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FRQ: Vertical Circular Motion — Inside the Loop

Assessments aligned to 2026 AP Physics 1 standards

Mathematical Routines (MR)  |  MID-LEVEL  |  10 points  |  25 min

▤ Scenario

A stunt rider rides a motorcycle around the inside of a vertical circular loop of track at a stunt show, as shown in Figure 1. The rider and the motorcycle together form the rider–motorcycle system, of mass m = 200 kg. Model the system as a single object that moves along a circle of radius r = 6.4 m.

The system enters the loop at the bottom moving at 20 m/s. It slows as it climbs, passes the top of the loop at 12 m/s, and completes the loop. From the moment the system enters the loop, the rider uses neither the engine nor the brakes. Rolling friction and air resistance are negligible, so no energy is lost to them.

Figure 1 — The stunt loop, with the system’s speed at the bottom and at the top of the loop.

Motion occurs in a vertical plane. Use an inertial reference frame fixed to the ground. On the radial axis at any point of the loop, take the direction toward the centre of the loop as positive. Use g = 10 m/s².

✎ Free Response Questions

Part A

i.    On the following dots, which represent the rider–motorcycle system at the top of the loop and at the bottom of the loop, draw and label the forces (not components) that are exerted on the rider–motorcycle system at each location. Each force must be represented by a distinct arrow starting on, and pointing away from, the dot.

Figure 2 — Dots representing the system at the top and at the bottom of the loop.

ii.   Starting with Newton’s second law, derive an expression for v_min, the minimum speed the system must have at the top of the loop to stay in contact with the track. Express your final answer in terms of m, r, and physical constants, as appropriate. Begin your derivation by writing the fundamental physics principle or an equation from the reference information.

iii.  Calculate the magnitude of the normal force exerted by the track on the system at the bottom of the loop. Then calculate the ratio of that magnitude to the magnitude of the gravitational force exerted on the system.

Part B

Rider A is the rider in the Scenario, with a rider–motorcycle system of mass 200 kg. Rider B rides a lighter machine on the same loop, with a rider–motorcycle system of mass 150 kg. Rider A claims: “My motorcycle and I are heavier, so I need a higher speed at the top to avoid falling.” Let v_A and v_B be the minimum speeds at the top of the loop for Rider A’s and Rider B’s systems, respectively.

Indicate whether v_B is greater than, less than, or equal to v_A by writing one of the following.

•   v_B > v_A

•   v_B < v_A

•   v_B = v_A

Justify your answer. In your justification, include qualitative reasoning beyond mathematical derivations or expressions.

❖ Answer Key & Scoring Guide

▸ earns credit  ⚠︎ common error, partial credit  ✗ common error, no credit

Points are independent and there are no deductions. ▸ criteria follow College Board’s published scoring standard; ⚠︎ and ✗ lines are TheScienceCube teaching commentary, not College Board rubric.

Part A (i) — Model Answer

Two objects interact with the system at each location. Earth exerts the gravitational force F_g, which points downward everywhere on the loop. The track exerts the normal force F_N, perpendicular to the track and pointing away from its surface, which on the inside of the loop means toward the centre. The rider uses neither the engine nor the brakes and rolling friction is negligible, so the track exerts no force along the direction of motion.

At the top of the loop the track is above the system, so the normal force points downward, in the same direction as the gravitational force; the two arrows are drawn side by side. At the bottom the track is below the system, so the normal force points upward.

Figure 3 — The forces on the system at the top and at the bottom of the loop, drawn to one scale.

At the top the system moves at 12 m/s, so the net force toward the centre must be mv²/r = (200 kg)(12 m/s)²/(6.4 m) = 4500 N. The gravitational force supplies 2000 N of it and the track supplies the remaining 2500 N; the magnitude at the bottom is found in part A (iii). At both locations the net force points toward the centre, and no separate “centripetal force” acts, because that name belongs to the net of the real forces already drawn.

Scoring (2 points):

▸ A1 — 1 point: For drawing, on the top-of-the-loop dot, the gravitational force and the normal force both directed downward, as two distinct labelled arrows starting on the dot and drawn side by side, and no other force arrow starting on the dot.

▸ A2 — 1 point: For drawing, on the bottom-of-the-loop dot, the gravitational force directed downward and the normal force directed upward, as distinct labelled arrows starting on the dot, and no other force arrow starting on the dot.

Scoring Note: Any label that names the force is accepted — F_g, mg, W, weight or gravitational force; F_N, N, normal force or force of the track. The response is not required to name the object exerting each force. Arrow lengths are not scored. An arrow for the velocity, the acceleration or a “centripetal force” drawn from the dot counts as another force arrow; a sign-convention arrow drawn beside the diagram does not.

⚠︎ Common error (partial credit): Draws the normal force at the top of the loop pointing upward, as if the system were resting on top of the track, with the bottom diagram correct — earns 1 of 2 points, A2. At the top the track is above the system, and a surface can only push away from itself.

