Question

Figure shows a uniform disc of radius $R$, from which a hole of radius $R/2$ has been cut out from left of the centre and is placed on the right of the centre of the disc. Find the COM of the resulting disc.

• A. $(\frac{R}{2},0)$
  
• B. $(\frac{R}{4},0)$
• C. $(\frac{R}{2},\frac{R}{2})$
• D. $(\frac{R}{4},\frac{R}{4})$

Answer 1

The correct option is B $(\frac{R}{4},0)$

Since the mass is distributed uniformally about x-axis, it can be said that y coordinate of centre of mass of the resulting disc will lie at the origin.
$\Rightarrow y_{com}=0$
Now to find the x-coordinate of centre of mass we need to consider each part of the resulting disc separately.

Thus taking the disc of diameter R and mass $m_1$ separately. It is evident that centre of mass of this disc will lie at its centre which is at a distance $R/2$ from the origin.
Now the centre of mass of the original disc of mass M without cutting the disc was at origin i.e (0,0) and since the mass was equally distributed, the cut out disc which is 1/4 th part of original disc (in terms of area) will have mass($m_1$) as $M/4$.

thus for original disc,
$\begin{array}{}0=\frac{m_1(-\frac{R}{2})+m_2\left(x\right)}{M}& \text{(1)}\end{array}$
Here $m_1=M/4$ and $m_2=M-\frac{M}{4}=\frac{3M}{4}$
Substituting these in $\left(1\right)$, we get
$x=R/6$
This is the centre of mass of the left out part of mass $m_2$.

Now the x-coordinate of centre of mass of resulting disc ($x_{com}$) will be,
$x_{com}=\frac{m_2\left(\frac{R}{6}\right)+m_1\left(\frac{R}{2}\right)}{M}$
$x_{com}=\frac{\frac{3M}{4}\left(\frac{R}{6}\right)+\frac{M}{4}\left(\frac{R}{2}\right)}{M}$
$\Rightarrow x_{com}=\frac{R}{4}$