2 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 2
An electric fan is switched on, and after a few seconds its blades turn at a constant angular velocity. Which statement about the torques exerted on the blades while they turn at this constant angular velocity is correct?
Answer and reasoning
AThe motor's torque is balanced by the opposing torques.Correct A constant angular velocity means rotational equilibrium, so the net torque on the blades is zero. Torques are still exerted on them: the motor's torque in the sense of rotation, and the opposing torques of air resistance and axle friction, which have the same total magnitude.
BThe motor's torque exceeds the torques that oppose the rotation. A student who thinks steady rotation needs a net torque in its direction picks this. If the motor's torque were larger, the net torque would not be zero and the blades would speed up. The motor's torque only has to balance the opposing torques.
CThe net torque is not zero, since each point on a blade accelerates. A student who treats the acceleration of points on the blades as a change in rotation picks this. Each point has a centripetal acceleration because its direction of motion changes, but the blades' angular velocity is constant, and rotational equilibrium needs only a zero net torque.
DNo torques are exerted, since the blades are in equilibrium. A student who thinks equilibrium means no torques are exerted picks this. The motor exerts a torque, and air resistance and friction exert opposing torques; they balance, so the net torque is zero, but none of them is zero.
The graph shows the angular velocity ω of a motor-driven platform as a function of time t, with counterclockwise taken as positive. Which claim about the torques exerted on the platform is supported by the graph?
Answer and reasoning
AThe net torque is not zero from 2 s to 5 s, as the platform turns steadily. A student who thinks a steady rotation needs a net torque picks this. From 2 s to 5 s the graph is level: ω is constant at 4 rad/s, so the platform is in rotational equilibrium and the net torque is zero.
BThe net torque is zero from 5 s to 7 s, while the platform slows down. A student who thinks a rotation slows down by itself picks this. From 5 s to 7 s the angular velocity falls from 4 rad/s to zero; a change in angular velocity, slowing as well as speeding up, requires a nonzero net torque.
CThe net torque is not zero at t = 7 s, when ω is momentarily zero.Correct At t = 7 s the graph crosses the time axis: ω is zero for an instant, but it is changing, from positive to negative, as it has been since t = 5 s. A changing angular velocity means the torques are unbalanced, so the net torque is not zero at t = 7 s even though the platform is momentarily at rest.
DNo torques are exerted on the platform between 2 s and 5 s. A student who thinks equilibrium means no torques are exerted picks this. From 2 s to 5 s the net torque is zero, which means only that the torques balance; the graph gives no evidence that no torques act, and a motor-driven platform turning steadily has a driving torque balanced by friction.
Working Read where ω is constant and where it changes. 0–2 s: ω rises from 0 to 4 rad/s (changing, net torque not zero). 2–5 s: ω constant at 4 rad/s (net torque zero; the graph cannot show whether individual torques act). 5–9 s: ω falls steadily from 4 rad/s through 0 at t = 7 s to −4 rad/s, so ω is changing throughout, including at t = 7 s, and the net torque is not zero there.
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 Translational equilibrium Fix
Translational equilibrium
The state of a system whose center-of-mass velocity is constant (including zero), which requires the net force exerted on the system to be zero. It says nothing about the system's rotation: a system can be in translational equilibrium while its angular velocity changes.
Rotational equilibrium
The state of a system whose angular velocity is constant (including zero), which requires the net torque exerted on the system to be zero: Σ τi = 0. It is independent of translational equilibrium: a system can be in either one without the other, in both, or in neither.
Force diagram (for a rigid system)
A diagram of a rigid system that shows each force exerted ON the system by another object as an arrow drawn from the point where the force is exerted, with its relative magnitude and direction. Unlike a free-body diagram, which draws the system as a single dot, it shows where each force acts relative to the axis of rotation, so the torques can be found from it.
Axis of rotation
The line about which a rigid system rotates or would rotate, such as a hinge, pivot or axle. Torques are calculated about a chosen axis; a force exerted at the axis has zero lever arm and so exerts no torque about it, which is why an unknown hinge or pivot force is removed by taking torques about that point.
Torque (τ, unit: N·m)
The rotational effect of a force about an axis: τ = rF⊥ = rF sin θ, where r is the distance from the axis to the point where the force is exerted and θ is the angle between the force and the position vector. Only the component of the force perpendicular to that position vector produces a torque.
Lever arm (unit: m)
The perpendicular distance from the axis of rotation to the line of action of a force. The magnitude of a torque equals the force multiplied by its lever arm, so the same force exerts a larger torque the farther its line of action is from the axis. It is measured from the axis, not from an end of the object.
Net torque and sign convention
The sum of all the torques exerted on a system about one axis, Σ τi. One sense of rotation, clockwise or counterclockwise, is taken as positive and the other as negative, so torques in opposite senses subtract. A uniform object's weight is treated as exerted at its center of mass.
