6 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 6
A student is asked to describe the rotational inertia of a rigid system about a fixed axis. Which description is correct?
Answer and reasoning
AHow rapidly it is rotating, so it grows as the system's angular velocity about the axis grows A student who thinks rotational inertia measures how fast a system spins picks this. How fast it rotates is the angular velocity; rotational inertia measures resistance to a change in angular velocity and, for a rigid system, does not depend on it.
BThe torque needed to keep the system turning at a steady angular velocity about that axis A student who thinks rotation needs a continual torque to keep going picks this. Rotational inertia measures a system's resistance to changes in its rotation, not a torque that keeps it turning, and it has units of kg·m², not the N·m of a torque.
CIts total mass, which resists rotation equally wherever on the system the mass sits A student who thinks rotational inertia depends only on mass picks this. The same mass placed farther from the axis gives a greater rotational inertia, so where the mass is matters.
DIts resistance to changes in angular velocity, set by its mass and how far that mass is from the axisCorrect Rotational inertia is the rotational counterpart of mass as a measure of inertia: the greater it is, the harder it is to start, stop, speed up or slow down the rotation about that axis. It depends on the amount of mass and on its distances from the axis, I = ∫ r² dm.
A small object of mass m is fastened to a light rod a distance r from the rod's axis of rotation. It is replaced by a small object of mass m/2 fastened a distance 3r from the same axis. By what factor does the rotational inertia of the system about the axis change?
Answer and reasoning
A×1.5 A student who takes I to be proportional to r, not r², gets ½ × 3 = 1.5. The distance is squared, so tripling it multiplies I by 9.
B×4.5Correct I = mr². Halving the mass multiplies I by ½ and tripling the distance multiplies it by 3² = 9, so the factor is ½ × 9 = 4.5.
C×0.5 A student who thinks rotational inertia depends only on mass gets a factor of ½ from the smaller mass. The object is also three times as far out, which multiplies I by 9.
D×9.0 A student who attends only to the distance gets 3² = 9. The mass has also been halved, so the factor is 9 × ½ = 4.5.
Working I = mr². New: I′ = (m/2)(3r)² = (9/2)mr². I′/I = 4.5.
The diagram shows three small objects fastened to a light rod that rotates about an axis through point O, perpendicular to the page. What is the rotational inertia of the system about this axis?
Answer and reasoning
A0.29 kg·m² A student who counts the object on the left of O as negative gets −0.040 + 0.080 + 0.25 = 0.29 kg·m². Each contribution is m r², and r² is positive on either side of the axis.
B0.37 kg·m²Correct Add mr² for each object about O: 1.0 × 0.20² + 0.50 × 0.40² + 1.0 × 0.50² = 0.040 + 0.080 + 0.25 = 0.37 kg·m². Every term is positive, whichever side of O the object is on.
C0.90 kg·m² A student who uses mr without squaring gets 0.20 + 0.20 + 0.50 = 0.90, which is in kg·m, not kg·m². Each distance must be squared.
D0.10 kg·m² A student who puts the total 2.5 kg at the center of mass, 0.20 m to the right of O, gets 2.5 × 0.20² = 0.10 kg·m². Because each distance is squared, Σ m r² is not M rcm².
Working Itot = Σ mi ri² = (1.0 kg)(0.20 m)² + (0.50 kg)(0.40 m)² + (1.0 kg)(0.50 m)² = 0.040 + 0.080 + 0.25 = 0.37 kg·m². The object on the left of O contributes a positive term like the others.
The diagram shows a thin, flat, uniform annular disk of mass M, with its inner and outer radii labeled. Treating it as a set of thin coaxial rings like the one outlined, what is its rotational inertia about the axis through its center, perpendicular to its face?
Answer and reasoning
A0.50 MR² A student who uses the solid-disk result ½MR², treating the hole as irrelevant, picks this. With the same mass and no material inside R/3, the mass is farther out on average, so I is larger than ½MR².
B1.00 MR² A student who puts all the mass at the outer radius R, as for a hoop, gets MR². Only the outermost ring is at R; the rings between R/3 and R contribute less.
