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]\sigma=m/A[/math]

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

[math]= \frac{\sigma}{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. On its way to the target the projectile breaks up into two EQUAL pieces. One piece lands 50m beyond the target.

Where does the second piece land

If one piece is at 150 m and the Center of mass is at 100 m

then

[math]100 = \frac{150m + xm}{2m}[/math]

[math]x = \frac{200 m -150 m }{m} = 50[/math]

The second piece is 50 m in front (towards the launch point) of the 100 m target.

Forest_UCM_MnAM#Center_of_Mass