4 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 4
The diagram shows a uniform rod swinging with small amplitude as a physical pendulum about a fixed horizontal axis through its upper end. Points P and Q are marked on the rod. How does the period TQ of the motion of point Q compare with the period TP of the motion of point P?
Answer and reasoning
ATQ = √2 TP A student who treats each point as a separate simple pendulum of its own length, with T ∝ √ℓ, picks this: Q is twice as far from the axis as P, giving √2 times the period. The points are joined rigidly, so they cannot swing independently; they share one period.
BTQ = 2TP A student who thinks the time for a cycle grows with the distance traveled picks this: Q's arc is twice as long as P's. Q also moves twice as fast as P, so both points complete a cycle in the same time.
CTQ = TPCorrect The rod is rigid, so every point on it turns through the same angle θ(t) at the same time. P and Q therefore return to their starting positions together, and the whole rod has a single period. Q moves along a longer arc than P, but at a proportionally greater speed.
DTQ = TP/2 A student who thinks points farther from the axis rotate faster picks this: Q's speed is twice P's. Q's greater linear speed comes from its greater distance from the axis; both points have the same angular velocity and the same period.
Working The rod is rigid, so θ(t) is the same for every point. A point at distance r from the axis is displaced along an arc rθ and moves with speed r|dθ/dt|, but it returns to its starting position whenever θ does. Hence TQ = TP (= 2π√(I/(mgd)) for the rod).
For small oscillations, the small-angle approximation and Newton's second law in rotational form give the following equation for a physical pendulum: d²θ/dt² = −ω²θ. Which statement correctly describes the quantity ω in this equation?
Answer and reasoning
AIt is the angular velocity, dθ/dt, which changes as the pendulum swings. A student who reads ω with its meaning from rotational kinematics picks this. The pendulum's angular velocity dθ/dt does change, from zero at the turning points to a maximum at equilibrium, but the ω in the SHM equation is the constant angular frequency √(mgd/I).
BIt is the angular frequency, 2π/T, which is greater for larger amplitudes. A student who thinks larger swings return sooner, so that ω increases with amplitude, picks this. ω² = mgd/I depends only on the pendulum, not on the amplitude, within the small-angle approximation that gives this equation.
CIt is the angular frequency, 2π/T, which stays constant throughout the motion.Correct ω is the angular frequency of the oscillation. Comparing Iα = −mgdθ with d²θ/dt² = −ω²θ gives ω² = mgd/I, which is fixed by the pendulum, so ω is constant; T = 2π/ω.
DIt is the frequency, 1/T, which is the number of oscillations completed each second. A student who treats angular frequency and frequency as the same quantity picks this. The frequency is f = 1/T, in hertz; the angular frequency is ω = 2πf = 2π/T, in radians per second.
Working Newton's second law in rotational form with sin θ ≈ θ: I d²θ/dt² = −mgdθ, so d²θ/dt² = −(mgd/I)θ. Comparing with d²θ/dt² = −ω²θ: ω² = mgd/I, a constant fixed by the pendulum's mass, rotational inertia and center-of-mass distance. ω is the angular frequency, T = 2π/ω, f = ω/(2π); the angular velocity dθ/dt = −θ₀ω sin(ωt + φ) varies.
Pendulum 1 is a simple pendulum: a small, dense ball on a light string of length ℓ. Pendulum 2 is a uniform rod of length ℓ that swings about a fixed horizontal axis through one end. The rotational inertia of a uniform rod of mass M and length ℓ about an axis through one end, perpendicular to the rod, is (1/3)Mℓ². Both pendulums swing with small amplitude. How does the period T₂ of pendulum 2 compare with the period T₁ of pendulum 1?
Answer and reasoning
AT₂ = 0.82 T₁Correct Pendulum 1 is modeled as a point mass at ℓ: T₁ = 2π√(ℓ/g). For the rod, I = (1/3)Mℓ² and d = ℓ/2, so T₂ = 2π√((Mℓ²/3)/(Mgℓ/2)) = 2π√(2ℓ/(3g)). T₂/T₁ = √(2/3) ≈ 0.82. The rod's mass is spread along its length, closer to the axis on average than the ball's.
BT₂ = 1.00 T₁ A student who uses 2π√(ℓ/g) for any pendulum of length ℓ picks this. That equation applies only when the hanging object can be modeled as a point mass at ℓ; the rod is an extended body with T₂ = 2π√(I/(Mgd)).
CT₂ = 0.71 T₁ A student who models the rod as a point mass at its center, ℓ/2 from the axis, gets 2π√(ℓ/(2g)), which is 0.71 T₁. The rod's rotational inertia about the axis, (1/3)Mℓ², is greater than M(ℓ/2)² = (1/4)Mℓ².
