## Section7.4Center of Mass

To fully understand mechanics of an object that consists of many moving parts, you will need to study motion of each part. For instance, if you wanted to understand motion of a person, you will need to follow motion of arms, legs, heart, etc.

Despite these complications, you can study overall translational motion of the entire body by replacing the entire body by a single point and placing the total mass at that point. This special point is called the center of mass. This mathematical point may or may not be part of the body. Below, you will learn to find the location of a center of mass. Near Earth, since gravity is uniform, center of mass coincides with the center of gravity.

### Subsection7.4.1Center of Mass of Discrete Masses

Consider a multipart system, such as your body. Suppose the body has $N$ parts with masses $m_1 \text{,}$ $m_2 \text{,}$ $\cdots \text{,}$ $m_N \text{.}$ These parts don't have to be joined together, e.g., when a projectile breaks up into ten parts, they can all go their separate ways, but for us, all ten parts will still be part of the combined system, whose overall motion we wish to study.

Let the centers of those parts be located at position vectors $\vec r_1 \text{,}$ $\vec r_2 \text{,}$ $\cdots \text{,}$ $\vec r_N \text{.}$ Then, the position of the center of mass, denoted by $\vec R_{\text{CM}}\text{,}$ is defined by

$$\vec R_{\text{CM}} = \dfrac{m_1\vec r_1 + m_2\vec r_2 + \cdots + m_N\vec r_N }{M},\label{eq-cm-position}\tag{7.4.1}$$

where $M$ is the total mass,

$$M = m_1 + m_2 + \cdots + m_N.\tag{7.4.2}$$

We can write the components of this equation to obtain the $X_{\text{CM}} \text{,}$ $Y_{\text{CM}} \text{,}$ and $Z_{\text{CM}}$ from the coordinates of the parts. For instance, $X_{\text{CM}} \text{,}$ will be:

$$X_{\text{CM}} = \dfrac{m_1 x_1 + m_2 x_2 + \cdots + m_N x_N }{M},\tag{7.4.3}$$

and similarly for $Y_{\text{CM}} \text{,}$ and $Z_{\text{CM}} \text{.}$ You might think of $\vec R_\text{cm}$ as mass-weighted position of the entire body.

Two blocks of masses $10\text{ kg}$ and $20\text{ kg}$ are attached at the ends of a $1\text{-meter}$ rod of negligible mass. Where is the CM located?

Hint

Use the formula.

The CM is located $\dfrac{1}{3}\text{ m}$ from the $20\text{-kg}$ block towards the other block.

Solution

Since we have only two masses, we place one of the masses at the origin and the other mass along the $x$ axis. Suppose we place the $20\text{-kg}$ block at the origin and the $10\text{-kg}$ block at $x = 1\text{ m} \text{.}$

Leaving the units out of the calculations we get the following from the $X_{\text{CM}}$ formula.

\begin{equation*} X_{\text{CM}} = \frac{ 20\times 0 + 10 \times 1}{30 } = \frac{1}{3}. \end{equation*}

Now, we put the units back in. Therefore, the CM is located $\dfrac{1}{3}\text{ m}$ from the $20\text{-kg}$ block towards the other block.

Three balls of masses $2\text{ kg}\text{,}$ $3\text{ kg}$ and $4\text{ kg}$ are located at the corners of a right-angles triangle as shown in the fogure below. The three masses may or may not be connected by any rods. Where is the CM located in a coordinate sytem with the 3-kg mass at the origin.

Hint

You will need both $x$ and $y\text{.}$

The CM is located at $(-\dfrac{8}{9}\text{ m}, \dfrac{4}{3}\text{ m})$ in the coordinate system in the figure. Note that these values will changes if we use another coordinate system, but the physical location of the CM will remain the same.

Solution

Since we have a right-angled triangle, two of those sides can be chosen to align with Cartesian axes as shown. This gives us $(x,y)$ coordinates of each of the masses. We use them in formulas for $X_{\text{CM}}$ and $Y_{\text{CM}} \text{.}$

\begin{align*} \amp X_{\text{CM}} = \frac{2\times (-4) + 3\times 0 + 4 \times 0}{2+3+4} = -\dfrac{8}{9} \\ \amp Y_{\text{CM}} = \frac{2\times 0 + 3\times 0 + 4 \times 3}{2+3+4} = \dfrac{4}{3} \end{align*}

Therefore, the CM is located at $(-\dfrac{8}{9}\text{ m}, \dfrac{4}{3}\text{ m})$ in the coordinate system in the figure. Note that these values will changes if we use another coordinate system, but the physical location of the CM will remain the same.