C0.56 MR²Correct Each ring has dm = σ(2πr dr) with σ = M/[π(R² − R²/9)], and contributes r² dm. Integrating 2πσr³ dr from R/3 to R gives (M/2)(R² + R²/9) = (5/9)MR² ≈ 0.56MR², more than a full disk's ½MR² because no mass is near the axis.
D0.44 MR² A student who places all the mass at the midpoint between the inner and outer radii, (R/3 + R)/2 = 2R/3, gets M(2R/3)² ≈ 0.44MR². Because each ring contributes r² dm, and more of the area (and mass) is at larger r, the average of r² is not the square of the average radius.
Working Surface density σ = M/[π(R² − R²/9)] = 9M/(8πR²). A ring of radius r and width dr has dm = σ(2πr dr) and contributes r² dm. I = ∫ from R/3 to R of 2πσ r³ dr = (πσ/2)(R⁴ − R⁴/81) = (πσ/2)(80R⁴/81). Substituting σ: I = (9M/(8πR²))(π/2)(80R⁴/81) = (5/9)MR² ≈ 0.56MR². Equivalently (M/2)(R² + R²/9).
The diagram shows a thin rod of length L whose linear mass density increases along its length as λ = λ₀x/L, where x is the distance from the rod's narrow left end. The rod can rotate about an axis perpendicular to the page through any one of the labeled points. About which point is the rod's rotational inertia least, and why?
Answer and reasoning
AR, because the rod's center of mass is at x = 2L/3Correct xcm = ∫x dm/M = 2L/3, at R. For rotation in a given plane, I is least about the axis through the center of mass: IR = ML²/18, compared with ML²/12 about Q, ML²/6 about S and ML²/6 about P.
BQ, because it is at x = L/2, the rod's middle A student who expects the least rotational inertia about the middle picks Q. That holds only when the middle is the center of mass; here more mass is toward S, so the center of mass is at 2L/3 and IQ = ML²/12 exceeds IR = ML²/18.
CS, because the densest part, at x = L, is then on the axis A student who expects the least value when the densest part is on the axis picks S. The rest of the rod is then up to L away: IS = ML²/6, three times IR.
DAny of them, because the rod and its mass do not change A student who treats rotational inertia as a property of the rod alone picks this. Moving the axis changes every element's distance from it, giving values from ML²/18 about R to ML²/6 about P and S.
Working xcm = ∫x dm/M = (λ₀L²/3)/(λ₀L/2) = 2L/3, at R. With M = λ₀L/2: Iend = ∫x²λ dx = ML²/2; IR = ML²/2 − M(2L/3)² = ML²/18; IP (L/3) = ML²/18 + M(L/3)² = ML²/6; IQ = ML²/12; IS = ML²/6. Least about R.
An irregularly shaped flat plate can rotate about either of two axes perpendicular to its face. Axis 1 and axis 2 are each a distance d from the plate's center of mass, on opposite sides of it, and axis 1 is nearer the heavier part of the plate. Which claim about the plate's rotational inertias about the two axes, I₁ and I₂, is correct?
Answer and reasoning
AI₁ < I₂, because axis 1 is nearer the plate's heavier part A student who expects a smaller rotational inertia when the axis is near the heavy part picks this. The mass distribution is already accounted for in the position of the center of mass; equal d gives equal I.
BI₁ = I₂, as both axes are equally far from the center of massCorrect Both axes are parallel to the axis through the center of mass and a distance d from it, so the parallel axis theorem gives each Icm + Md². The theorem holds for any rigid object, so the plate's irregular shape does not matter.
CI₁ > I₂, as the heavy part near axis 1 has less leverage A student who thinks mass far from an axis is easier to turn, by leverage, concludes that the plate is harder to turn about axis 1. Mass far from the axis increases I, and here the parallel axis theorem gives equal values.
DThey cannot be compared, as the plate is not symmetric A student who thinks the parallel axis theorem needs a symmetric object picks this. The theorem follows from the definition of the center of mass and holds for any rigid system.
In preparation: 0 of 6 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
5.4.A.1 Rotational inertia (I) Fix
Rotational inertia (I)
A measure of a rigid system's resistance to changes in its rotation about a given axis. It depends on the system's mass and on how that mass is distributed relative to the axis. SI unit: kg·m².