DT₂ = 0.58 T₁ A student who measures d to the bottom of the rod, d = ℓ, gets 2π√((Mℓ²/3)/(Mgℓ)) = 2π√(ℓ/(3g)), which is 0.58 T₁. The rod's weight acts at its center of mass, d = ℓ/2.
Working T₁ = 2π√(ℓ/g). T₂ = 2π√(I/(Mgd)) with I = Mℓ²/3, d = ℓ/2: T₂ = 2π√(2ℓ/(3g)). T₂/T₁ = √(2/3) = 0.82. Distractors: same formula → 1.00; I = M(ℓ/2)² → √(1/2) = 0.71; d = ℓ → √(1/3) = 0.58.
A torsion pendulum consists of a horizontal uniform disk suspended from a vertical wire attached to its center. The disk undergoes small rotational oscillations about the wire, for which Iα = −kΔθ. Which single change would increase the period of the oscillations?
Answer and reasoning
AReplacing the disk with a thin ring of equal mass and radiusCorrect The period is T = 2π√(I/k). A thin ring has all its mass at the radius R, so its rotational inertia, MR², is twice that of a uniform disk of the same mass and radius, (1/2)MR². The period becomes √2 times as long.
BTwisting the disk through a larger angle before it is released A student who thinks a wider swing takes longer because the disk turns farther picks this. The restoring torque is proportional to the angle, so a larger amplitude gives proportionally larger angular speeds and the period, 2π√(I/k), is unchanged.
CTaking the apparatus to a location where the value of g is smaller A student who thinks gravity provides the restoring torque, as for a simple pendulum, picks this. The torque comes from the twisted wire, so T = 2π√(I/k) does not depend on g.
DReplacing the disk with a heavier one of the same rotational inertia A student who treats the torsion pendulum like a block on a spring, with T = 2π√(m/k), picks this. For rotational oscillations the inertial quantity is the rotational inertia; with I unchanged, the period is unchanged.
In preparation: 0 of 4 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
7.5.A.1 Physical pendulum Fix
Physical pendulum
A rigid body that oscillates about a fixed axis. For it to oscillate under gravity, the axis must not pass through its center of mass: when the body is turned away from equilibrium, the gravitational force exerted on it, acting at its center of mass, then exerts a torque about the axis directed back toward the equilibrium orientation.
Motion of the points of a rigid pendulum
Because the body is rigid and turns about a fixed axis, every point of it turns through the same angle in the same time. All points therefore share one period and one angular frequency, although points farther from the axis move along longer arcs at greater speeds.
Students often think Each part of a physical pendulum swings like a separate simple pendulum of its own length, so points farther from the axis have longer periods. In fact No. A rigid body turns as a whole about the axis: every point turns through the same angle in the same time, so every point has the same period.
Students often think Points farther from the axis of a rotating rigid body move faster, so they complete their cycles in less time. In fact No. Points farther from the axis have greater linear speeds, but every point has the same angular velocity and completes each cycle in the same time.
7.5.A.2 Pivot-to-center-of-mass distance, d Fix
Pivot-to-center-of-mass distance, d
The distance from the axis of rotation to the pendulum's center of mass. The lever arm of the gravitational force about the axis is d sin θ, so d sets the size of the restoring torque. It is not the length of the object or the distance to its lowest point. SI unit: meter (m).
Rotational inertia about the pivot axis
The I in Tphys = 2π√(I/(mgd)) is the rotational inertia about the axis the pendulum actually swings about. For an axis a distance d from the center of mass it is I = Icm + md² (parallel axis theorem), which is greater than Icm. SI unit: kg·m².
Period of a physical pendulum, Tphys
For small amplitudes, Tphys = 2π√(I/(mgd)), where m is the pendulum's mass, d the distance from the axis to its center of mass and I its rotational inertia about the axis. It is found by applying Newton's second law in rotational form, with the small-angle approximation, and within that approximation it does not depend on the amplitude. SI unit: second (s).
Restoring torque on a physical pendulum
The torque about the axis exerted by the gravitational force on the pendulum, τ = −mgd sin θ. The minus sign shows that it is directed opposite to the angular displacement, back toward equilibrium. The force exerted by the axis acts at the axis and exerts no torque about it. SI unit: newton-meter (N·m).
Angular displacement, θ
The angle between the line from the axis to the center of mass and its equilibrium direction, vertically below the axis. Measured in radians when used in the small-angle approximation. SI unit: radian (rad).
Small-angle approximation
For a small angle θ measured in radians, sin θ ≈ θ, so the restoring torque becomes τ ≈ −mgdθ, proportional to the angular displacement. The approximation is within 1 percent for angles up to about 0.24 rad (about 14°) and worsens as the angle increases.
