Question

A circuit plate of diameter \[a\] is kept in contact with a square plate of edge \[a\] as shown. The density of the material and the thickness are the same everywhere. The centre of mass of the composite system will be:

A. Inside the circular plate.
B. Inside the square plate.
C. At the point of contact.
D. Outside the system.

Answer 1

Hint: We will redraw the diagram and introduce both x axis and y axis into it. We will need to use the expression given for finding the center of mass to find the coordinates of the centre of mass of the combined system. Also we will use the relation between density, mass and volume to find the mass of each body.

Formula used:
\[\begin{align}
  & x=\dfrac{{{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}}{{{m}_{1}}+{{m}_{2}}} \\
 & y=\dfrac{{{m}_{1}}{{y}_{1}}+{{m}_{2}}{{y}_{2}}}{{{m}_{1}}+{{m}_{2}}} \\
 & \rho =\dfrac{m}{V} \\
\end{align}\]

Complete step by step answer:
Firstly we will redraw the given diagram by introducing coordinate axes into it. We will take the centre of the circular plate as its origin. We will be taking the centre of mass of the circular plate as \[{{m}_{1}}\] and the centre of mass of the square plate as \[{{m}_{2}}\].

Now, we will find \[{{m}_{1}}\] and \[{{m}_{2}}\] by using the equation given for density. We will take density as \[\rho \] and it is equal for both bodies.
Density of the circular disc will be, \[{{\rho }_{c}}=\dfrac{{{m}_{1}}}{V}=\dfrac{{{m}_{1}}}{\pi {{r}^{2}}t}\]. Then, mass of the disc will be,
\[{{m}_{1}}=\rho \pi {{\left( \dfrac{a}{2} \right)}^{2}}t=\dfrac{\rho \pi {{a}^{2}}t}{4}\]
Here, \[\rho \]is the density, \[a\] is the diameter and \[t\] is the thickness.
Now, the density of the square plate will be, \[{{\rho }_{s}}=\dfrac{{{m}_{2}}}{V}=\dfrac{{{m}_{2}}}{{{a}^{2}}t}\]. Then, the mass will be,
 \[{{m}_{2}}=\rho {{a}^{2}}t\]
Here, \[\rho \]is the density, \[a\] is the side length and \[t\] is the thickness.
Now, to find the coordinates of the centre of mass, we will use the expression,
  \[\begin{align}
  & x=\dfrac{{{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}}{{{m}_{1}}+{{m}_{2}}} \\
 & y=\dfrac{{{m}_{1}}{{y}_{1}}+{{m}_{2}}{{y}_{2}}}{{{m}_{1}}+{{m}_{2}}} \\
\end{align}\]
But here, the mass of the disc is at origin and the mass of the plate is at \[a\] on the x-axis.
\[\begin{align}
  & \Rightarrow x=\dfrac{\left( \dfrac{\rho \pi {{a}^{2}}t}{4} \right)\times 0+\left( \rho {{a}^{2}}t \right)\times a}{\left( \dfrac{\rho \pi {{a}^{2}}t}{4} \right)+\left( \rho {{a}^{2}}t \right)}=\dfrac{\rho {{a}^{3}}t}{\rho {{a}^{2}}t\left( \dfrac{\pi }{4}+1 \right)} \\
 & \Rightarrow x=\dfrac{4a}{\pi +4}=\dfrac{4a}{7.14}=0.56a \\
\end{align}\]
So, the x-coordinate of the centre of mass is found to be at \[0.56a\], which is just outside the circular disc.
Now, to find the y-coordinate, the mass of the disc is at origin and mass of the plate is at zero on the y-axis,
\[\Rightarrow y=0\]
Therefore, the coordinates of the centre of mass of the combined body is found to be \[\left( 0.56a,0 \right)\] which is just outside the circular disc and inside the square plate. So, option b is the correct answer.

So, the correct answer is “Option A”.

Note:
We can predict the answer for this question by just analyzing the area of the two bodies. The area of the square plate is higher. So the centre of mass will be inside the square plate. We must know that the centre of mass of a circle will always be at its centre. In the question, the side of the square plate is given as \[a\] and the centre of mass of it will be also at its centre. That's the reason for taking the distance to its centre of mass as \[a/2\].