Continuous Center of Mass - Physics Q and A

« Back to AP Physics Guide / Unit C4: Momentum (Mechanics) / Continuous Center of Mass

Visualization of a glowing baseball bat balancing perfectly on a fulcrum, with mathematical integrals hovering nearby.

The Center of Mass is the unique point where a system’s entire mass can be considered to be concentrated.

In previous physics courses, if you needed to find the center of mass, you usually dealt with discrete point masses—like two separate blocks connected by a massless string. You simply multiplied each mass by its position, added them up, and divided by the total mass: $x_{cm} = \frac{\sum m_i x_i}{M_{total}}$ .

But what if the object is a solid, continuous rod? Or what if it’s a baseball bat that is thicker and heavier at one end? You can’t just add up a few blocks. You have to add up an infinite number of infinitesimally small pieces of mass ( $dm$ ). This requires Integration.

1. The Continuous Center of Mass Integral

$x_{cm} = \frac{1}{M} \int x \cdot dm$

Where $x_{cm}$ is the center of mass, $M$ is the total mass of the object, $x$ is the position of the tiny mass slice, and $dm$ is that infinitesimally small piece of mass.

There is a problem with the formula above: you cannot integrate position ( $x$ ) with respect to mass ( $dm$ ). You must convert $dm$ into $dx$ so the variables match. To do this, we use Density.

2. Linear Mass Density ( $\lambda$ )

If an object is a 1D rod or string, we describe its mass distribution using Linear Mass Density ( $\lambda$ ), which is the mass per unit length (kg/m).

If the rod is uniform, $\lambda$ is a constant number ( $\lambda = M/L$ ).
If the rod is non-uniform (heavier at one end), $\lambda$ will be given as a function of position, like $\lambda(x) = cx^2$ .

Because $\lambda = \frac{dm}{dx}$ , we can rearrange it to find our conversion factor:

$dm = \lambda dx$

Substitute this into the Center of Mass integral, and suddenly the math becomes easy to solve: $x_{cm} = \frac{1}{M} \int x (\lambda) dx$ .

Calculus Trap: If the problem does not give you the total mass ( $M$ ), you must find it yourself by integrating the density function over the length of the rod! Total Mass $M = \int \lambda dx$ .

3. Quick AP Practice

📚 Unit C4 Mastery Challenge

1. A rod of length $L$ lies on the x-axis from $x=0$ to $x=L$ . Its linear mass density is given by $\lambda = cx$ , where $c$ is a positive constant. What is the total mass $M$ of the rod?

Check Answer

Total mass is the integral of the density function:
$M = \int_{0}^{L} \lambda dx$
$M = \int_{0}^{L} cx dx$
$M = \left[ \frac{c x^2}{2} \right]_0^L = \mathbf{\frac{cL^2}{2}}$ .

2. Using the total mass from Question 1, calculate the Center of Mass ( $x_{cm}$ ) for the rod.

Check Answer

Set up the CoM integral: $x_{cm} = \frac{1}{M} \int_0^L x dm = \frac{1}{M} \int_0^L x (\lambda) dx$ .

$x_{cm} = \frac{1}{M} \int_0^L x (cx) dx$
$x_{cm} = \frac{c}{M} \int_0^L x^2 dx = \frac{c}{M} \left[ \frac{x^3}{3} \right]_0^L = \frac{cL^3}{3M}$ .

Now, substitute the mass we found in Q1 ( $M = cL^2 / 2$ ):
$x_{cm} = \frac{cL^3 / 3}{cL^2 / 2} = \mathbf{\frac{2}{3}L}$ .

1. The Continuous Center of Mass Integral

2. Linear Mass Density ()

3. Quick AP Practice

📚 Unit C4 Mastery Challenge

🔗 Next Steps

2. Linear Mass Density ( $\lambda$ )