Angular frequency, ω, of a pendulum
The constant in d²θ/dt² = −ω²θ. For a physical pendulum ω = √(mgd/I), and it is related to the period by T = 2π/ω and to the frequency by ω = 2πf. It is not the pendulum's angular velocity dθ/dt, which changes throughout each cycle. SI unit: rad/s.
Equation of motion for small oscillations
With the small-angle approximation, Newton's second law in rotational form, Iα = −mgdθ, gives d²θ/dt² = −(mgd/I)θ: the same form as d²x/dt² = −ω²x, with ω² = mgd/I. The angular displacement therefore varies sinusoidally with time, θ = θ₀ cos(ωt + φ), where θ₀ is the angular amplitude.
Students often think The time for one cycle grows with the distance traveled in it, so a wider swing, or a point that moves along a longer arc, takes longer to complete a cycle. In fact Not for simple harmonic motion. When the restoring torque is proportional to the angular displacement, a larger swing produces a proportionally larger restoring torque and greater speeds, and the period does not depend on the amplitude.
Students often think The rotational inertia in the physical-pendulum equations is the object's rotational inertia about its center of mass, the value usually tabulated. In fact No. I is the rotational inertia about the axis the pendulum swings about. For an axis a distance d from the center of mass, I = Icm + md², which is greater than Icm.
7.5.A.3 Simple pendulum Fix
Simple pendulum
A pendulum whose hanging object can be modeled as a point mass m at a distance ℓ from the pivot. Then I = mℓ² and d = ℓ, so Tphys = 2π√(I/(mgd)) reduces to Tp = 2π√(ℓ/g), which does not depend on the mass. The model is good when the object is small compared with ℓ and the string or rod holding it has negligible mass.
Length of a simple pendulum, ℓ
The distance from the pivot to the point mass, that is, to the center of mass of the hanging object, not to its lowest point. SI unit: meter (m).
Students often think The period of any pendulum is 2π√(ℓ/g), where ℓ is its length from the pivot to its lowest point. In fact No. Tp = 2π√(ℓ/g) applies to a simple pendulum, whose hanging object can be modeled as a point mass a distance ℓ from the pivot. For an extended rigid body, the period is Tphys = 2π√(I/(mgd)).
Students often think A bigger or heavier pendulum bob swings faster, so it has a shorter period. In fact No. For a simple pendulum the period is 2π√(ℓ/g), independent of the bob's mass. A bob that is not small compared with ℓ swings more slowly, not faster, because its rotational inertia about the pivot exceeds mℓ².
7.5.A.4 Torsion pendulum Fix
Torsion pendulum
A rotating system, such as a horizontal disk suspended at its center of mass from a vertical wire, that oscillates about the wire because the twisted wire exerts a restoring torque proportional to the angular displacement: Iα = −kΔθ. The restoring torque comes from the wire, not from gravity.
Torsion constant, k
The constant of proportionality between the restoring torque exerted by a twisted wire or fiber and the angle through which it is twisted, in Iα = −kΔθ. A stiffer wire has a larger k. SI unit: N·m/rad.
Period of a torsion pendulum
From Iα = −kΔθ, d²θ/dt² = −(k/I)Δθ, so ω = √(k/I) and T = 2π√(I/k). The period depends on the rotational inertia about the wire and on k; it does not depend on g, and it does not depend on the amplitude while the torque stays proportional to the angle. SI unit: second (s).
Students often think The period is the time for one swing from one extreme to the other, for example from a maximum to the next minimum on a θ–t graph. In fact No. The period is the time for one complete cycle, from one position and direction of motion back to the same position and direction: for example, from one maximum of the angular displacement to the next maximum, which is two swings.
Students often think A torsion pendulum behaves like a block on a spring: its period depends on its mass, T = 2π√(m/k), so a heavier object oscillates more slowly. In fact No. For rotational oscillations the inertial quantity is the rotational inertia about the axis, not the mass. A torsion pendulum's period is T = 2π√(I/k); two objects of different mass with the same I have the same period.
13 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 13
A uniform disk of radius R = 0.10 m hangs from a fixed horizontal axle that passes through a point on its rim, perpendicular to the face of the disk. The rotational inertia of a uniform disk of mass M about an axis through its center, perpendicular to its face, is (1/2)MR². The disk is displaced through a small angle and released. What is the period of its oscillations? Use g = 10 m/s².
Answer and reasoning
A0.77 sCorrect The axle is a distance d = R from the center of mass, so by the parallel axis theorem I = (1/2)MR² + MR² = (3/2)MR². Then T = 2π√(I/(Mgd)) = 2π√((3/2)R/g) = 2π√(0.015 s²) ≈ 0.77 s.
B0.44 s A student who uses the rotational inertia about the disk's center, (1/2)MR², gets 2π√(R/(2g)) ≈ 0.44 s. The disk rotates about the axle on its rim, so its rotational inertia about that axle, (3/2)MR², must be used.
