Forest UCM Ch3 CoM

The Center of mass

Definition of the Center of Mass

The position $\vec R$ of the center of mass is given by

$\vec{R} = \frac{\sum_i^N m_i \vec{r}_i}{\sum_i^N m_i}$

The center of mass is given as the sum of the position of each object in the system weighted by the objects mass.

For a rigid object the location of the center of mass is given by

$\vec{R} = \frac{1}{M} \int \vec{r} dm$

Example 1: CM of three particles

Calculate the location of the center of mass given the three particles below

$\vec{r}_1 = (1,1,0) = \hat i + \hat j$

$\vec{r}_2 = (1,-1,0) = \hat i - \hat j$

$\vec{r}_3 = (0,0,0) = \vec{0}$

when

$m_1 = m_2 = 3 m_3$

let $m_3 = M$

$\vec{R} = \frac{3M (1) + 3 M (1) + M(0)}{9M} \hat i + \frac{3M (1) + 3 M (-1) + M(0)}{9M} \hat j$

$\vec{R} = \frac{6}{9} \hat i + \frac{0}{9} \hat j$

Example 3: CM of a semicircle

A semicircle of radius R lies in the xy plane with its center at the origin and a diameter lying along the x axis. Use polar coordinates to locte the position of the center of mass of the semicircle.

Assume

$M =$ mass of the semicircle

$\sigma =$ the mass density

$\vec{R} = \frac{1}{M} \int \sigma \vec{r} dA$

Since the diameter of the semicircle lies along the x-axis, symmetry arguments can be used to locate the position of the center of mass in the x-direction as being on the x-axis.

For the Y-direction

$Y = \frac{1}{M} \int \sigma y dA$

$= \int \frac{\sigma}{M} y dA$

$= \int \frac{1}{A} y dA$

$= \int y \frac{dA}{A}$

$= \int r \sin \phi \frac{rdrd \phi}{\frac{1}{2}\pi R^2}$

$= \frac{1}{\pi R^2}\int_0^R r^2 dr \int_0^{\pi} \sin \phi d \phi$

$= \frac{1}{\pi R^2}\frac{R^3}{3}\int_0^{\pi} \sin \phi d \phi$

Problem 3-19

Forest_UCM_MnAM#Center_of_Mass