Forest UCM Ch3 CoM

The Center of mass

Definition of the Center of Mass

The position [math]\vec R[/math] of the center of mass is given by

[math]\vec{R} = \frac{\sum_i^N m_i \vec{r}_i}{\sum_i^N m_i}[/math]

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

[math]\vec{R} = \frac{1}{M} \int \vec{r} dm[/math]

Example 1: CM of three particles

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

[math]\vec{r}_1 = (1,1,0) = \hat i + \hat j[/math]

[math]\vec{r}_2 = (1,-1,0) = \hat i - \hat j[/math]

[math]\vec{r}_3 = (0,0,0) = \vec{0}[/math]

when

[math]m_1 = m_2 = 3 m_3[/math]

let [math] m_3 = M[/math]

[math]\vec{R} = \frac{3M (1) + 3 M (1) + M(0)}{9M} \hat i + \frac{3M (1) + 3 M (-1) + M(0)}{9M} \hat j[/math]

[math]\vec{R} = \frac{6}{9} \hat i + \frac{0}{9} \hat j[/math]

Example 2: CM of a flat disk

Find the center of mass for a disk of radius [math]R[/math] and area mass density [math]\rho=m/A[/math]

[math]x_{cm} = \frac{1}{m} \int x \rho dA = \frac{1}{m} \int x \rho r dr d\theta[/math]

[math]= \frac{\rho}{m}\int r \cos \theta r dr d\theta[/math]

[math]= \frac{1}{A}\int r^2 dr \int d(\sin \theta)[/math]

[math]= \frac{1}{A}\int r^2 dr \left . \sin \theta \right |_0^{2\phi}[/math]

[math]= \frac{1}{A}\int r^2 dr \left . \sin \theta \right |_0^{2\phi}[/math]

[math]= \frac{1}{A}\int r^2 dr (0) = 0[/math]

The center of mass is located along the x-axis at [math]x=0[/math]

Similarly,[math] y_{cm} = 0[/math]

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

[math]M =[/math] mass of the semicircle

[math]\sigma =[/math] the mass density

[math]\vec{R} = \frac{1}{M} \int \sigma \vec{r} dA[/math]

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

[math]Y = \frac{1}{M} \int \sigma y dA[/math]

[math]= \int \frac{\sigma}{M} y dA[/math]

[math]= \int \frac{1}{A} y dA[/math]

[math]= \int y \frac{dA}{A}[/math]

[math]= \int r \sin \phi \frac{rdrd \phi}{\frac{1}{2}\pi R^2}[/math]

[math]= \frac{2}{\pi R^2}\int_0^R r^2 dr \int_0^{\pi} \sin \phi d \phi[/math]

[math]= \frac{2}{\pi R^2}\frac{R^3}{3}\int_0^{\pi} \sin \phi d \phi[/math]

[math]= \frac{2R^3}{3\pi} \left . (-1) \cos \phi \right |_0^{\pi}[/math]

[math]= \frac{4R^3}{3\pi} [/math]

Problem 3-19 Projectile explodes in midair

Suppose a projectile of mass [math]M[/math] is fired to hit a target that is 100 m away.

Forest_UCM_MnAM#Center_of_Mass