C0.63 s A student who models the disk as a point mass at its center, a distance R from the axle, gets 2π√(R/g) ≈ 0.63 s. The disk's mass is spread out, so its rotational inertia about the axle is (3/2)MR², not MR².
D0.89 s A student who treats the disk as a simple pendulum whose length runs from the axle to the bottom of the disk, ℓ = 2R, gets 2π√(2R/g) ≈ 0.89 s. The disk is not a point mass; its period is 2π√(I/(Mgd)) with d = R.
Working d = R = 0.10 m. I = Icm + Md² = (1/2)MR² + MR² = (3/2)MR². T = 2π√(I/(Mgd)) = 2π√((3/2)MR²/(MgR)) = 2π√(3R/(2g)) = 2π√(3(0.10 m)/(2(10 m/s²))) = 2π√(0.015 s²) = 0.77 s.
The diagram shows a uniform rod of mass M and length L that swings about a fixed horizontal axis through the pivot shown, perpendicular to the rod. The rotational inertia of a uniform rod about an axis through its center, perpendicular to the rod, is (1/12)ML². What is the period T of the rod's small-amplitude oscillations? g is the acceleration due to gravity.
Answer and reasoning
AT = 2π√(2L/(3g)) A student who uses the rotational inertia about the rod's center, (1/12)ML², with d = L/8, gets 2π√((L²/12)/(gL/8)) = 2π√(2L/(3g)). The rod rotates about the pivot, which is L/8 from the center, so the term M(L/8)² must be added.
BT = 2π√(L/(8g)) A student who models the rod as a point mass at its center of mass, L/8 from the pivot, gets 2π√(d/g) = 2π√(L/(8g)). The rod's mass is spread along its length, so its rotational inertia about the pivot is far greater than M(L/8)².
CT = 2π√(19L/(120g)) A student who measures d from the pivot to the bottom of the rod, 5L/8, gets 2π√((19L²/192)/(g·5L/8)) = 2π√(19L/(120g)). The gravitational force on the rod acts at its center of mass, so d = L/8.
DT = 2π√(19L/(24g))Correct The center of mass is L/2 from the top end, so it is d = L/2 − 3L/8 = L/8 below the pivot. Parallel axis theorem: I = (1/12)ML² + M(L/8)² = (19/192)ML². T = 2π√(I/(Mgd)) = 2π√((19L²/192)/(gL/8)) = 2π√(19L/(24g)).
Working Center of mass: L/2 from the top end; pivot: 3L/8 from the top, so d = L/2 − 3L/8 = L/8 (center of mass below the pivot). I = Icm + Md² = ML²/12 + ML²/64 = 19ML²/192. Newton's second law in rotational form with sin θ ≈ θ: Iα = −Mgdθ, so T = 2π√(I/(Mgd)) = 2π√((19L²/192)/(gL/8)) = 2π√(19L/(24g)). Checked with sympy; distractors: I = ML²/12 → 2π√(2L/(3g)); I = Md² → 2π√(L/(8g)); d = 5L/8 with the correct I → 2π√(19L/(120g)).
A uniform rod of mass M and length L swings with small amplitude about a fixed horizontal axis through one end, with period T₀. A small object, also of mass M, is then fastened to the other end of the rod. The rotational inertia of a uniform rod about an axis through one end, perpendicular to the rod, is (1/3)ML². What is the new period of small-amplitude oscillations?
Answer and reasoning
A1.00 T₀ A student who thinks adding mass can never change a pendulum's period picks this. Mass cancels for a simple pendulum, but here the added object changes the rotational inertia by a factor of 4 and the gravitational torque by a factor of 3, so the period changes.
B1.15 T₀Correct Rotational inertia about the axis: from (1/3)ML² to (1/3)ML² + ML² = (4/3)ML², a factor of 4. Gravitational torque coefficient: from Mg(L/2) to Mg(L/2) + MgL = (3/2)MgL, a factor of 3. T ∝ √(I/(mgd)), so T = √(4/3) T₀ ≈ 1.15 T₀.
C2.00 T₀ A student who considers only the rotational inertia, which becomes 4 times as large, gets √4 = 2. The added object also triples the gravitational torque about the axis, which shortens the period; the ratio is √(4/3).
D0.58 T₀ A student who considers only the gravitational torque, which becomes 3 times as large, gets 1/√3 ≈ 0.58. The added object also makes the rotational inertia 4 times as large, which lengthens the period; the ratio is √(4/3).
