Rotational inertia (moment of inertia) is an object’s resistance to changes in rotational motion about a specific axis. Unlike “regular” inertia (mass), which measures resistance to linear acceleration, rotational inertia depends on both total mass and how that mass is distributed relative to the rotation axis: I = Σ m_i r_i^2 for point masses or I = ∫ r^2 dm for continuous bodies. So a hoop and a solid disk can have the same mass but very different I because more mass is farther from the axis. Key facts from the CED: I is smallest when the axis passes through the center of mass and you can use the parallel-axis theorem I′ = I_cm + M d^2 to shift axes. AP expectations: you should be comfortable using I = m r^2, summing/integrating I, and applying the parallel-axis theorem; calculus derivations are required for some shapes per the boundary. For a focused study guide and practice problems, check the Topic 5.4 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-5/4-rotational-inertia/study-guide/lSrDkHqB6EviD5CA) and the Unit 5 overview (https://library.fiveable.me/ap-physics-c-mechanics/unit-5). For lots of practice problems, see (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Because rotational inertia depends on how the mass is distributed, the hoop resists rotation more than the solid disk even if they have the same mass and radius. For a thin hoop I = MR^2; for a solid disk I = (1/2)MR^2. When either rolls without slipping, gravitational potential energy splits into translational KE (½Mv^2) and rotational KE (½Iω^2). More of the energy goes into rotation for the hoop (larger I), leaving less for translation, so it gains less linear speed and accelerates slower down the ramp. You can see this in the acceleration formula for rolling without slipping: a = g sinθ / (1 + I/(MR^2)). Plugging in I gives a_hoop = g sinθ/2 (≈0.50 g sinθ) and a_disk = g sinθ/(3/2) = (2/3) g sinθ (≈0.67 g sinθ). So the disk reaches the bottom faster. This is exactly the CED idea: rotational inertia depends on mass distribution (Topic 5.4). For a quick refresh, check the Topic 5.4 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-5/4-rotational-inertia/study-guide/lSrDkHqB6EviD5CA) and try practice problems (https://library.fiveable.me/practice/ap-physics-c-mechanics).