Mass distribution relative to the axis
How far the parts of a system's mass are from the axis of rotation. Of two systems with equal mass, the one with more of its mass farther from the axis has the greater rotational inertia: a hoop (MR²) has more than a uniform disk (½MR²) of the same mass and radius about their central axes.
Students often think Rotational inertia depends only on the mass of a system, so systems of equal mass have equal rotational inertia, and doubling the mass doubles it whatever else changes. In fact Not in general. Rotational inertia depends on how the mass is distributed relative to the axis as well as on how much mass there is, so a hoop has more rotational inertia than a uniform disk of the same mass and radius.
Students often think Rotational inertia measures how fast or how much a system is rotating, so it is larger when the system spins faster. In fact No. How fast a system rotates is its angular velocity. Rotational inertia measures how strongly the system resists a change in its angular velocity; for a rigid system about a fixed axis it does not change when the system speeds up or slows down.
5.4.A.2 Rotational inertia of a small object Fix
Rotational inertia of a small object
For an object small enough to be treated as a point, a perpendicular distance r from the axis: I = mr². Because r is squared, doubling the distance multiplies the rotational inertia by four.
Perpendicular distance from the axis (r)
The shortest distance from the axis of rotation to the object, measured along a line perpendicular to the axis (SI unit: m). For an object on a rod tilted at angle θ to the axis, r = ℓ sin θ, not the rod length ℓ.
Students often think Rotational inertia is proportional to distance from the axis (I = mr, or ∫ r dm, or Md in the parallel axis theorem), so doubling the distance doubles it. In fact No. I = mr², so doubling r multiplies I by four; in an integral the integrand is r² dm, not r dm.
Students often think Rotational inertia depends only on how far the mass is from the axis, so objects of the same size or at the same distance have the same rotational inertia. In fact No. I = mr² is proportional to m as well as to r², so two objects at the same distance have rotational inertias in the ratio of their masses.
5.4.A.3 Total rotational inertia of a collection of objects Fix
Total rotational inertia of a collection of objects
The sum of the rotational inertias of the objects about the same axis: Itot = Σ Ii = Σ mi ri². Every term is positive or zero, whichever side of the axis an object is on; an object on the axis contributes zero.
Light rod
A connecting rod whose mass is negligible, so that it adds nothing to the rotational inertia of the system it is part of.
Students often think Mass on one side of the axis counts negatively, so contributions from opposite sides partly cancel, as torques in opposite senses do. In fact No. Each object contributes m r², which is positive whichever side of the axis it is on; contributions from both sides add.
Students often think A system's rotational inertia equals M r², with r a single representative distance: the distance from the axis to the system's center of mass, or the midpoint between the nearest and farthest parts of its mass. In fact No. Because each distance is squared, Σ mi ri² (or ∫ r² dm) is not M times the square of an average distance. Two equal objects on opposite sides of an axis have their center of mass on the axis, yet their rotational inertia about it is not zero.
5.4.A.4 Rotational inertia as an integral Fix
Rotational inertia as an integral
For an extended solid treated as a collection of differential masses dm, I = ∫ r² dm, where r is the perpendicular distance from each dm to the axis. The limits run over the whole object, on both sides of the axis.
Linear mass density (λ)
Mass per unit length of a thin rod, λ = dm/dx (SI unit: kg/m), so dm = λ dx. For a nonuniform rod λ varies with position, and the total mass is M = ∫ λ dx.
Coaxial ring element
A thin ring of radius r and width dr, coaxial with the axis, used to build a disk or annular disk. All of its mass is at distance r, so it contributes r² dm; for a uniform flat disk of surface density σ, dm = σ(2πr dr).
Standard rotational inertias
Results of I = ∫ r² dm for uniform objects of mass M: thin hoop or cylindrical shell about its axis, MR²; disk about its axis, ½MR²; thin rod of length L about a perpendicular axis through its center, ML²/12, and through one end, ML²/3.