Working Before: I₀ = (1/3)ML², torque coefficient Mg(L/2), T₀ = 2π√((ML²/3)/(MgL/2)) = 2π√(2L/(3g)). After: I = (1/3)ML² + ML² = (4/3)ML²; total mass 2M with center of mass at (M·L/2 + M·L)/(2M) = 3L/4, so mgd = 2Mg(3L/4) = (3/2)MgL. T = 2π√((4ML²/3)/((3/2)MgL)) = 2π√(8L/(9g)). T/T₀ = √((8/9)/(2/3)) = √(4/3) = 1.15.
The diagram shows a pendulum made of a uniform rod with a small object fastened to its lower end. The pendulum swings about a fixed horizontal axis through the rod's upper end and is shown at an angle θ from the vertical. What is the magnitude of the torque about the axis exerted by the gravitational forces on the pendulum? g is the acceleration due to gravity.
Answer and reasoning
A|τ| = (M + m)gL sin θ A student who takes every weight to act at the lowest point of the pendulum, a distance L from the axis, picks this. The rod's weight acts at the rod's center of mass, L/2 from the axis, so its torque is only Mg(L/2) sin θ.
B|τ| = ½(M + m)gL sin θ A student who places the whole weight, (M + m)g, at the midpoint of the rod picks this. The object at the end shifts the center of mass below the rod's midpoint; the object's weight acts at L, giving it a torque mgL sin θ.
C|τ| = ½(M + 2m)gL sin θCorrect The rod's weight Mg acts at its center, L/2 from the axis, with lever arm (L/2) sin θ; the object's weight mg acts at L, with lever arm L sin θ. Total: Mg(L/2) sin θ + mgL sin θ = ½(M + 2m)gL sin θ.
D|τ| = ½(M + 2m)gLθ A student who treats sin θ ≈ θ as exact at any angle picks this. The torque is exactly ½(M + 2m)gL sin θ; replacing sin θ with θ is an approximation that holds only for small angles in radians, and at the angle drawn (about 30°) θ exceeds sin θ by about 5 percent.
Working Torque about the axis = Σ (weight × lever arm). Rod: Mg × (L/2) sin θ. Object: mg × L sin θ. |τ| = (M/2 + m)gL sin θ = ½(M + 2m)gL sin θ. (Equivalently (M + m)g d sin θ with d = (ML/2 + mL)/(M + m).)
A uniform meter stick is mounted on a fixed horizontal axle through a hole at its center, perpendicular to the stick; friction at the axle is negligible. The stick is held at rest at a small angle to the horizontal and then released. Which statement correctly predicts and explains what happens?
Answer and reasoning
AIt oscillates with a shorter period than about any other axle, because its rotational inertia is least. A student who reasons that the least rotational inertia, about the center of mass, gives the shortest period picks this. In Tphys = 2π√(I/(mgd)), d is zero here, so there is no restoring torque and no oscillation at all.
BIt oscillates about the horizontal, because the force from the axle exerts a restoring torque on it. A student who thinks the support provides the restoring torque picks this. The axle's force acts at the axle, so its lever arm about the axle is zero and it exerts no torque about it.
CIt rotates back to horizontal and stays there, because a stick balanced at its center settles back to level. A student who expects every balanced object to settle back to level, like a kitchen scale, picks this. Those devices have their pivot above the center of mass; here the center of mass is on the axle, so gravity exerts no torque at any angle and nothing turns the stick back.
DIt stays at rest at that angle, because the gravitational force exerts no torque on it about the axle.Correct The stick's center of mass is on the axle, so the gravitational force, acting at the center of mass, has zero lever arm about the axle in every orientation. The axle's force also acts at the axle. With zero net torque and zero initial angular velocity, the stick stays at rest: it is in equilibrium at any angle and does not oscillate.
A physical pendulum is released from rest at several different angular amplitudes θ₀, and its period T is measured each time. The graph shows the results. A student claims that the restoring torque exerted on the pendulum is proportional to its angular displacement at every amplitude shown. Which statement correctly uses the graph to evaluate the claim?
Answer and reasoning
ASupported: a torque proportional to θ makes wider swings take longer, because the pendulum travels farther, which is what the graph shows. A student who thinks a wider swing takes longer because the pendulum travels farther picks this. With a torque proportional to θ, the speeds grow in proportion to the amplitude, so the period would not change; the rise in the graph shows the torque is not proportional to θ.
BNot supported: the period rises with amplitude, but a torque proportional to θ gives SHM, whose period is independent of amplitude.Correct If the restoring torque were proportional to θ, the equation of motion would be d²θ/dt² = −ω²θ, which describes SHM, and the period would be the same at every amplitude. The graph shows the period increasing steadily, by about 7 percent at 60°, so at the larger amplitudes the torque is not proportional to θ: it is proportional to sin θ, which is less than θ.
CSupported: the period barely changes below 20°, so a torque proportional to θ, which fits the data there, fits at every amplitude shown. A student who assumes the small-angle approximation holds at any angle picks this. The graph supports the proportional torque only at small amplitudes; above about 20° the period rises clearly, so the proportionality does not hold at every amplitude shown.
