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5.4 Rotational Inertia

Syllabus
2024
Topic
5.4
Level
—

Learning objectives

5.4A—Describe the rotational inertia of a rigid system relative to a given axis of rotationDescribe the rotational inertia of a rigid system relative to a given axis of rotation.• Rotational inertia measures a rigid system’s resistance to changes in rotation and is related to the mass of the system and the distribution of that mass relative to the axis of rotation.• The rotational inertia of an object rotating a perpendicular distance r from an axis is described by the equation =Im r.2• The total rotational inertia of a collection of objects about an axis is the sum of the rotational inertias of each object about that axis: | AP Physics 1: Algebra-Based Course and Exam Description5.4B—Describe the rotational inertia of a rigid system rotating about an axis that does not pass through the system’s…Describe the rotational inertia of a rigid system rotating about an axis that does not pass through the system’s center of mass. BOUNDARY STATEMENT AP Physics 1 only expects students to calculate the rotational inertia for systems of five or fewer objects arranged in a two-dimensional configuration. Students do not need to know the rotational inertia of extended rigid systems, as these will be provided within the exam. Students should have a qualitative understanding of the factors that affect rotational inertia; for example, how rotational inertia is greater when mass is farther from the axis of rotation, which is why a hoop has more rotational inertia than a solid disk of the same mass and radius. AP Physics 1: Algebra-Based Course and Exam Description Torque and Rotational Dynamics UNIT 5 TOPIC 5.5 Rotational Equilibrium and Newton’s First Law in Rotational Form• A rigid system’s rotational inertia in a given plane is at a minimum when the rotational axis passes through the system’s center of mass.• The parallel axis theorem uses the following equation to relate the rotational inertia of a rigid system about any axis that is parallel to an axis through its center of mass:

Build rotational inertia from mass distribution

Connect mass distribution to rotational change

Rotational inertia measures a rigid system's resistance to a change in rotation about a specified axis. It depends on both how much mass the system has and how far that mass lies from the axis.

Calculate one object's contribution

I=mr2I=mr^2

Change for the same point mass New rotational inertia Effect
Double the mass 2I2I Doubles
Double the perpendicular distance 4I4I Quadruples

Add contributions about one axis

Itot=∑iIi=∑imiri2I_{\mathrm{tot}}=\sum_i I_i=\sum_i m_i r_i^2

Evaluate a two-object system

Two objects about one axis: a 2.0 kg2.0\,\text{kg} object is 0.30 m0.30\,\text{m} from the axis and a 1.0 kg1.0\,\text{kg} object is 0.60 m0.60\,\text{m} away.

Itot=(2.0)(0.30)2+(1.0)(0.60)2=0.18+0.36=0.54 kg m2I_{\mathrm{tot}}=(2.0)(0.30)^2+(1.0)(0.60)^2=0.18+0.36=0.54\,\text{kg}\,\text{m}^2.

The lighter object contributes more because its larger distance is squared.

Keep the axis and unit explicit

Rotational inertia is not fixed until the axis is specified. Use the perpendicular distance from each mass to that same axis, not the distance between masses. The SI unit is kg m2\text{kg}\,\text{m}^2.

Shift rotational inertia to a parallel axis

Start from the minimum axis

Among parallel axes in a given plane, a rigid system's rotational inertia is smallest for the axis through its center of mass. Shifting to a parallel axis moves the system's mass farther away overall, so rotational inertia increases.

Relate two parallel axes

I′=Icm+Md2I'=I_{\mathrm{cm}}+Md^2

Symbol Meaning
IcmI_{\mathrm{cm}} Inertia about a parallel axis through the center of mass
I′I' Inertia about the shifted parallel axis
MM Total mass of the rigid system
dd Perpendicular separation between the two parallel axes

Calculate the shifted-axis value

Shifted axis: a rigid body has Icm=0.50 kg m2I_{\mathrm{cm}}=0.50\,\text{kg}\,\text{m}^2 and total mass M=2.0 kgM=2.0\,\text{kg}. For a parallel axis 0.30 m0.30\,\text{m} away,

I′=0.50+(2.0)(0.30)2=0.68 kg m2I'=0.50+(2.0)(0.30)^2=0.68\,\text{kg}\,\text{m}^2.

The positive Md2Md^2 term confirms that the shifted-axis value exceeds the center-of-mass value.

Compare mass distributions

For equal mass and outer radius, a hoop has greater rotational inertia than a solid disk because more of the hoop's mass lies far from the axis. This comparison is qualitative unless the object's inertia formula is provided.

Apply the theorem within scope

The parallel axis theorem works only for parallel axes; dd is their perpendicular separation and MM is the total system mass. AP Physics 1 provides extended-body inertias and limits point-object calculations to five or fewer objects in two dimensions.