Rotational kinetic energy is the energy a rigid object has because it’s spinning. For a rigid body rotating about a fixed axis it’s given by K_rot = 1/2 I ω^2, where I is the moment (rotational) inertia and ω is angular speed (CED 6.1.A.1). Unlike “regular” (translational) kinetic energy, K_trans = 1/2 M v_CM^2 depends on the center-of-mass speed; K_rot depends on how the mass is distributed about the axis (I) and each point’s linear speed v = ωr. The total kinetic energy of a rolling or moving rigid body is the sum: K_total = 1/2 M v_CM^2 + 1/2 I_CM ω^2 (CED 6.1.A.1.ii). A rigid body can have rotational KE even when its center of mass is at rest because internal points still move (CED 6.1.A.2). Rotational KE is a scalar (CED 6.1.A.3). For a quick AP-aligned review, see the Topic 6.1 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-6/1-rotational-kinetic-energy/study-guide/OM0z7GYjhkcWoIlZ) and extra practice (https://library.fiveable.me/practice/ap-physics-1-revised).

Use the formula K_rot = 1/2 I ω^2. Here I is the rotational inertia (moment of inertia) about the rotation axis and ω is the angular speed in rad/s. For a system of point masses, I = Σ m_i r_i^2 (or for a continuous body compute the appropriate integral or use standard formulas: e.g., solid disk I = 1/2 MR^2, thin hoop I = MR^2). If the axis is parallel but not through the center of mass, use the parallel-axis theorem: I = I_cm + M d^2. Remember rotational kinetic energy is a scalar and the total kinetic energy of a rigid body can be written K_total = K_trans + K_rot = 1/2 M v_cm^2 + 1/2 I_cm ω^2. This is exactly what the AP CED expects for Topic 6.1 (see the study guide for worked examples: https://library.fiveable.me/ap-physics-1-revised/unit-6/1-rotational-kinetic-energy/study-guide/OM0z7GYjhkcWoIlZ). For extra practice, try problems at https://library.fiveable.me/practice/ap-physics-1-revised.

Even if an object’s center of mass isn’t changing position, its parts can be moving. Each tiny mass element at distance r from the axis has linear speed v = ωr, so it has kinetic energy ½ dm · v². Summing those ½ dm·(ωr)² over the whole rigid body gives Krot = ½ I ω², where I (rotational inertia) encodes how the mass is distributed. So the object “has” kinetic energy because its internal pieces are moving—even though the center of mass is at rest (CED 6.1.A.1–2). Rotational KE is a scalar and adds to any translational KE: Ktotal = ½MvCM² + ½Iω² (CED 6.1.A.1.ii). This idea shows up on the exam when you compare rotational and translational energy (Unit 6), so be ready to derive Krot from point-mass sums or use ½Iω² directly. For a concise review, check the Topic 6.1 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-6/1-rotational-kinetic-energy/study-guide/OM0z7GYjhkcWoIlZ) and try practice problems (https://library.fiveable.me/practice/ap-physics-1-revised).

Rotational inertia (I) and angular velocity (ω) play very different roles in Krot = ½ I ω². - Rotational inertia (I, sometimes called moment of inertia) is a geometry-and-mass property of the object. It tells you how the object’s mass is distributed relative to the rotation axis. A big I means the object “resists” changes in rotation more (e.g., a hoop has larger I than a solid disk of the same mass and radius). I depends on shape, axis, and mass (use standard formulas or the parallel-axis theorem). - Angular velocity (ω) is how fast the object is spinning (radians per second). It’s a kinematic quantity—higher ω means every mass element moves faster. In the energy formula, I scales how much kinetic energy you get for a given ω. For the same ω, an object with larger I stores more rotational KE; for the same I, increasing ω raises KE as ω². This matches CED 6.1.A (Krot = ½ I ω²). For review and examples, see the Topic 6.1 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-6/1-rotational-kinetic-energy/study-guide/OM0z7GYjhkcWoIlZ) and practice problems (https://library.fiveable.me/practice/ap-physics-1-revised).