⚠︎ Common error (partial credit): Draws only the gravitational force at the top of the loop, as if the track exerted no force there, with the bottom diagram correct — earns 1 of 2 points, A2. At 12 m/s the system needs a net force of 4500 N toward the centre, more than the 2000 N gravitational force, so the track still pushes on it.

⚠︎ Common error (partial credit): Draws the two downward forces at the top as one arrow labelled with both, or as two overlapping arrows, with the bottom diagram correct — earns 1 of 2 points, A2. Forces in the same direction are drawn as separate arrows, side by side.

✗ Common error (no credit): Adds a third arrow to each dot — one labelled “centripetal force”, pointing toward the centre, or a velocity arrow like those in Figure 1. Each diagram then carries an arrow that is not a force exerted by an object, so neither A1 nor A2 is earned. The centripetal force is the net of the real forces already drawn, not an additional force.

Part A (ii) — Model Answer

At the top of the loop both forces point downward, toward the centre, which is the positive radial direction. The system’s acceleration there is the centripetal acceleration v²/r, so Newton’s second law along the radial axis gives

F_net = F_N + F_g = mv²/r

F_N + mg = mv²/r

The track can push on the system but cannot pull it, so the system stays in contact only while F_N ≥ 0. Rearranged, F_N = mv²/r − mg, which shrinks as the speed at the top is lowered. The minimum speed is the one at which it reaches zero:

mg = m(v_min)²/r

v_min = √(gr)

At the minimum speed the track exerts no force, so the gravitational force is the only force causing the centripetal acceleration. For this loop v_min = √((10 m/s²)(6.4 m)) = 8.0 m/s, below the 12 m/s at which the system passes the top, so it does stay in contact with the track there. Below v_min the net force needed for the circle, mv²/r, would be smaller than the gravitational force alone; only an upward pull from the track could make up the difference, and a track cannot pull, so the system would leave the track.

Scoring (2 points):

▸ A3 — 1 point: For including Newton’s second law applied to the forces on the system at the top of the loop — for example, F_N + mg = ma or F_N + mg = mv²/r. This point is earned for the starting principle alone, even if the steps that follow are incorrect.

Scoring Note: Writing ΣF = ma or a_c = v²/r from the reference information, without applying it to the forces on the system at the top of the loop, does not meet the requirement for A3.

▸ A4 — 1 point: For indicating that at the minimum speed the gravitational force is the only force exerted on the system, so that it alone provides the centripetal acceleration — for example, mg = m(v_min)²/r or g = (v_min)²/r, or F_N = 0 in a second-law equation that then reduces to one of these.

Scoring Note: A correct, isolated, final expression for v_min earns points A3 and A4; a correct final expression preceded only by an unapplied equation from the reference information, or only by words, is treated as isolated. The normal force need not be named: Newton’s second law written for the minimum-speed case alone, mg = m(v_min)²/r or g = (v_min)²/r, meets the requirements for both A3 and A4. A4 carries no consistency credit. A3 and A4 do not depend on the response in part A (i).

⚠︎ Common error (partial credit): Writes F_N + mg = mv²/r correctly, then sets F_N = mg as for an object resting on level ground, reaching v = √(2gr) ≈ 11.3 m/s — earns 1 of 2 points, A3. At the minimum speed the track exerts no force at all; F_N = mg describes an object at rest on a horizontal surface, not one moving in a circle.

⚠︎ Common error (partial credit): Writes F_N + mg = mv²/r correctly, then argues that the track can hold the system in place at any speed, so v_min = 0 — earns 1 of 2 points, A3. A track can only push, so below √(gr) the system leaves the track.

✗ Common error (no credit): States that the minimum speed at the top is zero, because the system only has to reach the top and the track holds it there, and writes no equation. No second-law equation is written, so A3 is not earned, and gravity is never identified as the only force at the minimum speed, so A4 is not earned.

Part A (iii) — Model Answer

At the bottom of the loop the centre is directly above the system, so the positive radial direction is upward. The normal force points upward and the gravitational force points downward, and Newton’s second law along the radial axis gives

F_N − F_g = mv²/r

F_N = mg + mv²/r = m(g + v²/r)

F_N = (200 kg)(10 m/s² + (20 m/s)²/(6.4 m)) = (200 kg)(72.5 m/s²) = 14500 N

The gravitational force on the system is F_g = mg = (200 kg)(10 m/s²) = 2000 N, so

F_N / F_g = (14500 N)/(2000 N) = 7.25

The net force at the bottom points upward, toward the centre, so the normal force must exceed the gravitational force. The track pushes up on the system with 14500 N: the system’s apparent weight at the bottom of the loop is 7.25 times its weight.

Scoring (3 points):

▸ A5 — 1 point: For a correct application of Newton’s second law at the bottom of the loop, with the normal force upward, the gravitational force downward and a net force of magnitude mv²/r directed upward, toward the centre — F_N − mg = mv²/r, or any equivalent form such as F_N = m(g + v²/r). This point is earned for the correct equation alone.

