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FRQ: Conical Pendulum Period — Designing a Conical-Pendulum Timer

Assessments aligned to 2026 AP Physics 1 standards

Qualitative/Quantitative Translation (QQT)  |  CHALLENGE  |  8 points  |  20 min

▤ Scenario

An engineer is testing whether a conical pendulum can serve as the timing element of a slow clock. A small ball of mass m hangs from a fixed pivot on a light string of length L = 0.90 m. The ball is set moving so that it travels at constant speed in a horizontal circle, and the string sweeps out a cone. The cone half-angle θ is the angle between the string and the vertical, and it can be set to any value between 0° and 90°. The time the ball takes to complete one revolution is the period T, and the magnitude of the force exerted on the ball by the string is F_T.

The engineer runs the pendulum twice with the same ball and string, as shown in Figure 1. In Run 1 the cone half-angle is θ_1 = 30° and the period is T_1. In Run 2 the cone half-angle is θ_2 = 53° and the period is T_2.

Figure 1 — The two runs, drawn to one scale: the same ball and string at 30° and at 53°.

Use an inertial reference frame fixed to the ground. Treat the string as massless and the ball as a particle, and neglect air resistance. Angles are measured from the vertical. Use g = 10 m/s².

✎ Free Response Questions

Part A

Indicate whether T_2 is greater than, less than, or equal to T_1 by writing one of the following.

•   T_2 > T_1

•   T_2 < T_1

•   T_2 = T_1

Justify your answer using qualitative reasoning beyond referencing equations.

Part B

Consider the general case in which the ball, of mass m, moves at constant speed in a horizontal circle on a string of length L that makes an angle θ with the vertical.

Starting with Newton’s second law, derive an expression for the period T of the ball’s motion. Express your final answer in terms of m, L, θ, and physical constants, as appropriate. Begin your derivation by writing the fundamental physics principle or an equation from the reference information.

Part C

Justify how your derived equation in part B is or is not consistent with your reasoning in part A.

❖ 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 — Model Answer

T_2 < T_1

In each run the ball moves in a horizontal circle, so it has no vertical acceleration: the upward component of the string’s force balances the ball’s weight. The horizontal component, directed toward the centre of the circle, is then the net force on the ball, and it gives the ball its centripetal acceleration.

In Run 2 the string is tilted further from the vertical. For its upward component still to equal the same weight, the string must pull harder, and a larger share of that larger force points inward. So the net inward force, and with it the ball’s centripetal acceleration, is greater in Run 2.

The circle in Run 2 is also wider, but the inward pull grows by a larger factor than the radius does. The string’s force points along the string, so its inward component compared with its upward component is the circle’s radius compared with the circle’s depth below the pivot. From Run 1 to Run 2 the radius increases and the depth decreases, so the inward pull, the weight scaled by that ratio, grows by more than the radius: about 2.3 times, against 1.6 times (Figure 2). Had the period stayed the same, the ball’s motion in Run 2 would simply be Run 1’s scaled up 1.6 times at the same pace, so every speed and acceleration in it, the centripetal acceleration included, would be 1.6 times larger. The centripetal acceleration is 2.3 times larger, more than going round at that pace would need on the wider circle, so the ball goes round in less time. A wider cone raises the inward pull by a larger factor than it widens the circle, so each revolution takes less time.

Figure 2 — The two runs to one scale: the net inward force grows 2.3 times, the radius 1.6 times.

Scoring (3 points):

▸ A1 — 1 point: For indicating T_2 < T_1.

▸ A2 — 1 point: For a justification that indicates that the net force on the ball — the inward (horizontal) component of the string’s force — is greater in Run 2, or that the ball’s centripetal acceleration is greater in Run 2, or that in both runs the upward component of the string’s force balances the weight, so that the inward component is the weight scaled by the circle’s radius over its depth.

▸ A3 — 1 point: For a justification that indicates that the larger circle in Run 2 is more than made up for — that the inward pull or the centripetal acceleration grows by a greater factor than the radius, that the ball’s speed grows by a greater factor than the distance around the circle, or that the inward pull for each metre of radius, the weight divided by the circle’s depth, is larger in Run 2.

Scoring Note: A2 and A3 are content criteria, not route criteria. Stating that the Run 2 circle is larger but that the ball moves faster does not meet A3, because it does not compare the two. A greater tension alone does not meet A2. An expression for any quantity — a force, the centripetal acceleration, the speed or the period — quoted, or evaluated at the two angles, does not by itself meet A2 or A3; the response must also give in words the physical reason for what the expression is used to show (for example, that the more tilted string must pull harder to hold up the same weight, or that the string’s pull lies along it while its upward part balances the weight).

⚠︎ Common error (partial credit): Indicating T_2 < T_1 because the string pulls the ball inward harder in Run 2, so it goes round faster, without addressing that the Run 2 circle is also larger — earns 2 of 3 points, A1 and A2.