DNot supported: a torque proportional to θ is larger for wider swings, so it would make the period fall, not rise, as the amplitude increases. A student who thinks a larger restoring torque at the turning points makes the pendulum return sooner picks this. A torque proportional to θ is larger for wider swings, but the angle to be covered is larger in the same proportion, so the period would stay constant.
A physical pendulum of mass 0.80 kg swings about a fixed horizontal axis that is 0.25 m from its center of mass. Its rotational inertia about a parallel axis through its center of mass is 0.025 kg·m². At one instant the pendulum is at an angular displacement of 6.0° from its equilibrium position. Using the small-angle approximation, what is the magnitude of its angular acceleration at that instant? Use g = 10 m/s².
B8.4 rad/s² A student who divides by the rotational inertia about the center of mass, 0.025 kg·m², gets 8.4 rad/s². The pendulum rotates about an axis 0.25 m from its center of mass, so I = 0.025 kg·m² + (0.80 kg)(0.25 m)² = 0.075 kg·m².
C4.2 rad/s² A student who models the pendulum as a point mass at its center of mass, I = md² = 0.050 kg·m², gets 4.2 rad/s². The pendulum is an extended body: its rotational inertia about the axis includes Icm, so I = 0.075 kg·m².
D160 rad/s² A student who substitutes θ = 6.0 in degrees into τ = −mgdθ gets 160 rad/s². The approximation sin θ ≈ θ holds only for θ in radians: 6.0° = 0.105 rad.
Working I = Icm + md² = 0.025 + (0.80)(0.25)² = 0.075 kg·m². θ = 6.0° × π/180 = 0.105 rad. Small-angle form of Newton's second law in rotational form: |τ| = mgdθ = (0.80 kg)(10 m/s²)(0.25 m)(0.105) = 0.209 N·m. |α| = |τ|/I = 0.209 N·m / 0.075 kg·m² = 2.8 rad/s².
A pendulum consists of a light rigid rod of length L, pivoted about a fixed horizontal axis through its upper end, with two small objects, each of mass m, fastened to it: one at a distance L/4 from the axis and one at the rod's lower end. For small angular displacements θ, the small-angle approximation and Newton's second law in rotational form give d²θ/dt² = −ω²θ. What is ω? g is the acceleration due to gravity.
Answer and reasoning
A1.26√(g/L) A student who replaces the two objects by a point mass 2m at their center of mass, 5L/8 from the axis, gets ω² = g/(5L/8) = 8g/(5L), so ω ≈ 1.26√(g/L). The objects are at different distances, so the rotational inertia, (17/16)mL², is greater than 2m(5L/8)² = (25/32)mL².
B1.00√(g/L) A student who treats the system as a simple pendulum of length L, from the axis to its lowest object, gets ω = √(g/L). The object at L/4 contributes to both the torque and the rotational inertia, and the system is not a single point mass at L.
C5.79√(L/g) A student who gives the period, 2π√(I/(mgd)) = 2π√(17L/(20g)) ≈ 5.79√(L/g), picks this. That is T, in seconds; the ω in the equation is the angular frequency, √(20g/(17L)), in rad/s.
D1.08√(g/L)Correct Rotational inertia about the axis: m(L/4)² + mL² = (17/16)mL². Restoring torque for small θ: −mg(L/4)θ − mgLθ = −(5/4)mgLθ. So (17/16)mL² d²θ/dt² = −(5/4)mgLθ and ω² = (5/4)gL/((17/16)L²) = 20g/(17L), giving ω = √(20/17)√(g/L) ≈ 1.08√(g/L).
Working I = m(L/4)² + mL² = 17mL²/16. Restoring torque (sin θ ≈ θ): τ = −mg(L/4)θ − mgLθ = −(5/4)mgLθ. Iα = τ → d²θ/dt² = −[(5/4)mgL/(17mL²/16)]θ = −(20g/(17L))θ. ω = √(20g/(17L)) = √(20/17)√(g/L) = 1.08√(g/L). Checked with sympy; distractors: point mass 2m at 5L/8 → √(8/5) = 1.26; simple pendulum of length L → 1.00; T = 2π√(17L/(20g)) → 5.79√(L/g).
A pendulum is made from a uniform solid sphere of mass m and radius r fixed to the end of a light rigid rod. The pendulum swings with small amplitude about a fixed horizontal axis through the rod's other end, and the distance from the axis to the sphere's center is ℓ. The radius r is not small compared with ℓ. How does the pendulum's period compare with 2π√(ℓ/g), and why?