Consider a cylindrical object with a cap at the top as shown in Figure 7.4.7. Let $M_1$ be the mass of the cylidrical part and $M_2$ be the mass of the cap. Let $H$ be the height of the cylindrical part and $h$ the height of the cap. Find the center of mass of the full system.

Hint

Place mass of each part at the center of mass of that part. Then use the point mass formula for the center of mass.

$Y_{\text{CM}} = \dfrac{ M_1 \times H/2 + M_2 \times (H + h/3) }{M_1 + M_2}.$

Solution

The center of mass of a composite object can be obtained by replacing parts by point particles at their centers of mass and then using center of mass of multiparticle system.

We replace the cylinder and the conical cap by point masses at their corresponding CMs. The CM of the cylinder will be at the center of the cylinder, but the CM of a cone is at a height of $\frac{1}{3}h$ from the base. Since both CMs fall on one line, we need only one Cartesian axis. We place the $y$ coordinate axis for calculation purpose.

The $Y_{\text{CM}}$ of the two together gives

\begin{equation*} Y_{\text{CM}} = \dfrac{ M_1 \times H/2 + M_2 \times (H + h/3) }{M_1 + M_2}. \end{equation*}

Suppose you weigh $90\text{ kg}$ and your arm weighs $10\text{ kg}$ each. Suppose the center of mass of your arms is $18\text{ cm}$ from your shoulder. How much will your center of mass rise if you lift your arm from both down to both fully up and extended as shown in Figure 7.4.10.

Hint

Set up equations for the center of mass in the vertical direction in the two ssituations. Choosing origin at the center of mass when arms down may be helpful also.

$12\text{ cm}\text{.}$

Solution

Let us choose $y$ axis pointed up with origin at the center of mass when the arms are down. Let $y_0$ be the $y$-coordinate of the arms when they are down and $Y$ be the $y$ coordinate of the rest of the body. Let $M$ denote your entire mass and $m$ the mass of each arm.

Since in my construction, $y$ coordinate of the CM is zero when arms are down, from the left side of Figure 7.4.11, we will get

\begin{equation*} M\,y_\text{cm} = 2\, m\, y_0 + (M-2m_0)\, Y, \end{equation*}

which is zero on the left side, giving

$$2\, m\, y_0 + (M-2m_0)\, Y = 0.\label{eq-cm-when-hands-down}\tag{7.4.4}$$

Let $h$ be the $y$ coordinate of the CM when the arms are up and $y_1$ the $y$ coordinate of the arms when they are up. Then, the right side of Figure 7.4.11 will give

$$M\,h = 2\, m\, y_1 + (M-2m_0)\, Y.\label{eq-cm-when-hands-up}\tag{7.4.5}$$

Subtracting Eq. (7.4.4) from Eq. (7.4.5), we get

\begin{equation*} M\, h = 2\, m\, \left( y_1 - y_0 \right). \end{equation*}

Note that $y_1-y_0$ is twice the distance of the CM of the arm from the shoulder. Let us denote the distance of the CM of the arms from the shoulder by $d\text{.}$ Solving for $h$ we get

\begin{equation*} h = \dfrac{4\, m}{M}\, d = \dfrac{4\times 15}{90}\times 18 = 12\text{ cm}. \end{equation*}

### Subsection7.4.2(Calculus) Center of Mass of Continuous Objects

In a physical object, such as the human body, large number of discrete atoms are essentially distributed continuously. To take into account the possibility that different parts of the object may have different densities, we model a physical body by a density function, $\rho(x,y,z)\text{,}$ which is local density at position $(x,y,z)\text{.}$ An integral over the volume of the object gives the mass, $M$

$$M = \iiint\, \rho(x,y,z)\, dxdydz\label{eq-cm-continupius-object-mass}\tag{7.4.6}$$

The center of mass can then be stated as the following integral.

$$\vec R_\text{cm} = \dfrac{1}{M}\, \iiint\,\vec r\, \rho(x,y,z)\, dxdydz,\label{eq-cm-continupius-object-rcm}\tag{7.4.7}$$

where $\vec r = x \hat i + y \hat j + z \hat k \text{.}$

We often write the integral in a more conceptual form by denoting $\rho(x,y,z)\, dxdydz$ by $dm \text{,}$ which stands for a small mass at the infintesimal cubical cell located center at $(x,y,z) \text{.}$

$$\vec R_{\text{CM}} = \dfrac{1}{M}\, \int\, \vec r \, dm.\label{eq-cm-position-continuous}\tag{7.4.8}$$

The first step in calculating integral in Eq. (7.4.8) involves picking a representative infinitesimal element of the body guided by the symmetry of the body. Then, $dm$ can be cast into an integral that you can often do without much difficulty. The three steps of calculation of CM are illustrated for a cone in Figure 7.4.13.

