Unit 2 · Lesson 5 of 5

The balance point is a weighted average.

A rod can be heavier at one end even with a uniform shape. Its balance point depends on mass distribution.

Change the input. Follow the mathematics.

What you'll be able to do

Set up the model and calculate its center-of-mass position.
Connect the equation to the changing visual.
Explain the assumptions and identify a misleading interpretation.

Predict before exploring

What will change if you increase right edge b? Predict the direction, then explain the response.

A model you can changeMove one control at a time.

Density

Right edge b2 mDensity gradient k1 1/m

center-of-mass position

M=4\,\mathrm{kg},\quad\bar x=\frac{4.66667}{4}=1.16667\,\mathrm m

The center of mass is a density-weighted mean position. A heavier right side pulls it beyond the geometric midpoint.

Assumptions, units, and limits

A thin rod along the x axis with line density in kg/m; the constant density scale is 1 kg/m.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

From the situation to an equation

Integrate density to get total mass. Integrate position times density to get first moment. Divide moment by mass for the center of mass. For laminae the same weighted principle uses area strips and two coordinate moments.

\lambda(x)=1+kx,\quad M=\int_0^b\lambda\,dx=b+\frac{kb^2}2,\quad\bar x=\frac{\int_0^bx\lambda\,dx}{M}

Keep the reasoning visible

\bar x=\frac{\int x\,dm}{\int dm}

The center of mass is a density-weighted mean position. A heavier right side pulls it beyond the geometric midpoint.

What makes this model useful?

A thin rod along the x axis with line density in kg/m; the constant density scale is 1 kg/m.

A geometric centroid equals the center of mass only with an appropriate uniform-density model. Moment and mass have different units.

A worked example

Calculate the starting scenario

Use right edge b = 2 m; density gradient k = 1 1/m.
$M=4\,\mathrm{kg},\quad\bar x=\frac{4.66667}{4}=1.16667\,\mathrm m$
The center of mass is a density-weighted mean position. A heavier right side pulls it beyond the geometric midpoint. The checked quantity is center-of-mass position.

Make the idea your own

Calculate. Compare. Explain.

Numeric checks allow 1% rounding error. Values near zero use a small absolute tolerance. Work through the setup before opening a solution.

Calculate and explain

Rebuild the starting calculation

Calculate the center-of-mass position using right edge b 2 m, density gradient k 1 1/m. State the meaning of its sign and units.

Checked quantity: center-of-mass position · m. Follow the definition for finite-cutoff and numerical-model questions.

Hint

Explore the visual solution

Density

Right edge b2 mDensity gradient k1 1/m

center-of-mass position

M=4\,\mathrm{kg},\quad\bar x=\frac{4.66667}{4}=1.16667\,\mathrm m

The center of mass is a density-weighted mean position. A heavier right side pulls it beyond the geometric midpoint.

Assumptions, units, and limits

A thin rod along the x axis with line density in kg/m; the constant density scale is 1 kg/m.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

\lambda(x)=1+kx,\quad M=\int_0^b\lambda\,dx=b+\frac{kb^2}2,\quad\bar x=\frac{\int_0^bx\lambda\,dx}{M}

Calculate and explain

Apply the idea to a new scenario

For rod length $b=3$ m and gradient $k=0.5$ m⁻¹, find the center of mass.

Checked quantity: center-of-mass position · m. Follow the definition for finite-cutoff and numerical-model questions.

Hint

Explore the visual solution

Density

Right edge b3 mDensity gradient k0.5 1/m

center-of-mass position

M=5.25\,\mathrm{kg},\quad\bar x=\frac{9}{5.25}=1.71429\,\mathrm m

The center of mass is a density-weighted mean position. A heavier right side pulls it beyond the geometric midpoint.

Assumptions, units, and limits

A thin rod along the x axis with line density in kg/m; the constant density scale is 1 kg/m.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

\lambda(x)=1+kx,\quad M=\int_0^b\lambda\,dx=b+\frac{kb^2}2,\quad\bar x=\frac{\int_0^bx\lambda\,dx}{M}

Challenge the interpretation

Which explanation survives the model?

Choose the statement you can justify, then use the visual to explain why the others fail.

Test it with the visual

Density

Right edge b2 mDensity gradient k1 1/m

center-of-mass position

M=4\,\mathrm{kg},\quad\bar x=\frac{4.66667}{4}=1.16667\,\mathrm m

The center of mass is a density-weighted mean position. A heavier right side pulls it beyond the geometric midpoint.

Assumptions, units, and limits

A thin rod along the x axis with line density in kg/m; the constant density scale is 1 kg/m.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The reasoning. Integrate density to get total mass. Integrate position times density to get first moment. Divide moment by mass for the center of mass. For laminae the same weighted principle uses area strips and two coordinate moments.

Your learning path

Can you explain the connection?

Mark complete when you can set up the calculation and explain what the visual is showing. Opening a page never completes it automatically.

Completion is saved on this browser. This is separate from account practice progress.