Angular velocity (ω, unit: rad/s)
The rate at which a rigid system's angular position changes. All points of a rigid system share one angular velocity. Its sign shows the sense of rotation (clockwise or counterclockwise) about the chosen axis.
Newton's first law in rotational form
A system has a constant angular velocity only if the net torque exerted on it is zero. A steady rotation needs no net torque; torques may still be exerted on the system, as long as they balance.
Students often think If the net force on an object is zero, the net torque on it is also zero, so an object in translational equilibrium is in rotational equilibrium too. In fact No. Zero net force means the center of mass does not accelerate. Forces that add to zero can still exert a nonzero net torque if they act along different lines, so the object's angular velocity can change.
Students often think An object that is at rest, even momentarily, is in equilibrium: zero angular velocity means zero net torque. In fact No. Rotational equilibrium means the angular velocity stays constant, not that it is zero at an instant. An object that is momentarily at rest can have a nonzero net torque exerted on it, and its angular velocity is then changing.
5.5.A.2 Unbalanced torques Fix
Unbalanced torques
A set of torques whose sum about the axis is not zero. If the torques exerted on a rigid system are unbalanced, its angular velocity must be changing: speeding up, slowing down or reversing. This holds even at an instant when the angular velocity is zero.
Students often think Rotation dies away naturally: a net torque is needed to speed an object up, but slowing down happens by itself. In fact No. With a zero net torque, a rotating object keeps a constant angular velocity. Slowing down is a change in angular velocity and requires a nonzero net torque, usually from friction or air resistance.
5 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 5
Two identical uniform rods, X and Y, lie at rest on a horizontal surface with negligible friction. Two horizontal forces, each of magnitude F and perpendicular to the rod, are then exerted on each rod, one at each end, and they stay perpendicular to the rod. On rod X the two forces point in the same direction; on rod Y they point in opposite directions. Which rod or rods remain in rotational equilibrium while the forces are exerted?
Answer and reasoning
AOnly rod Y, since the two forces on it add up to zero A student who thinks that forces adding to zero guarantee rotational equilibrium picks this. The forces on rod Y do cancel, so its center of mass stays at rest, but they act along different lines and both turn the rod the same way: the net torque on Y is FL, not zero, so Y starts to rotate.
BOnly rod X, as its torques about the center cancelCorrect On rod X the forces act at equal distances on either side of the center and point the same way, so about the center one exerts a clockwise torque and the other an equal counterclockwise torque: the net torque is zero and X's angular velocity stays zero, even though its center of mass speeds up. On rod Y the forces add to zero but both turn the rod the same way, so Y starts to rotate. Rotational and translational equilibrium are independent.
CBoth rods, as each of the rods starts from rest A student who takes 'at rest' to mean 'in equilibrium' picks this. Both rods start with zero angular velocity, but rotational equilibrium means the angular velocity stays constant. The net torque on rod Y is not zero, so its angular velocity changes as soon as the forces are exerted.
DNeither rod, as forces exert torques on both of them A student who thinks equilibrium means that no torques are exerted picks this. Torques are exerted on both rods, but on rod X they are equal in magnitude and opposite in sense, so the net torque is zero and X does not start to rotate. Equilibrium needs a zero net torque, not the absence of torques.
A uniform board of weight W rests horizontally on two supports: support P at its left end and support Q one-quarter of the board's length from its right end. The diagram shows a student's force diagram for the board: FP and FQ are the forces exerted on the board by the supports, and W is drawn at the board's center. Which statement about the student's diagram is correct?
Answer and reasoning
AThe arrow at P should be twice the length of the arrow at Q. A student who multiplies each force by the other force's distance from the center picks this. It is the support nearer the center that must push harder: FP(L/2) = FQ(L/4) gives FQ = 2FP, not FP = 2FQ.
BIt is correct, as the upward forces add up to the weight. A student who checks only that the forces balance picks this. The two upward arrows do add up to W, so the board's center of mass would not accelerate, but their torques about the center do not cancel: a board with these forces would start to rotate clockwise.
CThe arrow at Q should be drawn twice as long as the arrow at P.Correct Take torques about the board's center, where W acts. FP acts L/2 from the center and FQ acts L/4 from it, on the other side, so zero net torque needs FP(L/2) = FQ(L/4), which gives FQ = 2FP; with FP + FQ = W, FP = W/3 and FQ = 2W/3. The student's arrows balance the forces but not the torques: as drawn, FP's clockwise torque about the center is twice FQ's counterclockwise torque.
DThe arrows at P and Q should point down, as the board presses on them. A student who draws the forces the board exerts on its supports picks this. A force diagram for the board shows only forces exerted ON the board. The supports push up on the board; the board's downward pushes act on the supports and belong on their diagrams.
A uniform horizontal rod of weight 40 N is attached to a wall by a hinge at its left end. A sign of weight 16 N hangs from the rod's right end, and a cable attached to the rod's right end holds the rod at rest, as shown in the diagram. Use sin 37° = 0.60 and cos 37° = 0.80. What is the tension in the cable?