Answer and reasoning
AEqual, because the gravitational force acts at the sphere's center, a distance ℓ from the axis. A student who treats the sphere as a point mass at its center picks this. That is correct for the gravitational torque, but not for the rotational inertia: the sphere's own rotational inertia about its center adds to mℓ², so the period is longer.
BLonger, because the sphere's rotational inertia about the axis is greater than mℓ².Correct The gravitational torque is mgℓ sin θ, the same as for a point mass at the sphere's center. But the sphere's mass is spread out, so its rotational inertia about the axis is Icm + mℓ², greater than mℓ². With T = 2π√(I/(mgℓ)), the period is longer than 2π√(ℓ/g).
CShorter, because a big, heavy bob swings faster than a small one on the same rod. A student who thinks bigger or heavier bobs swing faster picks this. The sphere's size increases its rotational inertia about the axis without changing the gravitational torque, so it swings more slowly, not faster.
DLonger, because the pendulum's length is ℓ + r, measured to the bottom of the sphere. A student who applies 2π√(ℓ/g) with the length measured to the lowest point picks this. The period is 2π√(I/(mgℓ)), with I = Icm + mℓ²; it is not the period of a simple pendulum of length ℓ + r.
A horizontal uniform disk of mass 1.2 kg and radius 0.30 m hangs from a vertical wire attached to its center and undergoes rotational oscillations about the wire, with Iα = −kΔθ. The graph shows the disk's angular displacement θ as a function of time t. The rotational inertia of a uniform disk about its central axis is (1/2)MR². What is the torsion constant k of the wire?
Answer and reasoning
A2.1 N·m/rad A student who reads the period as the time from a maximum to the next minimum, 1.0 s, gets 4π²(0.054)/(1.0)² ≈ 2.1 N·m/rad. That is half a cycle; the disk returns to its maximum displacement after 2.0 s.
B0.014 N·m/rad A student who takes the angular frequency to be 1/T = 0.50 s⁻¹ gets k = Iω² = (0.054)(0.50)² ≈ 0.014 N·m/rad. The angular frequency is ω = 2π/T = 3.14 rad/s.
C0.53 N·m/radCorrect From the graph, θ is at a maximum at t = 0 and again at t = 2.0 s, so T = 2.0 s. I = (1/2)(1.2 kg)(0.30 m)² = 0.054 kg·m². Iα = −kΔθ gives ω² = k/I, so k = Iω² = 4π²I/T² = 4π²(0.054 kg·m²)/(2.0 s)² ≈ 0.53 N·m/rad.
D12 N·m/rad A student who uses the spring–object equation T = 2π√(m/k) with the disk's mass gets 4π²(1.2)/(2.0)² ≈ 12 N·m/rad. For rotational oscillations the inertial quantity is the rotational inertia about the wire, 0.054 kg·m², not the mass.
Working Period from the graph: successive maxima at t = 0 and t = 2.0 s, so T = 2.0 s. I = (1/2)MR² = (1/2)(1.2 kg)(0.30 m)² = 0.054 kg·m². Iα = −kΔθ → d²θ/dt² = −(k/I)θ → ω = √(k/I) = 2π/T. k = 4π²I/T² = 4π²(0.054 kg·m²)/(2.0 s)² = 0.53 N·m/rad.
A torsion pendulum consists of a light horizontal rod, suspended at its center from a vertical wire, with two small spheres, each of mass m, fastened to the rod, each a distance r from the wire. The wire exerts a restoring torque proportional to the angle of twist, and the period of rotational oscillation is T₀. Both spheres are moved along the rod to a distance r/2 from the wire. What is the new period?
Answer and reasoning
A1.00 T₀ A student who thinks rotational inertia depends only on mass picks this. The spheres' mass is unchanged, but they are closer to the axis, and I = Σmr² falls to one-fourth of its value.
B0.71 T₀ A student who takes rotational inertia to be proportional to distance gets I halved and T multiplied by √(1/2) ≈ 0.71. For a small object I = mr², so halving r makes I one-fourth as large.
C0.25 T₀ A student who takes the period to be proportional to the rotational inertia gets 1/4. The period is proportional to √I, so a factor of 1/4 in I gives a factor of 1/2 in T.
D0.50 T₀Correct The rotational inertia is 2mr² before and 2m(r/2)² = (1/2)mr² after: one-fourth as large. The wire is unchanged, so T = 2π√(I/k) becomes √(1/4) = 1/2 as long: 0.50 T₀.
Working I₀ = 2mr²; I = 2m(r/2)² = mr²/2 = I₀/4. T = 2π√(I/k), k unchanged: T/T₀ = √(1/4) = 0.50.
A torsion pendulum consists of a horizontal uniform disk of mass M and radius R suspended from a vertical wire attached to its center. When the disk is rotated through an angle Δθ, the wire exerts a restoring torque, so that Iα = −kΔθ. A small object of mass m is fastened to the rim of the disk. The rotational inertia of a uniform disk about its central axis is (1/2)MR². What is the period of the rotational oscillations?
