Chapter 6, Problem 13P

To determine

To Find: The distance between the centres of the disk at which the cut should be made.

Answer to Problem 13P

The cut should be placed at the distance of $0.14\text{m}$ .

Explanation of Solution

Given:

The radius of the disk is $1\text{m}$ .

Calculation:

Consider the uniform disk with the radius of 1 m cut by the line such that the line $x=k$ .

Consider the centre of the mass of the smaller piece lies half of the way along the radius so the centre of mass is $\left(\frac{1}{2},0\right)$ .

The required diagram is shown in Figure 1

  

Figure 1

Here, the centre coordinate is $\overline{x}=\frac{1}{2}$ and $\overline{y}=0$ .

Consider the equation of the circle is given by,

  $\begin{array}{l}{x}^2+y^2=1\\ y=\pm \sqrt{1-x^2}\end{array}$

Then,

  $\begin{array}{l}\overline{x}=\frac{{\displaystyle {\int }_k^{1}xf\left(x\right)dx}}{{\displaystyle {\int }_k^{1}f\left(x\right)dx}}\\ \frac{1}{2}=\frac{{\displaystyle {\int }_k^{1}2x\sqrt{1-x^2}dx}}{{\displaystyle {\int }_k^{1}2\sqrt{1-x^2}dx}}\end{array}$

Consider the integral in the numerator is calculated as,

  $\begin{array}{c}{\displaystyle {\int }_k^{1}2x\sqrt{1-x^2}dx}={\left[-\frac{2}{3}{\left(1-x^2\right)}^{\frac{3}{2}}\right]}_k^1\\ =\frac{2}{3}{\left(1-k^2\right)}^{\frac{3}{2}}\end{array}$

Consider the integral in the denominator is calculated as,

  $\begin{array}{c}{\displaystyle {\int }_k^{1}2\sqrt{1-x^2}dx}=2{\left[\frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}{\sin }^{-1}\left(\frac{x}{1}\right)\right]}_k^1\\ =-k\sqrt{1-k^2}+{\sin }^{-1}\left(1\right)-{\sin }^{-1}k\end{array}$

Then,

  $\frac{1}{2}=\frac{\frac{2}{3}{\left(1-k^2\right)}^{\frac{3}{2}}}{-k\sqrt{1-k^2}+{\sin }^{-1}\left(1\right)-{\sin }^{-1}k}$

By the use of calculator or by computer system as,

  $\begin{array}{c}k=0.138\\ \cong 0.14\text{m}\end{array}$