Students often think An object's rotational inertia is fixed by its shape and total mass, so the standard formula for that shape applies even when the mass is not uniformly distributed. In fact No. ML²/3 is the result of ∫ r² dm for a uniform rod. If the density varies along the rod, the integral must be evaluated with the actual λ(x).
Students often think The distance r in the integral is measured from the end of the object, wherever the axis is, so the end-axis result applies to any axis. In fact Only when the axis is at that end. r is the distance from the axis, so for an axis inside the rod the integral runs over negative and positive positions measured from the axis.
5.4.B.1 Axis through the center of mass Fix
Axis through the center of mass
For rotation in a given plane, a rigid system's rotational inertia is least about the axis through its center of mass; about any parallel axis it is greater.
Students often think Rotational inertia is a fixed property of an object, like its mass, so it has the same value whichever axis the object rotates about. In fact No. Rotational inertia is defined relative to an axis; moving the axis changes every part's distance from it, and for parallel axes I′ = Icm + Md².
Students often think Rotational inertia is least about an axis through the middle of an object, whatever its mass distribution. In fact Only if the geometric middle is the center of mass. For rotation in a given plane, rotational inertia is least about the axis through the center of mass.
5.4.B.2 Parallel axis theorem Fix
Parallel axis theorem
I′ = Icm + Md², where Icm is the rotational inertia about an axis through the center of mass, M is the system's total mass and d is the perpendicular distance between that axis and the parallel axis of interest. The Md² term is never negative, so I′ is never less than Icm.
Students often think When the axis moves, the object can be treated as a single object at its center of mass, so I′ = Md² and the object's own Icm drops out. In fact No. I′ = Icm + Md²: Icm alone describes rotation about the center of mass only; the Md² term must be added for any parallel axis.
Students often think The d in the parallel axis theorem is the distance from the axis to the object itself, such as its nearest point, its point of attachment or its far edge. In fact No. d is the perpendicular distance between the new axis and the parallel axis through the center of mass.
10 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 10
The diagram shows three objects, each of which can rotate about a fixed axis through its center, perpendicular to the page. Each object's type and mass are labeled. Which ranking of their rotational inertias about these axes, I₁, I₂ and I₃, is correct?
Answer and reasoning
AI₃ > I₁ = I₂ A student who thinks rotational inertia depends only on mass ranks by mass: 2M first, then the two objects of mass M equal. The hoop's mass is all at the rim, so it has twice the rotational inertia of the disk of equal mass.
BI₃ > I₂ > I₁ A student who thinks a solid object has more rotational inertia than a hollow one of the same mass puts both disks above the hoop. With equal mass, the hoop's mass is farther from the axis, so I₁ = MR² is greater than I₂ = ½MR².
CI₁ = I₃ > I₂Correct The hoop has all its mass at distance R, I₁ = MR². A uniform disk has much of its mass nearer the axis, so I₂ = ½MR². Doubling the disk's mass doubles its rotational inertia: I₃ = ½(2M)R² = MR², equal to the hoop's.
DI₁ = I₂ = I₃ A student who thinks rotational inertia depends only on how far the mass extends from the axis gives all three the same value, since all have radius R. Mass and its distribution both matter: I₂ = ½MR² while I₁ = I₃ = MR².
Working All three have radius R. Hoop: all mass at R, I₁ = MR². Uniform disk: I₂ = ½MR². Uniform disk of mass 2M: I₃ = ½(2M)R² = MR². So I₁ = I₃ > I₂.
Two thin rods, X and Y, have the same mass and length. In rod X most of the mass is concentrated equally near the two ends; in rod Y most of the mass is concentrated near the middle. Each rod can rotate about an axis through its midpoint, perpendicular to the rod. Which claim about the rods' rotational inertias about these axes is correct?
Answer and reasoning
ARod X's is greater, because more of its mass is far from the axisCorrect Each bit of mass contributes r² dm, so mass far from the axis contributes much more than mass near it. Rod X has most of its mass near the ends, far from the midpoint axis, so its rotational inertia is greater.
BRod Y's is greater, because rod X's end masses give it more leverage A student who thinks mass far from the axis is easier to turn, as a force far from the axis gives more torque, picks this. Mass far from the axis increases the rotational inertia, so rod X is the harder one to set rotating.