Answer and reasoning
AT = 2π√((M + m)R²/(2k)) A student who thinks rotational inertia depends only on the total mass treats the disk and object together as a uniform disk of mass M + m, I = (1/2)(M + m)R². The object is at the rim, so its rotational inertia is mR², twice what it would contribute as part of a uniform disk.
BT = 2π√((M + 2m)R²/(2k))Correct The rotational inertia about the wire is (1/2)MR² + mR² = (M + 2m)R²/2. From Iα = −kΔθ, d²θ/dt² = −(k/I)Δθ, so ω = √(k/I) and T = 2π/ω = 2π√(I/k) = 2π√((M + 2m)R²/(2k)).
CT = 2π√((M + m)R²/k) A student who gives the disk the rotational inertia of a point mass at its rim, MR², gets I = MR² + mR² and picks this. A uniform disk has much of its mass closer to the axis than R, so its rotational inertia is (1/2)MR².
DT = √(2k/((M + 2m)R²)) A student who gives ω = √(k/I) when the period is asked for picks this. That is the angular frequency, in rad/s; the period is T = 2π/ω = 2π√(I/k).
Working I = Idisk + Iobject = (1/2)MR² + mR² = (M + 2m)R²/2. Iα = −kΔθ → d²(Δθ)/dt² = −(k/I)Δθ, SHM with ω = √(k/I). T = 2π√(I/k) = 2π√((M + 2m)R²/(2k)). Checked with sympy; dimensions: kg·m²/(N·m/rad) = s² (rad dimensionless). Distractors: I = (1/2)(M + m)R²; I = MR² + mR²; ω = √(2k/((M + 2m)R²)) (units s⁻¹) given as T.
A physical pendulum of mass m swings about a fixed horizontal axis. Its center of mass is a distance d from the axis, and its rotational inertia about the axis is I. The pendulum is released from rest at a small angular displacement θ₀, in radians. What is the maximum magnitude of its angular velocity dθ/dt during the motion? g is the acceleration due to gravity.
Answer and reasoning
Aθ₀√(mgd/I)Correct For small angles, Iα = −mgdθ gives d²θ/dt² = −ω²θ with ω = √(mgd/I). Released from rest at θ₀, the pendulum has θ = θ₀ cos(ωt), so dθ/dt = −θ₀ω sin(ωt), whose greatest magnitude is θ₀ω = θ₀√(mgd/I).
Bθ₀·√(2mgd/I) A student who takes the work done by the restoring torque as its value at release, mgdθ₀, multiplied by the whole angle θ₀ sets mgdθ₀² = (1/2)I(dθ/dt)² and gets θ₀·√(2mgd/I). The torque falls in proportion to θ as the pendulum swings down, so the work is the area under the straight-line τ–θ graph, (1/2)mgdθ₀², which gives θ₀√(mgd/I).
Cθ₀mgd/I A student who takes the maximum angular speed to be θ₀ω², the expression for the maximum angular acceleration, gets θ₀(mgd/I). Differentiating θ = θ₀ cos(ωt) once gives a greatest angular speed of θ₀ω; θ₀ω² is the greatest magnitude of d²θ/dt².
D√(2θ₀mgd/I) A student who takes the center of mass to fall through d sin θ₀ ≈ dθ₀, the lever arm in the torque, and sets mgdθ₀ = (1/2)I(dθ/dt)² gets √(2θ₀mgd/I). The center of mass falls only d(1 − cos θ₀); the SHM solution θ = θ₀ cos(ωt) gives a greatest angular speed of θ₀√(mgd/I).
Working Small angles: Iα = −mgdθ, so d²θ/dt² = −(mgd/I)θ, SHM with angular frequency ω = √(mgd/I). Released from rest at θ₀: θ = θ₀ cos(ωt), so dθ/dt = −θ₀ω sin(ωt), whose greatest magnitude is θ₀ω = θ₀√(mgd/I) (units s⁻¹; θ₀ in rad). Cross-check by energy: the restoring torque, of magnitude mgdθ, does work W = ∫τ dθ = (1/2)mgdθ₀² (the area under its straight-line torque–angle graph) = (1/2)I(dθ/dt)²max, the same result. Checked with sympy. Distractors: work of the restoring torque taken as its value at release times θ₀, mgdθ₀² = (1/2)I(dθ/dt)²max → θ₀·√(2mgd/I) (s⁻¹); maximum angular speed taken as θ₀ω² → θ₀mgd/I (s⁻²); center of mass taken to fall d sin θ₀ ≈ dθ₀, so mgdθ₀ = (1/2)I(dθ/dt)²max → √(2θ₀mgd/I) (s⁻¹).
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
