Question

A uniform metal disc of diameter 24 cm is taken out of it and a disc of diameter 8 cm is cut off from the right-side end. The center of mass of the remaining part will be at:
a) left side, 2 cm from the center
b) right side, 1 cm from the center
c) right side, 2 cm from the center
d) left side, 1 cm from the center

Answer 1

Hint: Let us assume that the surface density of the material of disc is $\rho $ . Calculate the mass of the complete disc as well as the remaining disc by using the formula: $\rho =\dfrac{m}{V}$ . Since the center of mass of the remaining piece and that of the removed portion combined must lie at the center of disc (0,0). So, by using the formula for center of mass: ${{x}_{cm}}=\dfrac{{{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}}{{{m}_{1}}+{{m}_{2}}}$ , find the center of mass of remaining part.

Formula used:
${{x}_{cm}}=\dfrac{{{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}}{{{m}_{1}}+{{m}_{2}}}$, where ${{x}_{cm}}$ is the center of mass of the whole body, ${{x}_{1}}$ and ${{x}_{2}}$ are the center of mass of the constituent parts, ${{m}_{1}}$ and ${{m}_{2}}$ are the masses of constituent parts.

Complete step by step answer:
We have a disc of diameter 24 cm and density $\rho $.
So, the volume of disc is:
 $\begin{align}
  & V=\pi {{r}^{2}} \\
 & =\pi {{\left( \dfrac{24}{2} \right)}^{2}} \\
 & =144\pi
\end{align}$
As we know that: $\rho =\dfrac{m}{V}$
$\Rightarrow m=\rho V$
So, the mass of complete disc is: $m=144\pi \rho $
A disc of diameter 8 cm is cut off from the right-side end.
So, the volume of disc is:
 $\begin{align}
  & V=\pi {{r}^{2}} \\
 & =\pi {{\left( \dfrac{8}{2} \right)}^{2}} \\
 & =16\pi
\end{align}$
As we know that: $m=\rho V$
So, the mass of disc removed is: ${{m}_{1}}=16\pi \rho $
Therefore, mass of the remaining disc is:
 $\begin{align}
  & {{m}_{2}}=\left( 144-16 \right)\pi \rho \\
 & =128\pi \rho
\end{align}$
As we know that, the center of mass of a disc lies at its center. So, the center of mass of complete disc lies at (0,0). And the center of mass of the disc removed lies at the right side, 8 cm from the center of complete disc, i.e. (8,0). Let us assume that the center of mass of the remaining disc lies at x cm from the center of complete disc.
So, by applying the formula: ${{x}_{cm}}=\dfrac{{{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}}{{{m}_{1}}+{{m}_{2}}}$, we get:
$\begin{align}
  & \Rightarrow 0=\dfrac{\left( 128\pi \rho \times x \right)+\left( 16\pi \rho \times 8 \right)}{128\pi \rho +16\pi \rho } \\
 & \Rightarrow 0=\dfrac{\pi \rho \left( 128x+128 \right)}{144\pi \rho } \\
 & \Rightarrow 0=x+1 \\
 & \Rightarrow x=-1cm \\
\end{align}$
Hence, the center of mass of the remaining disc is 1 cm to the left of center.

So, the correct answer is “Option D”.

Note:
Remember the definition and formula to calculate centre of mass of simple geometries. Here in this question the metal disc is circular in shape. Therefore, the formula of the centre of mass of circular shape must be applied. The centre of mass of both the discs will lie on the same point. And as we know that the centre of mass of a body is the average position of all the parts of the system, weighted according to their masses.