⚠︎ Common error (partial credit): Indicating T_2 < T_1 but justifying it only by an expression — for a force, the centripetal acceleration, the speed or the period — evaluated at the two angles (for example cos 53° < cos 30°, or mg tan θ larger at 53°) with no reason given in words, or only by saying that the ball moves faster in Run 2 — earns 1 of 3 points, A1.

✗ Common error (no credit): Indicating T_2 > T_1, reasoning only that the wider circle is a longer trip — earns 0 of 3 points. The ball does travel farther in Run 2, but its speed rises by a larger factor.

✗ Common error (no credit): Indicating T_2 = T_1, reasoning only that a pendulum’s period depends on its length and g — earns 0 of 3 points. That is the small-amplitude simple pendulum’s result; a ball moving on a cone has a period that depends on the cone angle.

Part B — Model Answer

Two forces act on the ball: the string’s force F_T, directed along the string toward the pivot, and the gravitational force mg, downward. The ball moves at constant speed on a horizontal circle, so its acceleration is horizontal and directed toward the centre, with magnitude a_c, and its vertical acceleration is zero. Resolve F_T into a vertical component F_T cos θ and a horizontal component F_T sin θ toward the centre (Figure 3). Newton’s second law in each direction gives

Vertical:  F_T cos θ = mg

Horizontal:  F_T sin θ = m·a_c

Dividing the horizontal equation by the vertical one eliminates F_T and m:

a_c = g tan θ

In one period the ball travels the circumference 2πr at speed v, so v = 2πr/T and a_c = v²/r = 4π²r/T². The radius of the circle is r = L sin θ. Then

T = 2π√(r/a_c) = 2π√(L sin θ/(g tan θ))

T = 2π√(L cos θ/g)

Figure 3 — The cone’s geometry and the forces on the ball: the two triangles have the same shape.

The period is set by the depth L cos θ of the circle below the pivot and by g, not by the mass. For Run 2 it gives T_2 = 2π√((0.90 m)(0.60)/(10 m/s²)) = 1.46 s, and for Run 1 T_1 = 1.75 s. As θ approaches 0°, the expression approaches 2π√(L/g) = 1.88 s, the period T_s of a simple pendulum of the same length swinging with small amplitude.

Scoring (3 points):

▸ B1 — 1 point: For a multistep derivation that includes Newton’s second law applied to the ball with its centripetal acceleration — any equation setting a force or force component on the ball equal to m times the centripetal acceleration (m·v²/r, m·4π²r/T² or m·a_c), or to its nonzero component in the chosen direction, such as mg sin θ = m·a_c cos θ perpendicular to the string. This point is earned for the starting principle alone, even if the steps that follow are incorrect; a final expression for T on its own is not a multistep derivation.

▸ B2 — 1 point: For indicating that the ball’s centripetal acceleration is g tan θ — equivalently, that the net force on it is mg tan θ, that tan θ = v²/(gr), or that mg sin θ = m·a_c cos θ perpendicular to the string. Writing F_T cos θ = mg together with F_T sin θ as the net force toward the centre indicates this relation. This point is not earned by a relation that contradicts the response’s own equations.

▸ B3 — 1 point: For a correct expression for the period, T = 2π√(L cos θ/g) or an equivalent form such as 2π√(L sin θ/(g tan θ)), consistent with the indicated centripetal acceleration, where the response indicates one. Where B2 was not earned, an expression that follows correctly from the response’s own incorrect centripetal acceleration, with r = L sin θ and a correct relation between the period and the centripetal acceleration (a_c = 4π²r/T², or v = 2πr/T with a_c = v²/r), also earns this point; an incorrect expression with any other error does not.

Scoring Note: A correct, isolated, final expression for T earns points B2 and B3. A correct final expression is treated as isolated when everything before it is one or more of the following: a_c = g tan θ or an equivalent stated on its own (tan θ = v²/(gr), or a net force of mg tan θ); the geometry r = L sin θ; kinematic relations such as v = 2πr/T or a_c = 4π²r/T²; unapplied equations from the reference information; words. B1 and B2 carry no consistency credit, and B1, B2 and B3 do not depend on the response in part A. Work, in equations or in words, that balances the real forces against an outward “centrifugal force” is not Newton’s second law: it does not meet B1, and a final expression that follows it is not treated as isolated. Nor is a final expression reached by treating the ball as a simple pendulum of some effective length, in equations or in words.

⚠︎ Common error (partial credit): Taking the string’s force equal to the ball’s weight, F_T = mg, in place of the vertical balance F_T cos θ = mg, so that θ cancels and the result is T = 2π√(L/g), a simple pendulum’s 1.88 s — earns 2 of 3 points, B1 and B3. Only the vertical component of the string’s force balances the weight.

