Chapter 6, Problem 9RCC

a.

To determine

To describe: The physical significance of center of mass of a thin plate.

Answer to Problem 9RCC

Center of mass of a thin plate is a point on which a thin plate balances horizontally.

Explanation of Solution

Given information :

The plate is thin.

Center of mass of a system is a point where we can assume entire mass is concentrated.

Hence, centre of mass of a thin plate is a point on which a thin plate balances horizontally.

b.

To determine

The expression for the coordinates of the center of mass of a given plate.

Answer to Problem 9RCC

The center of mass of the plate is located at the point $\left(\overline{x},\overline{y}\right)$ where,

  $\overline{x}=\frac{1}{A}{\displaystyle \underset{a}{\overset{b}{\int }}xf\left(x\right)dx}$

  $\overline{y}=\frac{1}{A}{\displaystyle \underset{a}{\overset{b}{\int }}\frac{1}{2}{\left[f\left(x\right)\right]}^{2}dx}$

Explanation of Solution

Given information :

The plate lies between $y=f\left(x\right)$ and

  $y=0$ , where $a\le x\le b$ .

Formula used :

Moment of the system about the Y-axis to be

  $\begin{array}{l}{M}_y={\displaystyle \sum _{i=1}^n{m}_i{x}_i}\\ \Rightarrow \overline{x}=\frac{M_y}{m}\end{array}$

Moment of the system about X-axis to be

  $\begin{array}{l}{M}_x={\displaystyle \sum _{i=1}^n{m}_i{y}_i}\\ \Rightarrow \overline{y}=\frac{M_x}{m}\end{array}$

The coordinates of center of mass is $\left(\overline{x},\overline{y}\right)$ .

Next we consider a flat plate (called a lamina) with uniform density $\rho$ that occupies a

Region $R$ of the plane. We wish to locate the center of mass of the plate, which is called

the centroid of $R$ . In doing so we use the following physical principles: The symmetry

principle says that if $R$ is symmetric about a line $l$ , then the centroid of lies on $l$ . (If $R$

is reflected about $l$ , then $R$ remains the same so its centroid remains fixed. But the only fixed points lie on $l$ .) Thus the centroid of a rectangle is its center. Moments should be

defined so that if the entire mass of a region is concentrated at the center of mass, then its

moments remain unchanged. Also, the moment of the union of two nonoverlapping regions

should be the sum of the moments of the individual regions.

  

Suppose that the region $R$ is of the type shown in Figure-a; that is, $R$ lies between

the lines $x=a$ and $x=b$ above the X -axis, and beneath the graph of $f$ , where is a

continuous function. We divide the interval into $\left[a,b\right]$ into $n$ subintervals with endpoints $x_{0,}{x}_1\mathrm{..........}{x}_n$

and equal width $\Delta x$ . We choose the sample point $x_i^{\ast }$ to be the midpoint $\overline{x_i}$ of the i^th

subinterval, that is,

  $\overline{x_i}=\left(x_{i-1}+x_i\right)/2$

This determines the polygonal approximation to $R$ shown in Figure-b.

  

The centroid of i^th approximating rectangle $R_i$ is its center $C_i\left(\overline{x_i},\frac{1}{2}f\left(\overline{x_i}\right)\right)$

. Its area is $f\left(\overline{x_i}\right)\Delta x$ ,

so its mass is $\rho f\left(\overline{x_i}\right)\Delta x$

The moment $R_i$ of about the Y-axis is the product of its mass and the distance from C to

The Y -axis, which is $\overline{x_i}$ . Thus

  $M_y\left(R_i\right)=\left[\rho f\left(\overline{x_i}\right)\Delta x\right]\overline{x_i}=\rho \overline{x_i}f\left(\overline{x_i}\right)\Delta x$

Adding these moments, we obtain the moment of the polygonal approximation to $R$ , and

then by taking the limit as $n\to \infty$ we obtain the moment of $R$ itself about the Y-axis:

  $M_y=\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\rho \overline{x_i}f\left(\overline{x_i}\right)\Delta x=\rho {\displaystyle {\int }_a^{b}xf\left(x\right)}}dx$

In a similar fashion we compute the moment of $R_i$ about the X-axis as the product of its

mass and the distance from $R_i$ to the X-axis:

  $M_y\left(R_i\right)=\left[\rho f\left(\overline{x_i}\right)\Delta x\right]\frac{1}{2}f\left(\overline{x_i}\right)=\rho \cdot \overline{x_i}\frac{1}{2}{\left[f\left(\overline{x_i}\right)\right]}^2\Delta x$

Again we add these moments and take the limit to obtain the moment of $R$ about the

X-axis:

  $M_x=\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\rho \cdot \frac{1}{2}{\left[f\left(\overline{x_i}\right)\right]}^2\Delta x}=\rho {\displaystyle {\int }_a^b\frac{1}{2}{\left[f\left(x\right)\right]}^{2}dx}$

Just as for systems of particles, the center of mass of the plate is defined so that $m\overline{x}=M_y$

and $m\overline{y}=M_x$ . But the mass of the plate is the product of its density and its area:

  $m=\rho A=\rho {\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}$

And so,

  $\overline{x}=\frac{M_y}{m}=\frac{\rho {\displaystyle \underset{a}{\overset{b}{\int }}xf\left(x\right)dx}}{\rho {\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}}=\frac{{\displaystyle \underset{a}{\overset{b}{\int }}xf\left(x\right)dx}}{{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}}$

  $\overline{y}=\frac{M_x}{m}=\frac{\rho {\displaystyle \underset{a}{\overset{b}{\int }}\frac{1}{2}{\left[f\left(x\right)\right]}^{2}dx}}{\rho {\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}}=\frac{{\displaystyle \underset{a}{\overset{b}{\int }}\frac{1}{2}{\left[f\left(x\right)\right]}^{2}dx}}{{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}}$

The center of mass of the plate is located at the point $\left(\overline{x},\overline{y}\right)$ where,

  $\overline{x}=\frac{1}{A}{\displaystyle \underset{a}{\overset{b}{\int }}xf\left(x\right)dx}$

  $\overline{y}=\frac{1}{A}{\displaystyle \underset{a}{\overset{b}{\int }}\frac{1}{2}{\left[f\left(x\right)\right]}^{2}dx}$