Although Eq. (7.4.8) is applicable to all systems, most of the time we will be working with regular solids of uniform density. In these situations, the density function $\rho(x,y,z) = \rho_0\text{,}$ the uniform density, comes out of integrals for $M$ and $\vec R_\text{CM}$ and cancels out.

$$\vec R_{\text{CM}} = \dfrac{1}{V}\,\iiint \vec r \, dxdydz;\ \ V = \iiint dxdydz,\tag{7.4.9}$$

where integrals are taken over the space occupied by the body. The integral over $\vec r$ locates the geometric centroid of the body. Thus, if a body has some symmetry, you do not need to do any integral since you can often locate the geometric centroid just by inspection.

Thus, CM can sometimes be guessed by symmetry. For instance, the center of mass of a spherical object will be at the center, the center of mass of a box will be at the center of the box, etc. Figure 7.4.14 shows the centers of a sphere, a cylinder, and a triangular shaped object of uniform density.

#### Subsubsection7.4.2.1Simplification for One- and Two-Dimensional Objects

Integrals in Eqs. (7.4.6) and (7.4.7) become simpler in the case where object is a wire (one-dimension) or a plate (two-dimension). In the case of wire, we work with linear density, denoted by $\lambda\text{,}$ which is mass per unit length, and in the case of plate, we work with surface density, denoted by $\sigma\text{,}$ which is mass per unit area. The corresponding integrals will be as follows.

Wire: Let $ds$ be a length elment on the wire. If the wire is along $x$ axis it will be $dx\text{,}$ but if the wire is curved, it may be more complicated. Then mass element $dm$ on the wire will be

\begin{equation*} dm = \lambda ds. \end{equation*}

The total mass and center of mass integrals will be

\begin{align*} \amp M = \int dm = \int \lambda\, ds.\\ \amp \vec R_\text{CM} = \dfrac{1}{M}\int \vec r dm = \dfrac{1}{M}\int \vec r\, \lambda\, ds. \end{align*}

Plate: Let $dA$ be an area elment on the surface. If the wire is flat on $xy$-plane it will be $dxdy\text{,}$ but if the plate is curved, it may be more complicated. Then mass element $dm$ on the wire will be

\begin{equation*} dm = \sigma dA. \end{equation*}

The total mass and center of mass integrals will be

\begin{align*} \amp M = \int dm = \int \sigma\, dA.\\ \amp \vec R_\text{CM} = \dfrac{1}{M}\int \vec r\, dm = \dfrac{1}{M}\int \vec r\, \sigma dA. \end{align*}

Consider a plate of uniform density and uniform thickness $t$ cut in the shape of right-angled triangle with base $b$ and height $h \text{.}$ Where is the center of mass?

Hint

Use a coodinate system with one corner of the base at origin and the end with $90^\circ$ corner at $x=b\text{.}$

$(\frac{2}{3}b, \frac{1}{3}h, 0)$ with respect to coordinate given in solution.

Solution

Let thickness of the plate be $t$ and the uniform density $\rho \text{.}$ Along the thickness of the plate, the CM will be in the plane half-way between the two faces of the plate. So, we only need to find the center of mass coordinates on the triangular shape surface. We will place the triangular face in the $xy$-plane as shown in the figure.

The $dm$ in Eq. (7.4.8) is

\begin{equation*} dm = \rho\, t\, dx\, dy. \end{equation*}

Then, the calculations for $X_{\text{CM}}$ goes as follows.

\begin{align*} X_{\text{CM}} \amp = \frac{1}{M} \int_{plate} x\ dm \\ \amp = \frac{1}{M} \int_{plate} x\ \rho\, t\, dx dy \\ \amp = \frac{\rho\ t}{M}\int_0^b dx \left[ x \int_0^{hx/b} dy \right] \\ \amp = \frac{2}{3}b,\ \ \text{ using } M = \dfrac{1}{2}bht\rho. \end{align*}

Similarly, $Y_{\text{CM}} = \dfrac{1}{3} h\text{.}$ By symmetry, $Z_{\text{CM}} = \dfrac{t}{2}\text{.}$

Density of material of a rod of uniform cross-section varies along its length. Let $x=0$ at one end and $x=L$ at the other end. Then, density $\rho(x) = \lambda_0\;(x/a)^2\text{,}$ where $\lambda_0$ and $a$ are constants. Find the location from the CM.