Short answer: an object can have both because parts of it move as a whole (translation of the center of mass) and also move about the center (rotation). The total kinetic energy of a rigid body is the sum of translational plus rotational parts: K_total = K_trans + K_rot = (1/2) M v_cm^2 + (1/2) I_cm ω^2. Why that makes sense: imagine a rolling wheel. Every small piece of the wheel has linear speed from the wheel’s center moving and extra speed from spinning around that center. Summing the kinetic energy of all those pieces gives the two-term expression above. If the center of mass is at rest (v_cm = 0) you can still have K_rot > 0 because the parts move around the center (a spinning disk). For a rolling object without slipping use v_cm = ωR to relate the two energies and find how energy splits between translation and rotation (useful on AP problems). For AP Topic 6.1, memorize K_rot = 1/2 I ω^2 and that total KE = translational + rotational. More examples and practice problems are in the Topic 6.1 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-6/1-rotational-kinetic-energy/study-guide/OM0z7GYjhkcWoIlZ) and in the unit practice set (https://library.fiveable.me/practice/ap-physics-1-revised).

If an object both translates (moves its center of mass) and rotates, the total kinetic energy is the sum of the translational KE of the center of mass and the rotational KE about the center of mass (CED 6.1.A.1.ii): K_total = (1/2) M v_cm^2 + (1/2) I_cm ω^2 - M is the object's total mass, v_cm is the center-of-mass speed, I_cm is the moment of inertia about the center-of-mass axis, and ω is the angular speed. Rotational KE is a scalar (CED 6.1.A.3). - For pure rolling without slipping, use v_cm = ωR to relate the two terms; for example a solid disk gives K_total = (1/2) M v_cm^2 + (1/4) M v_cm^2 = (3/4) M v_cm^2. - If rotation is about an axis not through the CM, use the parallel-axis theorem to get I about the rotation axis before using (1/2)Iω^2. This is exactly what AP expects—write both terms and add them (see Topic 6.1 study guide: https://library.fiveable.me/ap-physics-1-revised/unit-6/1-rotational-kinetic-energy/study-guide/OM0z7GYjhkcWoIlZ). For extra practice, try problems in the Unit 6 practice set (https://library.fiveable.me/practice/ap-physics-1-revised).

Saying rotational kinetic energy is a scalar means it has magnitude only—no direction. For a rigid body rotating about an axis its rotational kinetic energy is Krot = 1/2 I ω^2 (CED 6.1.A.1). Because ω is squared, Krot is always nonnegative and doesn’t change if the rotation reverses direction (ω → −ω). That contrasts with angular momentum (a vector) which does have direction. Practically for AP problems, treat Krot like any other scalar energy: you can add it to translational kinetic energy to get the total kinetic energy of a rigid system (CED 6.1.A.1.ii), and use it in energy-conservation statements without worrying about vector components (CED 6.1.A.3). For more review on this topic and examples, see the Topic 6.1 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-6/1-rotational-kinetic-energy/study-guide/OM0z7GYjhkcWoIlZ). For extra practice, check the unit practice set (https://library.fiveable.me/practice/ap-physics-1-revised).

They look alike because they come from the same idea: kinetic energy = 1/2 × (inertia) × (speed)^2. For a rigid body you can treat it as lots of point masses. Each point mass i has linear speed v_i = ω r_i when the body rotates about an axis, so its KE is 1/2 m_i v_i^2 = 1/2 m_i (ω^2 r_i^2). Factor ω^2 out and sum over all points: K_rot = 1/2 ω^2 Σ m_i r_i^2. That sum is the rotational inertia I, so K_rot = 1/2 I ω^2. That’s why the form mirrors 1/2 m v^2. Also remember the CED point that a rigid body’s total kinetic energy = translational (1/2 M v_cm^2) + rotational about the CM (1/2 I_cm ω^2). If you want more worked examples and AP-style practice on this, check the Topic 6.1 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-6/1-rotational-kinetic-energy/study-guide/OM0z7GYjhkcWoIlZ) and the practice problems collection (https://library.fiveable.me/practice/ap-physics-1-revised).

