9.6 Center of mass  (Page 6/13)

Center of mass of continuous objects

If the object in question has its mass distributed uniformly in space, rather than as a collection of discrete particles, then $m_j\to dm$ , and the summation becomes an integral:

In this context, r is a characteristic dimension of the object (the radius of a sphere, the length of a long rod). To generate an integrand that can actually be calculated, you need to express the differential mass element dm as a function of the mass density of the continuous object, and the dimension r . An example will clarify this.

Cm of a uniform thin hoop

Strategy

We replace dm with an expression involving the density of the hoop and the radius of the hoop. We then have an expression we can actually integrate. Since the hoop is described as “thin,” we treat it as a one-dimensional object, neglecting the thickness of the hoop. Therefore, its density is expressed as the number of kilograms of material per meter. Such a density is called a linear mass density    , and is given the symbol $\lambda$ ; this is the Greek letter “lambda,” which is the equivalent of the English letter “l” (for “linear”).

Since the hoop is described as uniform, this means that the linear mass density $\lambda$ is constant. Thus, to get our expression for the differential mass element dm , we multiply $\lambda$ by a differential length of the hoop, substitute, and integrate (with appropriate limits for the definite integral).

Solution

The center of mass is calculated with [link] :

We have to determine the limits of integration a and b . Expressing $\overrightarrow{r}$ in component form gives us

In the diagram, we highlighted a piece of the hoop that is of differential length ds ; it therefore has a differential mass $dm=\lambda ds$ . Substituting:

However, the arc length ds subtends a differential angle $d\theta$ , so we have

and thus

One more step: Since $\lambda$ is the linear mass density, it is computed by dividing the total mass by the length of the hoop:

giving us

Notice that the variable of integration is now the angle $\theta$ . This tells us that the limits of integration (around the circular hoop) are $\theta =\text{0 to}\theta =2\pi$ , so $a=0$ and $b=2\pi$ . Also, for convenience, we separate the integral into the x - and y -components of ${\overrightarrow{r}}_{\text{CM}}$ . The final integral expression is

as expected.

Center of mass and conservation of momentum

How does all this connect to conservation of momentum?