2 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 2
A uniform disk rests on a horizontal frictionless surface. The diagram shows the two horizontal forces then exerted on it, of equal magnitude, at opposite points on its edge. Which claim about the disk's subsequent motion is correct?
Answer and reasoning
AIt starts rotating, while its center of mass stays at rest A student who thinks forces exerted at the edge produce rotation instead of translation picks this. Every external force contributes to acm, here 2F/M to the right, and the two torques cancel, so the disk does not rotate.
BIt starts rotating, and its center of mass accelerates to the right A student who thinks each off-center force makes the disk rotate on its own picks this. The top force tends to turn the disk clockwise and the bottom force counterclockwise, with equal torques, so the net torque is zero.
CIts center of mass accelerates to the right; it does not rotateCorrect The forces add to 2F to the right, so the center of mass accelerates. Their torques about the center are equal in magnitude, FR, and opposite in sense, so the net torque is zero and the disk's angular velocity stays zero: it is in rotational equilibrium but not translational equilibrium.
DIt moves to the right at a constant velocity, and it does not rotate A student who thinks a constant net force produces a constant velocity picks this. The net force is 2F, so the center of mass accelerates; its velocity does not stay constant.
Working Net force = 2F to the right, so the center of mass accelerates (acm = 2F/M). About the center, the top force's torque is clockwise and the bottom force's is counterclockwise, each of magnitude FR, so the net torque is zero and ω stays zero.
The graph shows the angular velocity of a wheel as a function of time. During which time intervals is the net torque exerted on the wheel not zero?
Answer and reasoning
AFrom 0 s to 7 s, the whole time that the wheel is turning A student who thinks a turning wheel needs a net torque that grows with its angular velocity picks this. From 2 s to 5 s ω is constant, so the net torque is zero even though the wheel is turning fastest.
BFrom 0 s to 5 s, up until the wheel begins to slow A student who thinks a net torque is needed to keep the wheel turning, and that the wheel slows once that torque stops, picks this. From 2 s to 5 s ω is constant, so the net torque is zero; from 5 s to 7 s ω decreases, which requires a net torque opposite to the rotation.
CFrom 0 s to 2 s, while the wheel is speeding up A student who thinks a wheel slows down on its own picks only the speeding-up interval. From 5 to 7 s ω decreases, which also requires a net torque, opposite to the rotation.
DFrom 0 s to 2 s and from 5 s to 7 s, while ω changesCorrect If the torques on a rigid system are not balanced, its angular velocity must be changing, and if ω is constant the net torque is zero. ω changes (nonzero slope) from 0 to 2 s and from 5 to 7 s, so those are the intervals with a nonzero net torque.
Working Net torque is nonzero exactly when ω is changing, where the graph's slope is not zero: from 0 to 2 s (increasing) and from 5 to 7 s (decreasing).
In preparation: 0 of 2 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
5.5.A.1 Rotational equilibrium Fix
Rotational equilibrium
The state of a system whose angular velocity is constant (including zero), which requires the net torque on it to be zero: Σ τi = 0.
Translational equilibrium
The state of a system whose center-of-mass velocity is constant, which requires the net external force on it to be zero. A system can be in either kind of equilibrium without the other.
Couple
Two forces of equal magnitude and opposite direction whose lines of action do not coincide. Their vector sum is zero, but their torques are in the same sense, so they change a system's rotation without accelerating its center of mass.
Force diagram of a rigid system
A diagram showing each force exerted on the system, its relative magnitude and direction, and the point where it is exerted, so that both the net force and the net torque about an axis can be found from it.
Net torque
The sum of the torques exerted on a system about a chosen axis, counting one sense of rotation (clockwise or counterclockwise) as positive: Σ τi. SI unit: N·m.
Choice of axis in equilibrium problems
For a system in equilibrium with both ΣF = 0 and Στ = 0, the net torque is zero about every axis, so the axis can be chosen to pass through the point where an unknown force acts, removing that force from the torque equation.
Rotational form of Newton's first law
A system has a constant angular velocity only if the net torque exerted on it is zero; no torque is needed to keep it rotating at a steady rate.
Students often think If the forces on an object are balanced, the object is in equilibrium in every sense, so its rotation cannot change either. In fact No. Forces that sum to zero can still exert a nonzero net torque if they act along different lines, as two equal and opposite forces at the ends of a rod do. Translational and rotational equilibrium are separate conditions.
Students often think A nonzero net torque makes the object's center of mass accelerate as well as changing its rotation. In fact No. The center of mass accelerates only if the net force is nonzero. A couple exerts a net torque with zero net force, so the object's rotation changes while its center of mass keeps a constant velocity.
5.5.A.2 Unbalanced torques Fix
Unbalanced torques
If the torques exerted on a rigid system do not sum to zero, its angular velocity must be changing: speeding up, slowing down or reversing its sense of rotation.
