Center of mass  (Page 3/3)

Com of rigid body

Rigid body is composed of very large numbers of particles. Mass of rigid body is distributed closely. Thus, the distribution of mass can be treated as continuous. The mathematical expression for rigid body, therefore, is modified involving integration. The integral expressions of the components of position of COM in three mutually perpendicular directions are :

$$\begin{array}{l}\Rightarrow x_{\mathrm{COM}}=\frac{1}{M}\int xđm\\ \Rightarrow y_{\mathrm{COM}}=\frac{1}{M}\int yđm\\ \Rightarrow z_{\mathrm{COM}}=\frac{1}{M}\int zđm\end{array}$$

Note that the term in the numerator of the expression is nothing but the product of the mass of particle like small volumetric element and its distance from the origin along the axis. Evidently, this terms when integrated is equal to sum of all such products of mass elements constituting the rigid body.

Evaluation of above integrals is simplified, if the density of the rigid body is uniform. In that case,

$$\begin{array}{l}\rho =\frac{M}{V}=\frac{đm}{đV}\\ \Rightarrow \frac{đm}{M}=\frac{đV}{V}\end{array}$$

Substituting,

$$\begin{array}{l}\Rightarrow x_{\mathrm{COM}}=\frac{1}{V}\int xđV\\ \Rightarrow y_{\mathrm{COM}}=\frac{1}{V}\int yđV\\ \Rightarrow z_{\mathrm{COM}}=\frac{1}{V}\int zđV\end{array}$$

We must understand here that once we determine COM of a rigid body, the same can be treated as a particle at COM with all the mass assigned to that particle. This concept helps to find COM of a system of rigid bodies, comprising of many rigid bodies. Similarly, when a portion is removed from a rigid body, the COM of the rigid body can be obtained by treating the "portion removed" and the "remaining body" as particles. We shall see the working of this concept in the example given in the next section.

Symmetry and com of rigid body

Evaluation of the integrals for determining COM is very difficult for irregularly shaped bodies. On the other hand, symmetry plays important role in determining COM of a regularly shaped rigid body. There are certain simplifying facts about symmetry and COM :

1. If symmetry is about a point, then COM lies on that point. For example, COM of a spherical ball of uniform density is its center.
2. If symmetry is about a line, then COM lies on that line. For example, COM of a cone of uniform density lies on cone axis.
3. If symmetry is about a plane, then COM lies on that plane. For example, COM of a cricket bat lies on the central plane.

The test of symmetry about a straight line or a plane is that the body on one side is replicated on the other side.

Problem : A small circular portion of radius r/4 is taken out from a circular disc of uniform thickness having radius "r" and mass "m" as shown in the figure. Determine the COM of the remaining portion of the uniform disc.

Com of remaining portion of circular disc

Solution : If we start from integral to determine COM of the remaining disc portion, then it would be a really complex proposition. Here, we shall make use of the connection between symmetry and COM. We note that the COM of the given disc is "O" and COM of the smaller disc removed is "O'". How can we use these fact to find center of mass of the remaining portion ?

The main idea here is that we can treat regular bodies with known COM as particles, which are separated by a known distance. Then, we shall employ the expression of COM for two particles to determine the COM of the remaining portion. We must realize that when a portion is removed from the bigger disc on the right side, the COM of the remaining portion shifts towards left side (heavier side).

The test of symmetry about a straight line or a plane is that the body on one side is replicated on the other side. We see here that the remaining portion of the disc is not symmetric about y-axis, but is symmetric about x-axis. It means that the COM of the remaining portion lies on the x-axis on the left side of the center of original disc. It also means that we need to employ the expression of COM for one dimension only.

While employing expression of COM, we use the logic as explained here. The original disc (with known COM) is equivalent to two particles system comprising of (i) remaining portion (with unknown COM) and (ii) smaller disc (with known COM). Now, x-component of the COM of original disc is :

Com of remaining portion of circular disc

$$\begin{array}{l}{x}_{\mathrm{COM}}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}\end{array}$$

But, x-component of the COM of original disc coincides with origin of the coordinate system. Further, let us denote the remaining disc by subscript "r" and the smaller circular disk removed by subscript "s".

$$\begin{array}{l}\Rightarrow 0\mathrm{=}\frac{m_r{x}_r+m_s{x}_s}{m_r+m_s}\end{array}$$

$$\begin{array}{l}\Rightarrow m_r{x}_r+m_s{x}_s=0\\ \Rightarrow x_r=-\frac{m_s{x}_s}{m_r}\end{array}$$

As evident from figure, $x_s=\frac{r}{2}$ . We, now, need to find mass of smaller disc, $m_s$ , and mass of remaining portion, $m_r$ , using the fact that the density is uniform. The density of the material is :

$$\begin{array}{l}\rho =\frac{m}{V}=\frac{m}{\pi r^{2}t}\end{array}$$

where "t" is thickness of disc.

$$\begin{array}{l}\Rightarrow m_s=V\rho =\left(\frac{\pi r^2}{16}\right)tx\frac{m}{\pi r^{2}t}=\frac{m}{16}\\ \Rightarrow m_r=m-m_s=m-\frac{m}{16}=\frac{15m}{16}\end{array}$$

Putting these values in the expression of the position of COM, we have :

$$\begin{array}{l}\Rightarrow x_r=-\frac{m_s{x}_s}{m_r}=-\frac{\left(\frac{m}{16}\right)\left(\frac{r}{2}\right)}{\frac{15m}{16}}=-\frac{r}{30}\end{array}$$

Acknowledgment

Author wishes to thank Aladesanmi Oladele A for making suggestion to remove error in the module.