⚠︎ Common error (partial credit): Setting the whole string force equal to m·a_c, alongside F_T cos θ = mg, which gives T = 2π√(L sin θ cos θ/g) — earns 2 of 3 points, B1 and B3. Only the horizontal component of the string’s force points toward the centre.

⚠︎ Common error (partial credit): Measuring θ from the horizontal, so that sin θ and cos θ are exchanged throughout, which gives T = 2π√(L sin θ/g) — 1.69 s for Run 2 instead of 1.46 s — earns 1 of 3 points, B1. B3’s consistency clause does not apply, because the radius L cos θ is a second error, not a consequence of the centripetal acceleration.

⚠︎ Common error (partial credit): Balancing the inward component of the string’s force against an outward “centrifugal force” mv²/r, alongside F_T cos θ = mg, and reaching T = 2π√(L cos θ/g) — earns 2 of 3 points, B2 and B3. In the ground’s frame nothing pushes the ball outward; the inward component is the net force, and it produces the centripetal acceleration.

⚠︎ Common error (partial credit): Treating the ball as a simple pendulum whose length is the depth, in words or with T_p = 2π√(ℓ/g) and ℓ = L cos θ, with no Newton’s second law relation — earns 1 of 3 points, B3. The expression is right, but the pendulum formula describes a swing in a vertical plane; that the depth sets the period has to come out of the forces.

✗ Common error (no credit): Writing the reference-information pendulum period, T_p = 2π√(ℓ/g) with ℓ = L, as the answer, with no Newton’s second law relation — earns 0 of 3 points. It describes a small-amplitude swing in a vertical plane, not a ball moving on a cone.

Part C — Model Answer

The derived equation is consistent with the reasoning in part A.

In T = 2π√(L cos θ/g), L and g are the same in both runs. T depends on θ only through cos θ under the square root, so every widening of the cone shortens the period. From 30° to 53°, cos θ falls from 0.87 to 0.60, so T_2 < T_1, as part A concluded (Figure 4).

The equation also carries part A’s reasoning. Written as T = 2π√(r/a_c), it makes the period shorter when the centripetal acceleration is larger compared with the radius, and with r = L sin θ and a_c = g tan θ that ratio, a_c/r, is g/(L cos θ), which grows as the depth shrinks. That is the comparison part A made: widening the cone raises a_c = g tan θ by a larger factor than r = L sin θ.

Figure 4 — The derived period for L = 0.90 m: T falls as θ grows and reaches T_s only at 0°.

Can a conical pendulum keep time? (Beyond what part C asks.) Not as well as a simple pendulum. A clock needs a period that stays fixed while the pendulum slowly loses energy. Each cone angle goes with one steady speed, v = √(gL sin θ tan θ), which rises with θ: 1.61 m/s at 30°, 3.10 m/s at 53°. The speed does not appear in T = 2π√(L cos θ/g), but it enters through θ: a ball moving steadily at a higher speed must run on a wider cone, with a shorter period. A real ball, slowed by air resistance, therefore settles onto narrower cones, and its period lengthens toward T_s = 1.88 s, the small-amplitude period of a simple pendulum of the same length. A simple pendulum swinging with small amplitude keeps T_s as its swing dies away, so it makes the better timer. Nor can the two ever match: T = T_s only if cos θ = 1, at θ = 0°, where there is no circle at all, so at every real cone angle T < T_s.

Scoring (2 points):

▸ C1 — 1 point: For attempting to address the functional dependence between T and θ in the equation derived in part B.

Scoring Note: It is not necessary to use the functional dependence correctly to earn this point. The response only needs functional-dependence language — such as proportional, inversely proportional, related, increases, decreases, numerator or denominator — to relate T and θ.

▸ C2 — 1 point: For correctly using functional dependence to evaluate how T depends on θ in the expression derived in part B, and stating the verdict — consistent or not consistent — that follows from it together with the response’s own part A answer or reasoning. For the correct expression, T decreases as θ increases because cos θ decreases, which is consistent with T_2 < T_1.

Scoring Note: The dependence may be read from the structure of the expression — for example, that cos θ, under the square root, decreases as θ increases — or shown by evaluating the expression at the two angles and stating how T changes between them.

⚠︎ Common error (partial credit): Stating how T depends on θ, but giving no verdict on whether this agrees with part A — earns 1 of 2 points, C1.

⚠︎ Common error (partial credit): Reading the dependence backwards — cos θ increases as θ increases, so T increases — and calling the equation not consistent with T_2 < T_1 — earns 1 of 2 points, C1.

✗ Common error (no credit): Stating that the two parts agree because both give T_2 < T_1, or quoting the two calculated periods, with no words relating T to θ — earns 0 of 2 points. A consistency argument must say how the equation makes T change with θ.