Students often think Rotating objects slow down on their own, so a decreasing angular velocity needs no net torque; torque is needed only to speed rotation up. In fact No. A decreasing angular velocity is a change in angular velocity, so a nonzero net torque opposite to the rotation must be acting.
Students often think The net torque on a rotating object is greatest when it rotates fastest, because a larger torque is needed for a larger angular velocity. In fact No. Net torque is related to how the angular velocity changes, not to its value; a wheel spinning fast at constant ω has zero net torque.
6 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 6
A rod at rest on a horizontal frictionless surface is pushed by two horizontal forces of equal magnitude in opposite directions, each perpendicular to the rod, one at each end. A student claims that the rod's center of mass stays at rest while the rod's angular velocity changes. Which reasoning correctly supports the claim?
Answer and reasoning
AThe forces sum to zero, so acm is zero, but their torques about the center have the same sense and addCorrect Newton's second law for the center of mass uses the net force, which is zero, so the center of mass stays at rest. About the center, both forces turn the rod in the same sense, so their torques add and the net torque is not zero; the angular velocity therefore changes.
BForces exerted at the ends of a rod can only rotate it, not move its center, since no force acts at the center A student who thinks off-center forces produce rotation instead of translation picks this. An off-center force does accelerate the center of mass; here the center stays at rest only because the two forces sum to zero.
CThe rod's angular velocity is proportional to the net torque, which is nonzero, while the net force is zero A student who links torque to angular velocity itself picks this. A nonzero net torque changes the angular velocity; it does not fix its value, so this reasoning does not show that ω changes.
DA torque is needed to keep any object turning, and the two forces supply it while their sum keeps the center still A student who thinks a torque is needed to maintain rotation picks this. With zero net torque the angular velocity would stay constant; the net torque here changes the angular velocity, which is what the claim states.
Working ΣF = F − F = 0, so acm = 0 and the center of mass, initially at rest, stays at rest. About the center, each force has lever arm L/2 and both turn the rod in the same sense: Στ = FL ≠ 0, so ω changes.
The force diagram shows all the forces exerted on a rod of length L that is initially at rest, drawn at the points where they are exerted. Which claim about the rod's motion just after this instant is correct?
Answer and reasoning
AThe net force is zero, so the net torque is also zero, and the rod stays at rest without rotating A student who treats balanced forces as complete equilibrium picks this. The diagram also shows where the forces act; because they act along different lines, their torques do not cancel.
BThe net force is zero but the net torque is not, so it starts rotating while its center of mass stays at restCorrect The forces sum to 2F − F − F = 0, so the center of mass does not accelerate. The diagram shows where each force acts: about the left end, the two downward forces exert torques FL/2 and FL in the same sense, so the net torque is 3FL/2 and the angular velocity changes.
CThe 2F force is the largest force exerted, so the rod's center of mass accelerates upward A student who decides the motion by the largest single force picks this. The two forces of magnitude F together balance the 2F force, so the net force, and acm, is zero.
DThe net torque is not zero, so the rod starts rotating and its center of mass also accelerates A student who thinks a net torque also accelerates the center of mass picks this. The center of mass responds only to the net force, which is zero here.
Working ΣF = 2F − F − F = 0, so acm = 0. Torques about the left end: the 2F force has zero lever arm; the middle force F has lever arm L/2 and the end force F has lever arm L, both clockwise: Στ = FL/2 + FL = 3FL/2 ≠ 0. The rod begins to rotate while its center of mass stays at rest.
A uniform board of mass 3.0 kg and length 2.0 m rests horizontally on a pivot 0.60 m from its left end. What mass must be placed at the left end of the board for it to remain balanced?
Answer and reasoning
A5.0 kg A student who measures the board's lever arm from its end, 1.0 m, gets m = 3.0 × 1.0/0.60 = 5.0 kg. Lever arms are measured from the pivot: the center is 0.40 m from it.
B3.0 kg A student who thinks a balance needs equal masses on each side picks the board's own mass. Torque depends on lever arm as well as on weight, and the lever arms here are 0.60 m and 0.40 m.
C4.5 kg A student who makes the masses proportional to their distances gets 3.0 × 0.60/0.40 = 4.5 kg. Balance needs m₁d₁ = m₂d₂, so the mass farther from the pivot must be the smaller one.
D2.0 kgCorrect Take torques about the pivot. The board's weight acts at its center, 0.40 m to the right of the pivot; the added mass is 0.60 m to the left. m g (0.60 m) = (3.0 kg) g (0.40 m) gives m = 2.0 kg.