CThey are equal, because the rods have the same mass and length A student who thinks rotational inertia is set by mass (and size) alone picks this. The distribution of the mass also matters, and the two rods distribute their equal masses differently.
DIt depends on where each rod is pushed, not on the rods alone A student who mixes up torque and rotational inertia picks this. Where a rod is pushed affects the torque on it, not its rotational inertia, which depends only on its mass and the mass's distances from the axis.
The diagram shows a small object fastened to the end of a light rod. The rod's other end is fixed to a vertical axis, and the system rotates about that axis with the rod at the angle shown. What is the rotational inertia of the system about the axis?
Answer and reasoning
A0.020 kg·m²Correct r is the perpendicular distance from the axis: (0.40 m) sin 30° = 0.20 m. Then I = mr² = (0.50 kg)(0.20 m)² = 0.020 kg·m².
B0.080 kg·m² A student who uses the rod's length as r gets (0.50)(0.40)² = 0.080 kg·m². The object moves in a circle whose radius is its perpendicular distance from the axis, (0.40 m) sin 30° = 0.20 m.
C0.060 kg·m² A student who finds the distance with cos 30° gets r ≈ 0.35 m and (0.50)(0.35)² ≈ 0.060 kg·m². With the angle measured from the axis, the perpendicular distance is the opposite side, (0.40 m) sin 30°.
D0.100 kg·m² A student who uses I = mr without squaring gets (0.50)(0.20) = 0.10, which has units of kg·m, not kg·m². The distance must be squared: (0.50)(0.20)² = 0.020 kg·m².
Working The perpendicular distance from the object to the axis is r = ℓ sin θ = (0.40 m)(sin 30°) = 0.20 m. I = mr² = (0.50 kg)(0.20 m)² = 0.020 kg·m². The light rod contributes nothing.
Four small objects are fastened at the corners of a square frame of light rods, as shown in the diagram. The frame rotates about an axis through point P, perpendicular to the page. What is the rotational inertia of the system about this axis?
Answer and reasoning
A5.0 mL² A student who measures the top objects' distance along the frame, L/2 + L = 1.5L, gets 0.5mL² + 2m(1.5L)² = 5.0mL². The distance in mr² is the straight-line distance to the axis, √5 L/2 ≈ 1.12L.
B1.0 mL² A student who puts the total mass 4m at the center of the square, L/2 from P, gets 4m(L/2)² = 1.0mL². Each object contributes its own m r², and the sum of squares is not M rcm².
C4.0 mL² A student who uses the side length L as every object's distance from P gets 4mL². Only the top objects are more than L/2 from P, and their distance is √5 L/2, not L.
D3.0 mL²Correct The bottom objects are L/2 from P, contributing 2m(L/2)² = 0.5mL². Each top object is a straight-line distance √(L² + L²/4) from P, so each contributes m(5L²/4); together 2.5mL². The total is 3.0mL².
Working The two bottom objects are L/2 from P. The two top objects are √(L² + (L/2)²) = (√5/2)L from P, so r² = 5L²/4. I = 2m(L²/4) + 2m(5L²/4) = (1/2 + 5/2)mL² = 3.0 mL².
A thin rod of length L and total mass M has linear mass density λ = λ₀x/L, where x is the distance from end A and λ₀ is a positive constant. What is the rod's rotational inertia about an axis through end A, perpendicular to the rod?
Answer and reasoning
A(1/3)ML² A student who uses the uniform-rod result for an end axis picks this. That result assumes λ = M/L everywhere; here the mass is concentrated toward the far end, which raises I to (1/2)ML².
B(1/4)ML² A student who evaluates I = λ₀L³/4 correctly but then sets λ₀ = M/L gets (1/4)ML². λ₀ is the density at the far end; integrating λ gives M = λ₀L/2, so λ₀ = 2M/L.
C(1/2)ML²Correct First find the mass: M = ∫₀ᴸ (λ₀x/L) dx = λ₀L/2. Then I = ∫₀ᴸ x²(λ₀x/L) dx = λ₀L³/4 = (λ₀L/2)(L²/2) = (1/2)ML². More of the mass is near the far end than for a uniform rod, so I exceeds ML²/3.