I = m r^2 is the moment of inertia for a point mass m at a perpendicular distance r from the rotation axis. Use it like this: - Single point mass: I = m r^2. Example: a 2.0 kg mass at r = 0.50 m gives I = 2.0·(0.50)^2 = 0.50 kg·m^2. - Several point masses: add them: I_tot = Σ m_i r_i^2 (CED 5.4.A.3). - Continuous bodies: treat the object as dm elements and integrate I = ∫ r^2 dm (CED 5.4.A.4). That’s how you get standard formulas (e.g., disk I = 1/2 m r^2). - Axis not through CM: use the parallel-axis theorem I' = I_cm + M d^2 (CED 5.4.B.1). d is distance between parallel axes. On the AP C exam you may need to do the dm integral for rods, disks, shells (see the Boundary Statement). Want worked examples and practice? Check the Topic 5.4 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-5/4-rotational-inertia/study-guide/lSrDkHqB6EviD5CA) and try problems at (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Use I' = I_cm + M d^2 whenever you need the moment of inertia of a rigid object about an axis that is parallel to an axis through the object’s center of mass. Key points to check before using it: - The two axes must be parallel. - d is the perpendicular distance between the CM axis and the new axis. - I_cm is the object's moment of inertia about the axis through its center of mass (same orientation). - M is the object’s total mass. Why it works: the theorem shifts every dm’s r^2 by adding the same Md^2 term that comes from the CM’s offset (see CED 5.4.B.2 and EK 5.4.A.1–4). Use it for rigid bodies and composite systems (sum I' of parts). Example: a uniform rod length L: I_cm = (1/12)ML^2, so about one end I_end = (1/12)ML^2 + M(L/2)^2 = (1/3)ML^2. Don’t use it if axes aren’t parallel or if you only need point-mass distances (then I = Σ m r^2). For a concise refresher, see the Topic 5.4 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-5/4-rotational-inertia/study-guide/lSrDkHqB6EviD5CA). For more practice, check the Unit 5 overview (https://library.fiveable.me/ap-physics-c-mechanics/unit-5) and 1000+ practice problems (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Rotational inertia (I) measures how hard it is to change an object’s rotation—it depends on mass and how far that mass is from the rotation axis (I = ∫ r² dm or for point masses I = Σ m r²). When the axis goes through the center of mass (CM), I is at its minimum for that plane. If the axis is shifted so it doesn’t pass through the CM, the object’s I increases because every mass element is, on average, farther from the axis. The parallel-axis theorem gives the exact change: I′ = I_cm + M d², where M is the total mass and d is the distance between the two parallel axes. So you can quickly get I about any parallel axis from I_cm. For more examples, derivations, and AP-aligned practice on Topic 5.4 see the Fiveable study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-5/4-rotational-inertia/study-guide/lSrDkHqB6EviD5CA) and the Unit 5 overview (https://library.fiveable.me/ap-physics-c-mechanics/unit-5). Practice problems: (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Think of rotational inertia I as the rotational equivalent of mass: it measures how hard it is to change an object's rotation. For a point mass the formula is I = m r^2, so moving the same mass farther from the axis (bigger r) multiplies its contribution by r^2. For a composite object you add those contributions I_tot = Σ m_i r_i^2 (or I = ∫ r^2 dm for a continuous body), so weight far from the axis dominates I. Physically: for the same torque τ, angular acceleration α = τ/I, so a larger I (mass farther out) gives smaller α—it “resists” spin more. Also in energy terms, more mass at large r stores more rotational kinetic energy ½Iω^2 for the same ω, so gravity/torque must do more work to spin it up. That’s why a hoop (mass at r) has larger I than a solid disk (mass spread near center). For more on AP C expectations (including using calculus for disks/rods) see the Topic 5.4 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-5/4-rotational-inertia/study-guide/lSrDkHqB6EviD5CA). For extra practice, try problems on Unit 5 (https://library.fiveable.me/ap-physics-c-mechanics/unit-5) or the practice bank (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Start with I = ∫ r^2 dm and model the uniform thin rod (mass M, length L) as a line of linear mass density λ = M/L. Choose the axis perpendicular to the rod through its center. Take a small element at position x (measured from center), width dx, mass dm = λ dx. Then I = ∫ r^2 dm = ∫_{-L/2}^{L/2} x^2 (λ dx) = λ ∫_{-L/2}^{L/2} x^2 dx. Evaluate the integral: ∫_{-L/2}^{L/2} x^2 dx = 2∫_0^{L/2} x^2 dx = 2[(L/2)^3/3] = L^3/12. So I_center = λ (L^3/12) = (M/L)(L^3/12) = (1/12) M L^2. Axis through one end: either integrate from 0 to L, I_end = (M/L) ∫_0^L x^2 dx = (M/L)(L^3/3) = (1/3) M L^2, or use the parallel-axis theorem: I_end = I_cm + M d^2 = (1/12)ML^2 + M(L/2)^2 = (1/3)ML^2. This follows AP CED Topic 5.4 (use calculus for thin rods). For a quick review, see the Topic 5.4 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-5/4-rotational-inertia/study-guide/lSrDkHqB6EviD5CA) and tons of practice problems at (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Rotational inertia I = ∫ r^2 dm depends on how far each bit of mass is from the rotation axis. The center of mass (CM) is the point that minimizes the average squared distance r^2 of the mass distribution, so putting the axis through the CM makes those r values as small (on average) as possible—hence I is minimal. Mathematically the parallel-axis theorem shows this directly: for any axis a distance d from the CM, I′ = Icm + M d^2. The extra term M d^2 is always positive (unless d = 0), so any axis not through the CM increases I by exactly M d^2. Intuitively, moving the axis away from the CM shifts every mass element slightly farther on average, raising the sum of m r^2 and so increasing resistance to angular acceleration. If you want more practice or the CED-aligned explanation, see the Topic 5.4 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-5/4-rotational-inertia/study-guide/lSrDkHqB6EviD5CA) and try related problems at (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Saying “rotational inertia measures resistance to changes in rotation” is just the rotational analogue of mass in linear motion. For a given torque τ, an object’s angular acceleration α is α = τ / I, so larger I means smaller α for the same torque—it’s harder to speed up or slow down the spin. I depends on total mass and how that mass is distributed about the rotation axis: a point mass at distance r gives I = m r^2, and for many pieces I_tot = Σ m_i r_i^2 or I = ∫ r^2 dm for continuous bodies (CED 5.4.A.2–4). Rotational inertia is smallest when the axis goes through the center of mass, and shifts by M d^2 for a parallel axis (parallel-axis theorem, 5.4.B.2). So a hoop (mass far from axis) has bigger I than a solid disk of the same mass and radius—it “resists” changes in rotation more. For AP review and worked examples, see the Topic 5.4 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-5/4-rotational-inertia/study-guide/lSrDkHqB6EviD5CA) and practice problems (https://library.fiveable.me/practice/ap-physics-c-mechanics).

If several objects rotate about the same axis, just add each object’s moment of inertia about that axis. For point-mass pieces use I_tot = Σ m_i r_i^2, where r_i is the perpendicular distance to the axis (CED 5.4.A.2–3). For continuous bodies use I = ∫ r^2 dm (CED 5.4.A.4). If a body's standard I is given about its center of mass but the rotation axis is parallel and displaced by d, use the parallel-axis theorem: I_axis = I_cm + M d^2 (CED 5.4.B.2). Quick tips: treat thin rods/disks with the calculus derivations the exam expects only when required (see CED boundary). For a composite system, compute each part’s I about the common axis (use point-mass formula, known formulas, or integrals), then sum. Need more worked examples? Check the Topic 5.4 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-5/4-rotational-inertia/study-guide/lSrDkHqB6EviD5CA) and practice problems (https://library.fiveable.me/practice/ap-physics-c-mechanics).

