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SCIENCE · INVESTIGATE AND EXPLAIN

Build rotational inertia from five planar particles

Rotational inertia describes how strongly a rigid mass arrangement resists angular acceleration about a specified axis. For a point particle the contribution is mr², with r the perpendicular distance to that axis. For several particles, add the contributions: I=Σmr². Its unit is kg m². Total mass alone does not determine it.

Learning outcome: Compare (a,b)=(0.2,0.2) with (0.4,0.4), then with (0.4,0.2). Make a five-row mass-coordinate table and calculate every contribution before checking. Explain why only the first comparison gives an exact factor of four.

Starting recap

Review signed graph slopes and areas, extended-body force diagrams, Newton’s laws and center-of-mass coordinates. State the rotation axis.

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Try

With a=0.6 m and b=0.8 m, each corner has r²=0.36+0.64=1 m² and contributes 1 kg m². The total is 4 kg m². The 2 kg center particle adds no inertia about this axis. Moving only that center particle to (0.5,0) would add 2×0.5²=0.5 kg m², though the center of mass and support needs would then change.

Open the experiment

Predict

Can adding mass leave rotational inertia about one specified axis unchanged?

Experiment

Compare (a,b)=(0.2,0.2) with (0.4,0.4), then with (0.4,0.2). Make a five-row mass-coordinate table and calculate every contribution before checking. Explain why only the first comparison gives an exact factor of four.

Sum inertia for five particles in a plane

Worked teaching model · synthetic values · Four 1 kg point particles occupy (±a,±b), with a fifth 2 kg particle at the origin. Massless rigid connectors. Axis is perpendicular to the coordinate plane through the center. Particle size is neglected.

Reset keeps your prediction, captures and practice history.

Five-particle mass distribution

Four 1 kg particles at (+a,+b), (+a,−b), (−a,+b), (−a,−b), with a=0.6 m and b=0.8 m. A 2 kg particle at (0,0) lies on the perpendicular rotation axis.

1 kg1 kg1 kg1 kg2 kg / axis

Application locations and directions are meaningful; force arrows are schematic unless explicitly labeled otherwise. Exact quantities and units follow in the ledger.

Calculated evidence ledger — synthetic teaching model
QuantityValueUnit
Total mass6kg
Squared corner radius1m²
Each corner inertia1kg m²
Center particle inertia0kg m²
Total rotational inertia4kg m²
Accessible plot coordinates
Primary plotted points
time (s)angular velocity (rad/s)

Mass at the axis contributes mass but zero point-particle inertia about that axis.

I=Σm(x²+y²)=4(a²+b²). Doubling both geometric offsets multiplies inertia by four at unchanged mass. Changing just one offset need not quadruple the total.

Mass at the axis contributes mass but zero point-particle inertia about that axis.

I=Σm(x²+y²)=4(a²+b²). Doubling both geometric offsets multiplies inertia by four at unchanged mass. Changing just one offset need not quadruple the total.

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

1. Frame the question

Rotational inertia describes how strongly a rigid mass arrangement resists angular acceleration about a specified axis. For a point particle the contribution is mr², with r the perpendicular distance to that axis. For several particles, add the contributions: I=Σmr². Its unit is kg m². Total mass alone does not determine it.

2. Build the mechanism

For an axis perpendicular to the xy plane through the origin, each particle has r²=x²+y². The signs of x and y disappear on squaring; particles on opposite sides contribute positive inertia rather than canceling. A particle exactly on the axis has zero point-particle contribution even though it adds to total mass.

The model places four 1 kg particles at (±a,±b), plus a 2 kg particle at the center, joined by massless rigid connectors. It therefore has total mass 6 kg and I=4(a²+b²). Doubling all distances from the axis multiplies inertia by four at unchanged mass. Moving only part of the distribution requires recomputing that part rather than scaling the entire total.

For extended bodies, use a supplied expression rather than treating every kilogram as if it were on the rim. A thin hoop has supplied I=MR² and a uniform disk I=½MR² about their central perpendicular axes. At equal M and R the hoop has greater inertia because more of its mass lies farther out. These supplied formulas are not a requirement to derive continuous-body integrals or memorize shape tables.

3. Work an example

With a=0.6 m and b=0.8 m, each corner has r²=0.36+0.64=1 m² and contributes 1 kg m². The total is 4 kg m². The 2 kg center particle adds no inertia about this axis. Moving only that center particle to (0.5,0) would add 2×0.5²=0.5 kg m², though the center of mass and support needs would then change.

4. Predict and investigate

Predict first: Can adding mass leave rotational inertia about one specified axis unchanged?

Compare (a,b)=(0.2,0.2) with (0.4,0.4), then with (0.4,0.2). Make a five-row mass-coordinate table and calculate every contribution before checking. Explain why only the first comparison gives an exact factor of four. Retain settings and raw evidence in the lesson notebook.

5. Transfer and explain

Try a new practice case

Two 2 kg point particles are each 0.5 m from a fixed axis. Find their combined inertia.

This extra practice is not saved and does not award mastery. Record your first answer, explanation and whether you revealed help in the lesson notebook. Use the lesson’s saved question separately.

6. Reflect and retain evidence

For equal total mass and outer radius, why must the axis and internal mass distribution be specified before comparing rotational response?

Retain your first response, an extended-body diagram or labeled graph, units, assumptions and a changed-case prediction. Public worked checks are practice, not independent mastery evidence. Synthetic model captures are not a performed physical experiment.

Framework and source notes

Original teaching prose and numerical cases. College Board AP Physics 1 CED, Unit 5 topic mapping: 5.4. Checked 30 October 2026. Planar clockwise/counterclockwise analysis; supplied formulas for extended bodies, no required calculus or three-dimensional torque-vector direction. Framework cross-reference, not AP authorization.

Apply

Four 1 kg particles are at (±0.6,±0.8) m and a 2 kg particle at the origin. About the perpendicular origin axis, what is I?

Public practice checks a specific claim. It does not establish independent mastery.

For equal total mass and outer radius, why must the axis and internal mass distribution be specified before comparing rotational response?

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

Skipping retains experiments and awards no mastery.

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