Chapter 7, Problem 52P

Solution

The position of the center of mass of the plate.

  $\left(-0.27R,0\right)$

Given:

Radius of uniform circular plate is 2R.

A circular hole of radius R is cut out.

Smaller circle center C^’is at a distance of 0.8R from plate center C.

Formula used:

Center of mass of plate on x-axis is,

  $C.M.=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}$

Calculation:

Consider mass of ring is m, mass of circular hole is $m\text{'}$

If origin is the center of mass of the big plate then x = -0.80R

The center of mass of plate on the x-axis is,

  $C{M}_X=\frac{mx+m\text{'}x\text{'}}{m+m\text{'}}$

Location of center of mass on x-axis is,

  $\begin{array}{l}C{M}_X=\frac{mx+m\text{'}x\text{'}}{m+m\text{'}}\\ x\text{'}=\frac{C{M}_X\left(m+m\text{'}\right)-mx}{m\text{'}}\end{array}$

The y-coordinate of center of mass of circle is located at $y=0$

The x-coordinate of the circle at $C{M}_X=0$ is,

  $\begin{array}{c}x\text{'}=\frac{0\left(m+m\text{'}\right)-m\left(0.8R\right)}{m\text{'}}\\ =\frac{-0.8Rm}{M-m}\left(\because M=m+m\text{'}\right)\\ =\frac{-0.8R\left(d\times \pi R^2\right)}{d\times \pi \times \left({\left(2R\right)}^2-R^2\right)}\\ =\frac{-0.8R^3}{3R^2}\\ =-0.266666R\\ \approx -0.27R\end{array}$

So, the center of mass is located at $\left(-0.27R,0\right)$