I = ∫ r² dm is just the extension of I = m r² from a single point mass to a continuous solid. For a rigid body you can’t treat all the mass as one point at one distance r from the axis unless it actually is (like a thin hoop). So you break the object into many tiny pieces dm, each at its own perpendicular distance r, sum their contributions I = Σ m_i r_i², and take the limit → integral: I = ∫ r² dm. Practically you use the body’s density (dm = ρ dV, or λ dx for a rod, or σ dA for a shell) and integrate over the shape to get standard formulas (disk, sphere, rod)—which is exactly what the CED says (5.4.A.4) and what you’ll derive for rods/disks in this course. If the axis isn’t through the CM, use I′ = I_cm + M d² (parallel-axis theorem, 5.4.B.2). For worked derivations and practice, see the Topic 5.4 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-5/4-rotational-inertia/study-guide/lSrDkHqB6EviD5CA) and plenty of problems at (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Think of r^2 in I = ∫ r^2 dm as a “levered mass” weight: mass farther from the axis resists changes in rotation much more than mass close in. Two ways to see it physically: - Kinetic-energy view (most direct for AP C): a small mass dm at distance r has tangential speed v = ωr, so its rotational kinetic energy is dK = ½ dm v^2 = ½ dm (ω^2 r^2). Factor out ½ω^2 and the remaining ∫ r^2 dm is the coefficient that converts ω into rotational energy—that coefficient is I. That r^2 comes from v being proportional to r, so energy (and resistance) scale with r squared. - Geometric / distribution view: I is the mass-weighted mean of r^2. Moving mass outward increases I rapidly (quadratically), which is why a hoop (mass at large r) has bigger I than a solid disk of same mass. AP C expects you to use I = ∑ m r^2 and I = ∫ r^2 dm (and the parallel axis theorem I′ = Icm + Md^2); some derivations use calculus per the CED. For a quick refresher, check the Topic 5.4 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-5/4-rotational-inertia/study-guide/lSrDkHqB6EviD5CA) and hit the practice problems (https://library.fiveable.me/practice/ap-physics-c-mechanics) to see this used in examples.

Rotational inertia (moment of inertia) depends on how mass is distributed relative to the rotation axis: I = ∑ m r^2 for point masses or I = ∫ r^2 dm for a continuous body. Moving the axis changes each mass element’s perpendicular distance r, so I changes—mass farther from the axis increases I; mass closer decreases I. For a rigid body, I is smallest when the axis passes through the center of mass (CED 5.4.B.1). If you shift to a parallel axis a distance d away, use the parallel-axis theorem: I′ = Icm + M d^2 (CED 5.4.B.2). Practically: a hoop (mass at large r) has larger I than a solid disk of the same M and R; shifting either axis off-center increases I by M d^2. For AP exam work, be ready to apply I = ∑ m r^2, I = ∫ r^2 dm, and the parallel-axis theorem (see Topic 5.4 study guide: https://library.fiveable.me/ap-physics-c-mechanics/unit-5/4-rotational-inertia/study-guide/lSrDkHqB6EviD5CA) and try practice problems (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Yes—it’s directly about rotational inertia. Rotational inertia (I) measures how hard it is to change an object’s rotation; for point masses I = Σ m r^2, so mass farther from the axis increases I. A spinning skater has (nearly) no external torque, so angular momentum L = I ω is conserved. When the skater pulls their arms in, the average distance r of their mass from the spin axis decreases → I decreases. To keep L constant, ω (angular speed) must increase: ωf = ωi (Ii / If). That’s why they spin faster. This is exactly the CED idea that mass distribution controls rotational inertia and why a hoop (mass far out) has larger I than a solid disk. For a quick review, see the Topic 5.4 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-5/4-rotational-inertia/study-guide/lSrDkHqB6EviD5CA). For more practice applying L = Iω and I = Σ m r^2, check the Unit 5 resources and practice problems (https://library.fiveable.me/ap-physics-c-mechanics/unit-5 and https://library.fiveable.me/practice/ap-physics-c-mechanics).

Rotational inertia matters everywhere you want to control how hard it is to start/stop rotation—it’s about mass and how far that mass sits from the rotation axis (I = ∫ r² dm, or for point masses I = m r²). Real-world examples: - Engineering: Flywheels (energy storage) use large I to smooth power output; shafts and turbine rotors are designed to minimize I where quick response is needed. Use the parallel-axis theorem (I′ = Icm + M d²) when components aren’t centered. - Automotive: Brake rotors and wheel/tire combinations tuned so rims (farther from axis) increase rotational inertia and affect acceleration. - Sports: Ice skaters pull in arms to reduce I and spin faster (conserve angular momentum); a diver tucks to rotate quicker; baseball bats and golf clubs balance mass distribution to adjust swing “moment of inertia” for control vs. power (hoop: I = mR² vs. solid disk: I = ½ mR²). On the AP exam you’re expected to use I = ∫ r² dm and point-mass sums, know qualitative effects of mass distribution, and (for calculus derivations) handle thin rods, disks, shells per the CED. For a focused review, see the Topic 5.4 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-5/4-rotational-inertia/study-guide/lSrDkHqB6EviD5CA) and grab practice problems at (https://library.fiveable.me/practice/ap-physics-c-mechanics).