Yes—numerically they can have the same kinetic energy, but you must be careful what “same” means. Rotational KE for a rigid body is Krot = 1/2 I ω^2 (CED 6.1.A.1). A wheel spinning in place (center of mass at rest) can have nonzero Krot because points in the wheel still move (CED 6.1.A.2). A car driving has translational KE of the whole car Ktrans = 1/2 M v_cm^2 plus rotational KE of its wheels; total K = Ktrans + ΣKrot (CED 6.1.A.1.ii). So it’s possible for a single spinning wheel to have the same numeric KE as a moving car if 1/2 I ω^2 equals the car’s total KE. For a thin hoop I = MR^2, Krot = 1/2 M(ωR)^2 = 1/2 M v^2, so a hoop spinning with ω that gives v = ωR has the same KE as a mass M translating at speed v. For rolling without slipping, relate ω and v by v = ωR (CED keywords: rotational-translational equivalence, moment of inertia). Want practice comparing scenarios like this? See the Topic 6.1 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-6/1-rotational-kinetic-energy/study-guide/OM0z7GYjhkcWoIlZ) and try problems at Fiveable practice (https://library.fiveable.me/practice/ap-physics-1-revised).

Use rotational KE whenever parts of the object are moving in a circle about an axis—then K_rot = 1/2 I ω^2 applies (I about that axis, ω the angular speed). Use translational KE (K_trans = 1/2 Mv_cm^2) for motion of the object’s center of mass. For rigid bodies the total kinetic energy is the sum: K_total = 1/2 M v_cm^2 + 1/2 I_cm ω^2 (so you often need both, e.g., a rolling wheel: it has translational KE from v_cm and rotational KE about its center). If the CM is at rest but parts still move (a spinning disk fixed at its center), only K_rot appears. On AP tasks you should state which axis and which I you use and add energies (CED 6.1.A). Want more worked examples and practice? See the Topic 6.1 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-6/1-rotational-kinetic-energy/study-guide/OM0z7GYjhkcWoIlZ) and plenty of practice problems (https://library.fiveable.me/practice/ap-physics-1-revised).

Rotational inertia (moment of inertia, I) is a measure of how hard it is to change an object’s rotational motion about a given axis. It plays the same role for rotation that mass plays for linear motion. In equations: rotational kinetic energy K_rot = 1/2 I ω^2 (CED 6.1.A.1). Key differences from “regular” inertia (mass): - Mass is a scalar that resists changes in linear velocity; I depends on both mass and how that mass is distributed relative to the rotation axis (point masses at larger radii contribute more: I = Σ m r^2 for point masses). - I depends on the chosen axis (use the parallel-axis theorem to shift axes). - Both are measures of resistance to acceleration, but mass resists linear acceleration (F = ma) while I resists angular acceleration (τ = I α). - Rotational inertia is used in calculating rotational kinetic energy and total kinetic energy (translational + rotational) of rigid bodies (CED 6.1.A.1.ii, 6.1.A.2). For a quick AP-aligned review, see the Topic 6.1 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-6/1-rotational-kinetic-energy/study-guide/OM0z7GYjhkcWoIlZ). For lots of practice problems, check Fiveable practice (https://library.fiveable.me/practice/ap-physics-1-revised).

When an object spins, every bit of its mass moves in a circle about the rotation axis—that motion is linear (tangent) motion even though the object as a whole is "rotating." For a rigid body turning with angular speed ω, a point at distance r from the axis has linear speed v = ωr (units: m/s). That means each little mass element mi has kinetic energy 1/2 mi v^2 = 1/2 mi (ωr)^2. Adding those energies for all the bits gives Krot = 1/2 I ω^2, where I = Σ mi r^2 is the rotational inertia (CED 6.1.A.1.i & 6.1.A.2). So even if the center of mass is at rest, the points around it have nonzero v and thus the object has rotational kinetic energy. For a concrete example: if ω = 2 rad/s and r = 0.5 m, v = 1.0 m/s and that point’s KE per kg is 0.5*(1.0)^2 = 0.5 J/kg. Review this derivation and examples in the Topic 6.1 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-6/1-rotational-kinetic-energy/study-guide/OM0z7GYjhkcWoIlZ). For more practice problems, check Fiveable’s AP Physics 1 practice set (https://library.fiveable.me/practice/ap-physics-1-revised).