▸ A6 — 1 point: For a normal force of magnitude 14500 N. This point is not earned by a value that follows from an incorrect equation.

▸ A7 — 1 point: For a ratio F_N/F_g consistent with the response’s value of F_N and a gravitational force of 2000 N (1960 N with g = 9.8 m/s²) — 7.25 for the correct values.

Scoring Note: A6 accepts 14500 N, or 14460 N with g = 9.8 m/s², and those values rounded to two significant figures (1.4 × 10⁴ N or 1.5 × 10⁴ N); A7 is judged against the F_N the response reports or its unrounded working. A5 requires the equation to be shown, symbolically or with numbers substituted; a correct value of F_N alone does not meet it. A5, A6 and A7 do not depend on the responses in parts A (i) and A (ii).

⚠︎ Common error (partial credit): Takes the normal force to be the whole net force, F_N = mv²/r = 12500 N, leaving out the gravitational force, and reports the ratio 6.25 — earns 1 of 3 points, A7. The net force is the sum of both forces: the track must first cancel the gravitational force before it can provide the upward net force.

⚠︎ Common error (partial credit): States that the normal force equals the gravitational force, 2000 N, because the track holds the system up, and reports a ratio of 1 — earns 1 of 3 points, A7. The system is accelerating toward the centre at the bottom, so the forces on it are not balanced.

Part B — Model Answer

v_B = v_A

At the minimum speed the track exerts no force, so the gravitational force is the only force available to provide the net force toward the centre, and it is proportional to mass: Rider B’s 150 kg system has 1500 N of it, three-quarters of the 2000 N on Rider A’s, just as 150 kg is three-quarters of 200 kg. But the net force needed to keep a system on a circle of radius r at speed v, mv²/r, is proportional to mass as well, so Rider B’s system also needs only three-quarters as much. At 8.0 m/s the requirement is 2000 N for Rider A’s system and 1500 N for Rider B’s — exactly the gravitational force on each.

Figure 4 — Both systems at the top at 8.0 m/s: gravity alone supplies the net force needed.

Equivalently, the gravitational force alone gives every object the same acceleration, g, whatever its mass, and moving on a circle of radius r at speed v requires the same centripetal acceleration, v²/r, whatever the mass. Both the gravitational force and the net force needed are proportional to mass, so the mass cancels and every system needs the same minimum speed at the top of a given loop. The part A (ii) expression agrees: √(gr) contains no mass, and gives 8.0 m/s for both riders.

Scoring (3 points):

▸ B1 — 1 point: For indicating v_B = v_A.

▸ B2 — 1 point: For indicating one of the following: the gravitational force exerted on a system is proportional to its mass; or the gravitational force gives every system the same acceleration, g, whatever its mass.

▸ B3 — 1 point: For indicating one of the following: the net force needed to keep a system on a circle of a given radius at a given speed, mv²/r, is proportional to the system’s mass; or the centripetal acceleration needed at a given speed on a given radius, v²/r, does not depend on the system’s mass.

Scoring Note: B1, B2 and B3 are scored independently of the responses in part A and of one another. The minimum requirement for B2 and for B3 is a correct statement of how a single force or acceleration depends on mass, in words or as a proportionality — F_g ∝ m; mv²/r ∝ m; the acceleration the gravitational force gives, g, or the centripetal acceleration needed, v²/r, is the same for any mass. Values meet it only when related to the masses: 1500 N is three-quarters of 2000 N, as 150 kg is of 200 kg. It is not met by an argument that the combined condition or the result lacks m — that m cancels from mg = m(v_min)²/r, or that g = (v_min)²/r or √(gr) contains no m — nor by stating only that Rider A’s system is heavier or that the gravitational force on Rider B’s system is smaller.

⚠︎ Common error (partial credit): Indicates v_B = v_A and justifies it only with equations — that m cancels from mg = m(v_min)²/r, or that m does not appear in √(gr) — earns 1 of 3 points, B1. The prompt asks for reasoning beyond the expression: why the mass cancels.

⚠︎ Common error (partial credit): Indicates v_B = v_A and states that the gravitational force on each system is proportional to its mass, but says nothing about the net force or the acceleration the circle needs — earns 2 of 3 points, B1 and B2.

⚠︎ Common error (partial credit): Indicates v_B < v_A, agreeing with Rider A, reasoning that the net force needed to stay on the circle, mv²/r, is proportional to mass, so the heavier system needs more — earns 1 of 3 points, B3. The heavier system does need more net force; the error is forgetting that the gravitational force on it is larger by the same factor.

✗ Common error (no credit): Indicates v_B < v_A, agreeing with Rider A, reasoning that a heavier object falls faster and so drops away from the track sooner. No correct statement of how a force or an acceleration depends on mass is made, so neither B2 nor B3 is earned, and B1 is not earned. The gravitational force gives every object the same acceleration, g, whatever its mass.