D(4/9)ML² A student who finds the center of mass, xcm = 2L/3, and puts all the mass there gets M(2L/3)² = (4/9)ML². Each element contributes x² dm, and the integral of x² is not M times xcm².
Working M = ∫₀ᴸ (λ₀x/L) dx = λ₀L/2. I = ∫ x² dm = ∫₀ᴸ x²(λ₀x/L) dx = λ₀L³/4 = (λ₀L/2)(L²/2) = (1/2)ML².
A thin uniform rod of mass 2.0 kg and length 1.2 m rotates about an axis perpendicular to the rod through a point 0.30 m from one end. Using I = ∫ r² dm, what is the rod's rotational inertia about this axis?
Answer and reasoning
A0.39 kg·m² A student who subtracts the short piece on the other side of the axis gets λ(0.90³ − 0.30³)/3 = 0.39 kg·m². Every element contributes x² dm, which is positive on both sides of the axis.
B0.42 kg·m²Correct Put the origin on the axis: the rod runs from −0.30 m to +0.90 m with λ = 2.0/1.2 kg/m. I = λ∫x² dx = λ(0.90³ + 0.30³)/3 = 0.42 kg·m². The short piece on the far side of the axis adds a positive amount, since x² > 0 there.
C0.96 kg·m² A student who measures r from the end of the rod uses the end-axis result ML²/3 = (2.0)(1.2)²/3 = 0.96 kg·m². The axis is 0.30 m in from the end, so the distances must be measured from there.
D0.24 kg·m² A student who treats rotational inertia as the same about any axis uses the center result ML²/12 = 0.24 kg·m². The axis is 0.30 m from the center, which adds M(0.30 m)² = 0.18 kg·m².
Working λ = M/L = 2.0 kg / 1.2 m = 1.67 kg/m. With x measured from the axis, the rod runs from x = −0.30 m to x = +0.90 m. I = ∫ x² λ dx = λ[x³/3] from −0.30 to 0.90 = (1.67 kg/m)(0.729 + 0.027)/3 m³ = 0.42 kg·m². (Check: ML²/12 + M(0.30 m)² = 0.24 + 0.18 = 0.42 kg·m².)
A student claims that, for rotation in a given plane, any rigid object's rotational inertia is least about the axis through its center of mass. Which reasoning correctly supports the claim?
Answer and reasoning
AContributions r² dm on opposite sides of the center of mass cancel, leaving only the unbalanced mass A student who thinks contributions from opposite sides of an axis cancel picks this. Every element contributes r² dm > 0, so nothing cancels; the minimum comes from the Md² term, not from cancellation.
Bxcm is at the object's middle, where its mass is closest to the axis on average A student who identifies the center of mass with the geometric middle picks this. For a nonuniform object they are different points, and the claim is about the center of mass, wherever it is.
CGravity's torques balance about the center of mass, Σ τ = 0, so it turns most easily there A student who mixes up rotational inertia with the torques exerted on an object picks this. Rotational inertia depends only on the mass distribution relative to the axis, not on gravity or any other force.
DAbout a parallel axis a distance d away, I′ = Icm + Md², and Md² is positive for any nonzero dCorrect The parallel axis theorem relates the rotational inertia about any parallel axis to Icm. Md² can only add to Icm, and it is zero only when the axis passes through the center of mass, so Icm is the least value in that plane.
Working Parallel axis theorem: I′ = Icm + Md². Since M > 0 and d² ≥ 0, I′ ≥ Icm, with equality only for d = 0, the axis through the center of mass.
A uniform disk of mass M and radius R rotates about an axis through its center, perpendicular to its face. The axis is moved, parallel to itself, to a point on the disk's rim. By what factor does the disk's rotational inertia change?
Answer and reasoning
A×3Correct I′ = Icm + Md² with Icm = ½MR² and d = R gives (3/2)MR², three times ½MR².
B×1 A student who treats rotational inertia as the same about any axis picks this. Moving the axis to the rim puts most of the disk's mass farther from the axis, and the parallel axis theorem adds MR².
