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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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.
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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.
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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.
Application locations and directions are meaningful; force arrows are schematic unless explicitly labeled otherwise. Exact quantities and units follow in the ledger.
| Quantity | Value | Unit |
|---|---|---|
| Total mass | 6 | kg |
| Squared corner radius | 1 | m² |
| Each corner inertia | 1 | kg m² |
| Center particle inertia | 0 | kg m² |
| Total rotational inertia | 4 | kg m² |
Accessible plot coordinates
| 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.
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For equal total mass and outer radius, why must the axis and internal mass distribution be specified before comparing rotational response?
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