Use Krot = 1/2 I ω^2. To compare two spinning objects, plug in each object’s moment of inertia I (depends on mass and how it’s distributed about the axis) and their angular speeds ω. Key points: - If two objects have the same ω, the one with larger I has more rotational kinetic energy. Example: a thin hoop (I = MR^2) at the same ω has twice the Krot of a solid disk (I = 1/2 MR^2) with the same mass and radius. - If two objects have the same I, the one with larger ω has more Krot (K ∝ ω^2), so doubling ω quadruples Krot. - For moving/rolling rigid bodies, total kinetic energy = translational Kcm + rotational Krot about the center of mass (CED 6.1.A.2). Rotational KE is a scalar (CED 6.1.A.3). - Use the parallel-axis theorem if the rotation axis isn’t through the center of mass to get the correct I before comparing. For more examples and quick reference (moments of inertia and comparisons), check the Topic 6.1 study guide on Fiveable (https://library.fiveable.me/ap-physics-1-revised/unit-6/1-rotational-kinetic-energy/study-guide/OM0z7GYjhkcWoIlZ) and practice problems (https://library.fiveable.me/practice/ap-physics-1-revised). On the exam, be ready to justify comparisons by citing I and ω and possibly using parallel-axis or total-energy statements.

Think of rotational kinetic energy as the sum of the translational kinetic energies of every little piece of a rigid body. Each small mass dm at radius r moves with speed v = ωr, so its KE is (1/2) dm · v² = (1/2) dm · (ω² r²). Integrate over the whole object: K = (1/2) ω² ∫ r² dm. That integral ∫ r² dm is the rotational inertia I about the rotation axis, so K = (1/2) I ω². So you need both factors because they answer two different questions: - I (the moment of inertia) encodes how much mass is involved and how it’s distributed relative to the axis—masses far from the axis contribute much more. - ω² gives the square of the angular speed, turning that geometry into actual linear speeds (v = ωr) and therefore kinetic energy. This matches the CED essential knowledge (6.1.A.1 and 6.1.A.1.i): rotational KE is equivalent to the sum of translational KEs and depends on rotational inertia and angular velocity. For more practice and the topic study guide, see Fiveable’s Topic 6.1 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-6/1-rotational-kinetic-energy/study-guide/OM0z7GYjhkcWoIlZ) and unit overview (https://library.fiveable.me/ap-physics-1-revised/unit-6).

Rotational kinetic energy depends on the axis because Krot = 1/2 I ω^2 and the moment of inertia I changes with the axis. Use the parallel-axis theorem: Iaxis = Icm + M d^2 (d = distance from center-of-mass axis). So if the body has the same angular speed ω about a new axis a distance d from the CM, Krot(axis) = 1/2 (Icm + M d^2) ω^2 = 1/2 Icm ω^2 + 1/2 M (d ω)^2. Interpretation: Krot about the CM (1/2 Icm ω^2) is the “pure” rotation part; the extra 1/2 M(d ω)^2 is just the translational kinetic energy of the center of mass when it moves with speed vcm = d ω. So changing the axis (increasing d) increases Krot because I grows—rotational KE is a scalar and changes with the chosen rotation axis unless you account separately for CM translation. This idea shows up on the AP exam (CED 6.1.A and the parallel-axis theorem); review Topic 6.1 on Fiveable (https://library.fiveable.me/ap-physics-1-revised/unit-6/1-rotational-kinetic-energy/study-guide/OM0z7GYjhkcWoIlZ) and practice more at (https://library.fiveable.me/practice/ap-physics-1-revised).