14.2 Center of mass (check your understnading)

The COM of the portion removed is the geometric center of the triangle. Note here that the geometric center of the triangle coincides with the geometric center of the square - even though the triangle is asymmetrically oriented. Thus, removal of the triangular portion does not alter the COM. The COM of the remaining part of the square plate still lies at the center of the square.

Hence, option (d) is correct.

In order that the COM of the system of particles lies at the center, the masses of the particles on opposite corners should be equal such that masses are symmetrically distributed about the planner axes. Thus, for the given set of particle masses, the COM does not lie at the center of the square.

Hence, option (d) is correct.

The center of mass coincides with geometric center when density of the rigid body is uniform. However, a rigid body can have non-uniform density as well. In that case, center of mass does not coincide with geometric center. As discussed in the text in the modules on the topic, only choice (b) is correct.

Hence, option (b) is correct.

The height of the triangle is :

System of particles

$$\begin{array}{l}\mathrm{AC}=\mathrm{OB}\sin {60}^0=1x\frac{\surd 3}{2}\\ \Rightarrow \mathrm{AC}=\frac{\surd 3}{2}=0.866m\end{array}$$

The COM of the system of particles is :

$$\begin{array}{l}{x}_{\mathrm{COM}}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2+m_3}\\ \Rightarrow x_{\mathrm{COM}}=\frac{1x0+2x1+3x0.5}{6}=0.75m\end{array}$$

and

$$\begin{array}{l}{y}_{\mathrm{COM}}=\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3}{m_1+m_2+m_3}\\ \Rightarrow x_{\mathrm{COM}}=\frac{1x0+2x0+3x0.866}{6}=0.43m\end{array}$$

Hence, option (b) is correct.

If the particles are uniformly distributed, then COM lies at the center. On the other extreme, if there is only one particle in the system, then COM lies at a linear distance "r" from the center. For other possibilities, COM should lie between these two extremes.

Hence, option (a) is correct.

When density increases or decreases continuously from one end to another, the COM should lie on heavier side. In these conditions, there is no possibility that the rod is balanced at the geometric center.

Hence, options (b) and (c) are correct.

The COMs of each body is its geometric center. They form two particles system separated by a distance "a" in x - direction. Let subscripts "s" and "c" denote square and circular plates respectively. Let "σ" be the area surface density. Then the masses are :

$$\begin{array}{l}{m}_s=a^2\sigma \end{array}$$

and

$$\begin{array}{l}{m}_c=\left(\frac{\pi a^2\sigma }{4}\right)\end{array}$$

We see that the system of bodies is symmetric about x-axis, but not about y-axis. Thus, COM lies on x - axis. Now, the x-component of COM is :

$$\begin{array}{l}{x}_{\mathrm{COM}}=\frac{m_s{x}_s+m_c{x}_c}{m_s+m_c}\end{array}$$

$$\begin{array}{l}\Rightarrow x_{\mathrm{COM}}=\frac{\left(a^2\sigma \right)(-\frac{a}{2})+\left(\frac{\pi a^2\sigma }{4}\right)\left(\frac{a}{2}\right)}{\left(a^2\sigma \right)+\left(\frac{\pi a^2\sigma }{4}\right)}\end{array}$$

$$\begin{array}{l}\Rightarrow x_{\mathrm{COM}}=\frac{-(1-\frac{\pi }{4})}{2(1-\frac{\pi }{4})}=-0.06a\end{array}$$

Hence, option (d) is correct.