Hint

Set up integrals $\int dm$ and $\int x dm\text{.}$

$\frac{3 L}{4}\text{.}$

Solution

Let $A$ be the area of cross-section. Consider an element between $x$ and $x+dx\text{.}$ The $X_\text{cm}$ will be

$$MX_\text{cm} = \int x dm.\label{eq-Rod-of-Varying-Density-1}\tag{7.4.10}$$

Here,

\begin{equation*} dm = \rho A dx = \lambda_0\;(x/a)^2 A dx. \end{equation*}

Therefore, total mass will be

\begin{equation*} M = \int dm = \int_0^L \lambda_0\;(x/a)^2 A dx = \frac{\lambda_0}{3a^2}\; A L^3. \end{equation*}

Using this and $dm$ in Eq. (7.4.10), we get

\begin{align*} X_\text{cm} \amp = \frac{1}{M}\;\int x dm\\ \amp = \frac{3a^2}{\lambda_0 A L^3}\;\int_0^L \lambda_0\;(x/a)^2 A x dx\\ \amp = \frac{3a^2}{\lambda_0 A L^3}\; \frac{\lambda_0}{4a^2} AL^4 = \frac{3 L}{4}. \end{align*}

A uniform wire of length $L$ is bent into an equilateral triangle. Find its center of mass.

Hint

Use symmetry and place the triangle with one side along $y$-axis.

$\frac{L}{6\sqrt{3}}\text{.}$

Solution

Let $\lambda$ be the mass per unit length of the wire and let us denote the total mass by $3m\text{.}$ Let $2a=L/3$ denote the length of each side. Making use of symmetry, we will place the triangle so that the center of mass will be on $x$ axis as shown in Figure 7.4.19.

Let us find $x_0$ by working on the side in the first quadrant. The equation of the line of this side is

\begin{equation*} y = a - \frac{1}{\sqrt{3}}\, x. \end{equation*}

A mass element of the side will be

\begin{align*} dm \amp = \lambda\, ds = \lambda\, \sqrt{dx^2 + dy^2}\\ \amp = \lambda\, \sqrt{1 + (dy/dx)^2}\, dx = \frac{2}{\sqrt{3}}\,\lambda\, dx. \end{align*}

Now, we get $x_0\text{,}$ the $x$-cm of the two slanted sides.

\begin{align*} x_0 \amp = \frac{1}{ \frac{2}{\sqrt{3}}\,\lambda \int dx}\, \frac{2}{\sqrt{3}}\,\lambda \int x dx\\ \amp = \frac{\sqrt{3}}{2}\,a. \end{align*}

Therefore, CM of the triangular structure will be at

\begin{equation*} X_\text{CM} = \frac{1}{3m}\left(m\times 0 + 2m\times \frac{\sqrt{3}}{2}\,a \right) = \frac{a}{\sqrt{3}}. \end{equation*}

Writing this in terms of $L$ we have

\begin{equation*} X_\text{CM} = \frac{L}{6\sqrt{3}}. \end{equation*}

Find the location of the center of mass of a uniform semi-circular ring of radius $R\text{.}$

Hint

Use mass element of projected angle $d\theta\text{.}$ Project their mass on symmetric line. Integrate.

$\dfrac{2}{\pi}R\text{.}$

Solution

Refer to Figure 7.4.21 for calculations. I have placed the ring symmetrically about $y$ axis. Therefore, CM will be on the $y$ axis.

Let $\lambda$ [kg/m] be linear density of the ring. The mass in the arc element shown in the figure is

\begin{equation*} dm = \lambda\; ds = \lambda Rd\theta. \end{equation*}

We want to evaluate two integrals.

\begin{align*} \amp(1)\ \ M = \int dm = \int_0^\pi \lambda Rd\theta\\ \amp(2)\ \ \int y\; dm = \int_0^\pi (R\sin\theta) (\lambda Rd\theta) \end{align*}

The first one (1) is not even needed. We know the integral already since we have length of the semicircular ring. This will give

\begin{equation*} M= \pi R \lambda. \end{equation*}

The second integral (2) will be

\begin{equation*} \int y\; dm = R^2 \lambda \int_0^\pi \sin\theta\, d\theta = 2 R^2\lambda. \end{equation*}

Therefore,

\begin{equation*} Y_\text{cm} = \frac{1}{M}\int y\; dm = \frac{2}{\pi}R. \end{equation*}

A semi-disk of radius $R$ has uniform density. Find its CM.

Solution

No Solution provided.

A spherical bowl of radius $R$ and height $R$ has uniform density. Find its CM.

Solution

No Solution provided.

A hemisphere block of radius $R$ has uniform density. Find its CM.

$\frac{3}{8} R$ above center.