C×2 A student who treats the disk as a point object at its center, so that I′ = Md² = MR², gets MR²/(½MR²) = 2. The disk still rotates about its own center as well, so Icm must be kept: I′ = ½MR² + MR².
D×9 A student who measures d to the far edge of the disk, d = 2R, gets ½MR² + 4MR² = 4.5MR², nine times Icm. d is the distance from the axis to the center of mass, which is R.
Working Icm = ½MR². About the rim axis, d = R: I′ = ½MR² + MR² = (3/2)MR². I′/Icm = 3.
A uniform solid sphere of mass M and radius R has rotational inertia (2/5)MR² about an axis through its center. It is fixed to the end of a light rod whose other end is at a pivot; the rod meets the sphere's surface, and the sphere's center is a distance 2R from the pivot. What is the rotational inertia of the system about an axis through the pivot, perpendicular to the rod?
Answer and reasoning
A4.0 MR² A student who treats the sphere as a point object at its center gets M(2R)² = 4.0MR². The sphere also rotates about its own center as it swings, so its (2/5)MR² must be added.
B1.4 MR² A student who measures d from the pivot to where the rod meets the sphere, a distance R, gets (2/5)MR² + MR² = 1.4MR². d must be measured to the sphere's center of mass, 2R from the pivot.
C1.6 MR² A student who puts the pivot distance 2R in place of R in (2/5)MR² gets (2/5)M(2R)² = 1.6MR². R in that result is the sphere's own radius; the distance to the pivot enters through Md².
D4.4 MR²Correct The center of mass of the sphere is 2R from the pivot, so I′ = Icm + Md² = (2/5)MR² + M(2R)² = 4.4MR².
Working Parallel axis theorem with d = 2R (pivot to the sphere's center): I′ = (2/5)MR² + M(2R)² = (0.4 + 4)MR² = 4.4MR². The light rod adds nothing.
A thin, flat disk of radius R and total mass M has surface mass density σ = σ₀r/R, where r is the distance from the disk's center and σ₀ is a positive constant. Treating the disk as a set of thin coaxial rings, what is its rotational inertia about the axis through its center, perpendicular to its face?
Answer and reasoning
A0.50 MR² A student who uses the standard result for a disk, (1/2)MR², picks this. That result is ∫ r² dm for a uniform disk; here more of the mass is near the rim, so I is larger.
B0.40 MR² A student who takes σ₀ to be M/(πR²), the average surface density, picks this: (2/5)πσ₀R⁴ becomes (2/5)MR². σ₀ is the density at the rim; integrating gives M = (2/3)πσ₀R², so σ₀ = 3M/(2πR²).
C0.60 MR²Correct Each ring has mass dm = σ₀(r/R)2πr dr and contributes r² dm. Integrating from 0 to R, M = (2/3)πσ₀R² and I = (2/5)πσ₀R⁴, so I = (3/5)MR² = 0.60 MR². It exceeds the uniform disk's (1/2)MR² because the density increases toward the rim.
D1.00 MR² A student who places all of the mass at the outer radius picks this: MR², the result for a hoop. The rings inside the rim are closer to the axis and each contributes less than its mass times R².
Working A ring of radius r and width dr has area 2πr dr and mass dm = σ₀(r/R)2πr dr; every part of it is a distance r from the axis. M = ∫₀ᴿ 2πσ₀r²/R dr = (2/3)πσ₀R². I = ∫₀ᴿ r² dm = ∫₀ᴿ 2πσ₀r⁴/R dr = (2/5)πσ₀R⁴. I/M = (3/5)R², so I = (3/5)MR² = 0.60 MR². Distractors: the uniform-disk result (1/2)MR² = 0.50 MR²; σ₀ taken as M/(πR²), so (2/5)πσ₀R⁴ = 0.40 MR²; all the mass at radius R, 1.00 MR². Key and distractors checked with sympy.
Compiled from the AP Physics C: Mechanics Course and Exam Description (effective Fall 2024, 2026 reissue) and our question bank · Specialist review in progress. How these pages are made · Free, no account