Working The board's weight acts at its center, 1.0 m from the left end, so its lever arm about the pivot is 1.0 m − 0.60 m = 0.40 m. Στ = 0 about the pivot: m g (0.60 m) = (3.0 kg) g (0.40 m), so m = 3.0 × 0.40/0.60 = 2.0 kg.
A uniform rod of mass M is hinged to a wall at one end and held horizontal by a light cable attached to its other end. The cable makes an angle of 37° with the rod. What is the tension in the cable? Take g as the acceleration due to gravity, and use sin 37° = 3/5 and cos 37° = 4/5.
Answer and reasoning
A0.625 Mg A student who uses the cable's component along the rod, T cos 37°, for the torque gets T = Mg/(2 × 0.80) = 0.625Mg. Only the component perpendicular to the rod, T sin 37°, exerts a torque about the hinge.
B1.000 Mg A student who takes the cable's tension to equal the weight it holds up picks this. The tension is fixed by the torque balance about the hinge, which depends on the angle and on where the weight acts.
C0.833 MgCorrect Taking torques about the hinge removes the hinge force. The weight's torque is Mg(L/2); the cable's is T L sin 37°. Setting them equal: T = Mg/(2 sin 37°) = Mg/1.2 ≈ 0.833Mg.
D1.667 Mg A student who thinks the hinge exerts no force sets the cable's vertical component equal to the whole weight: T sin 37° = Mg, so T ≈ 1.667Mg. The hinge supports part of the weight; about the hinge, the cable balances only the weight's torque, Mg(L/2).
Working Torques about the hinge: the weight Mg acts at L/2, giving Mg(L/2); the cable's torque is T L sin 37°. Στ = 0: T L sin 37° = MgL/2, so T = Mg/(2 sin 37°) = Mg/(2 × 0.60) = 0.833Mg.
A wheel turns on a fixed horizontal axle that exerts a frictional torque of magnitude 0.50 N·m on it. A block of mass 3.0 kg hangs from a string wound around a drum of radius 0.20 m on the wheel. A second string, wound the opposite way around a drum of radius 0.25 m on the same wheel, is pulled straight down so that the block rises at a constant speed. What is the magnitude of the force pulling on the second string? Use g = 10 m/s².
Answer and reasoning
A30 N A student who thinks the force on one string equals the tension in the other picks 30 N. The strings act at different radii and the axle exerts a frictional torque, so the torque balance gives 26 N.
B26 NCorrect Constant speed means the block's string has tension mg = 30 N, and constant angular velocity means zero net torque on the wheel. The pull's torque must balance both the block's torque and the opposing friction: F(0.25) = 30(0.20) + 0.50, so F = 26 N.
C22 N A student who takes the axle's frictional torque to help the rotation gets F(0.25) = 6.0 − 0.50, so F = 22 N. Friction opposes the rotation, so the pull must supply an extra 0.50 N·m.
D40 N A student who expects the force at the larger radius to be larger, swapping the lever arms, gets F(0.20) = 30(0.25) + 0.50, so F = 40 N. Each force's torque uses its own radius: the pull acts at 0.25 m and the block's string at 0.20 m.
Working The block moves at constant velocity, so the tension in its string is mg = 30 N. The wheel's angular velocity is constant, so Στ = 0 about the axle: F(0.25 m) = (30 N)(0.20 m) + 0.50 N·m = 6.5 N·m, the frictional torque opposing the rotation. F = 6.5/0.25 = 26 N.
A uniform beam of mass M and length L hangs horizontally from two vertical strings, one at each end. A block of mass M sits at the center of the beam. The block is moved to a point L/4 from the left end, and the beam stays horizontal and at rest. By what factor does the tension in the left string change?
Answer and reasoning
A×1.25Correct Take torques about the right string. With the block at the center, TL = Mg/2 + Mg/2 = Mg. With the block L/4 from the left end, its lever arm about the right end is 3L/4: TL = Mg/2 + 3Mg/4 = 1.25Mg. The tension increases by a factor of 1.25.
B×1.00 A student who thinks the strings always share the load equally picks this. Moving the block toward the left string increases that string's share: the torque balance about the right end gives TL = 1.25Mg.
C×0.75 A student who thinks the farther string carries more of the block's weight gives the left string only Mg/4 from the block, so TL = Mg/2 + Mg/4 = 0.75Mg. The nearer string carries the larger share: 3Mg/4 of the block's weight.
D×1.50 A student who leaves the beam's own weight out of the torque balance compares Mg/2 with 3Mg/4, a factor of 1.5. The beam's weight adds Mg/2 to the tension in both cases, which makes the factor 1.25.
Working Torques about the right end: TL L = Mg(L/2) + Mg(xR), where xR is the block's distance from the right end. Center: xR = L/2, TL = Mg. At L/4 from the left: xR = 3L/4, TL = Mg/2 + 3Mg/4 = 5Mg/4. Factor 1.25.
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
