\right)}}$ which is found to be ${\displaystyle \frac{-\text{R}}{6}}$ here the negative sign indicates that the centre of gravity of the resulting flat body gets shifted towards the left point O.