Answer and reasoning
A60 NCorrect Taking torques about the hinge removes the unknown hinge force. Only the component of the tension perpendicular to the rod, T sin 37°, has a torque about the hinge, with lever arm L. Balancing it against the rod's weight (at L/2) and the sign (at L): 0.60T × L = 40 N × L/2 + 16 N × L = 36 N × L, so T = 60 N.
B93 N A student who puts the rod's whole weight at its far end picks this: 0.60T = 40 N + 16 N gives 93 N. A uniform rod's weight acts at its center, so its lever arm about the hinge is L/2, not L.
C36 N A student who treats the whole tension as perpendicular to the rod picks this: T × L = 36 N × L gives 36 N. Only the perpendicular component, T sin 37°, has a torque about the hinge; the component along the rod points through the hinge.
D45 N A student who takes the perpendicular component as T cos 37° picks this: 0.80T = 36 N gives 45 N. The diagram shows that the 37° angle is between the cable and the rod, so T cos 37° is the component along the rod and the perpendicular component is T sin 37° = 0.60T.
Working Take torques about the hinge, so the unknown hinge force has zero lever arm. Let the rod's length be L. Counterclockwise: the cable's perpendicular component T sin 37° = 0.60T at lever arm L. Clockwise: the rod's weight 40 N at its center (lever arm L/2) and the sign's weight 16 N at the end (lever arm L). Zero net torque: 0.60T × L = 40 N × L/2 + 16 N × L = 36 N × L, so T = 36 N ÷ 0.60 = 60 N.
A uniform plank of mass M and length L rests on a pivot a distance d from its left end, where d < L/2. A block of mass m hangs from the left end of the plank, and the plank is horizontal and at rest. Which expression gives m?
Answer and reasoning
AML/(2d) A student who measures the plank's lever arm from its left end, L/2, instead of from the pivot picks this. A lever arm is measured from the axis, here the pivot, so the plank's weight has lever arm L/2 − d. This expression also fails the check: it gives m = M, not zero, when the pivot is under the center.
B2Md/(L−2d) A student who multiplies each weight by the other weight's distance from the pivot picks this: mg(L/2 − d) = Mgd. Each force must be multiplied by its own lever arm. This expression would need an infinite mass when the pivot is under the center, where no block at all is needed.
CM(L−d)/d A student who places the plank's weight at its right end, L − d from the pivot, picks this: mgd = Mg(L − d). A uniform plank's weight acts at its center, so its lever arm about the pivot is L/2 − d; this expression gives m = M, not zero, when the pivot is under the center.
DM(L/(2d)−1)Correct About the pivot, the block's weight mg has lever arm d, and the plank's weight Mg acts at the plank's center, L/2 − d from the pivot on the other side. Zero net torque gives mgd = Mg(L/2 − d), so m = M(L − 2d)/(2d) = M(L/(2d) − 1). As a check, m = 0 when d = L/2, when the pivot is under the center.
Working Torques about the pivot. The block's weight mg has lever arm d (counterclockwise torque mgd). The plank's weight Mg acts at its center, L/2 − d to the right of the pivot (clockwise torque Mg(L/2 − d)). Zero net torque: mgd = Mg(L/2 − d), so m = M(L − 2d)/(2d) = M(L/(2d) − 1). Check: when d = L/2 the pivot is under the center and m = 0, as it must be.
A student holds a loaded wheelbarrow at rest by lifting its handles vertically. The wheelbarrow's own weight is negligible. Measured horizontally from the wheel's axle, the handles are 1.5 m away and the load's center of mass is 0.60 m away. The load is then moved so that its center of mass is 0.30 m from the axle. By what factor does the force the student must exert on the handles change?
Answer and reasoning
A1.0 A student who thinks the force needed depends only on the weight being held picks this. The load's weight is unchanged, but its lever arm about the axle is halved, so the torque the student must balance is halved.
B0.5Correct About the axle, the student's force F has lever arm 1.5 m and the load's weight W has lever arm d, so F × 1.5 m = Wd and F is proportional to d. Moving the load from 0.60 m to 0.30 m halves d, so it halves the force: the factor is 0.5.
C2.0 A student who multiplies each force by the other's distance, F × 0.60 m = W × 1.5 m, picks this; that makes F inversely proportional to the load's distance. Each force goes with its own lever arm: F × 1.5 m = Wd, so a smaller d needs a smaller force.
D1.3 A student who measures the load's distance from the handles instead of from the axle picks this: (1.5 − 0.30)/(1.5 − 0.60) = 1.2/0.90 ≈ 1.3. Lever arms are measured from the axis of rotation, the axle.
Working Torques about the axle (the ground's force on the wheel acts there): F × 1.5 m = W × d, so F = Wd/(1.5 m) and F is proportional to d. Halving d from 0.60 m to 0.30 m multiplies F by 0.30/0.60 = 0.5.
Compiled from the AP Physics 1 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
