Circular motion and conical pendulums: worked solutions

Answer: θ = 2.62 rad; arc length = 1.05 m; f = 1.50 Hz; ω = 9.42 rad/s; v = 3.77 m/s; a = 35.5 m/s² inward.

1. θ = 150π/180 = 5π/6 = 2.62 rad;
   s = rθ = 0.400(5π/6) = 1.05 m.
2. f = 90.0/60 = 1.50 Hz;
   ω = 2πf = 9.42 rad s⁻¹.
3. v = ωr = 3.77 m s⁻¹;
   a = ω²r = 35.5 m s⁻², towards the axis.
4. Speed is constant, but the velocity direction changes continuously, so the marker accelerates.

Check: 150° is 5/12 of a turn, so the arc is 5/12 of circumference. v²/r independently gives the same acceleration.

Mark checklist · 6 marks

• 1 mark — Correct radian angle, approximately 2.62 rad.
• 1 mark — Uses s = rθ, approximately 1.05 m; carry the radian angle forward.
• 1 mark — Converts to 1.50 revolutions per second/Hz.
• 1 mark — Uses 2πf, approximately 9.42 rad/s.
• 1 mark — Uses ωr, approximately 3.77 m/s; carry ω forward.
• 1 mark — Uses ω²r or v²/r, approximately 35.5 m/s²; carry prior values forward.

Common mistake: Using 90 as Hz or 150 as radians changes the physical scale. Convert before substituting; do not replace arc length with the chord.

Exam technique: Identify the rotation axis and real forces first. These are original teaching values; the checklist is PhysicsUK’s own marking guidance.

Related examiner evidence: AQA 7408/1, Autumn 2021 — Q04.2–04.3; report p. 4

Answer: Velocity is downward in the top view; acceleration and static friction point left towards the axis. fmax = 0.631 Hz. At 0.500 m and 0.500 Hz it slips: the required friction exceeds its limit.

1. Instantaneous velocity is downward along the tangent in the top view. Acceleration points left towards the axis. The inward horizontal force is static friction, not a separate extra centripetal interaction.
2. At the threshold: m(2πf_max)²r = μmg;
   f_max = [1/(2π)]√(μg/r) = [1/(2π)]√[(0.400 × 9.81)/0.250] = 0.631 Hz.
3. At f = 0.500 Hz, r = 0.500 m:
   F_required = 0.0800(2π × 0.500)²(0.500) = 0.395 N;
   F_max = 0.400(0.0800)(9.81) = 0.314 N.
4. At fixed frequency, required force doubles when r doubles. The available friction has not changed; the puck cannot maintain that circular path.

Check: At the original radius and frequency, required friction is only 0.197 N, below 0.314 N. The threshold does not depend on puck mass in this model.

Mark checklist · 6 marks

• 1 mark — Velocity downward along the tangent in the stated top view.
• 1 mark — Acceleration left, towards the axis.
• 1 mark — Names inward static friction as the actual horizontal resultant.
• 1 mark — Equates m(2πf)²r to μmg and rearranges.
• 1 mark — Finds approximately 0.631 Hz with units.
• 1 mark — Explains doubled requirement/0.395 N exceeds 0.314 N, or the new threshold 0.446 Hz is below 0.500 Hz.

Common mistake: Static friction is not always μN: that is its maximum. Use the required inward force until the limit is reached; do not draw an additional centripetal-force arrow.

Exam technique: Identify the rotation axis and real forces first. These are original teaching values; the checklist is PhysicsUK’s own marking guidance.

Related examiner evidence: AQA 7408/1, June 2025 — Q07.4; report pp. 5–6 · AQA 7408/1, Autumn 2021 — Q04.1; report p. 4

Answer: r = 0.459 m; v = 1.78 m/s; P = 1.62 s; S = 1.80 N. Vertical forces balance; horizontal tension supplies the inward resultant.

1. Vertical acceleration is zero: S cos θ = mg.
   Radially inward: S sin θ = mv²/r.
2. r = L sin θ = 0.800 sin 35.0° = 0.459 m.
   Dividing the force equations gives tan θ = v²/(rg).
   v = √(rg tan θ) = 1.78 m s⁻¹.
3. P = 2πr/v;
   P² = 4π²r/(g tan θ) = 4π²L sin θ/(g tan θ) = 4π²L cos θ/g;
   P = 2π√(L cos θ/g) = 1.62 s.
4. S = mg/cos θ = (0.150 × 9.81)/cos 35.0° = 1.80 N.
5. This is steady conical rotation, sometimes called conical oscillation: the bob has fixed height and a constant string angle. It is different from the back-and-forth motion of a simple pendulum. A horizontal projection is sinusoidal, but the bob’s complete motion is a circle; do not replace this period with the planar small-angle formula at a finite θ.

Check: The calculated S exceeds mg = 1.47 N. ω² = g/(L cos θ) independently gives P. As θ approaches zero, the formula tends to the small-angle pendulum period; a zero-radius bob itself is not moving in a nonzero circle.

Mark checklist · 8 marks

• 1 mark — Vertical balance S cos θ = mg.
• 1 mark — Radial equation S sin θ = mv²/r.
• 1 mark — Uses L sin θ to find approximately 0.459 m.
• 1 mark — Divides the force equations or uses them consistently to obtain v² = rg tan θ.
• 1 mark — Finds approximately 1.78 m/s; carry their radius forward.
• 1 mark — Shows the algebra from P = 2πr/v and r = L sin θ to the stated period expression.
• 1 mark — Finds approximately 1.62 s with units.
• 1 mark — Finds approximately 1.80 N from the vertical balance.

Common mistake: Using L as the circle radius, swapping sine and cosine, or equating the whole tension to mv²/r misses the geometry. Resolve the single real tension into components; the components are not extra forces.

Exam technique: Identify the rotation axis and real forces first. These are original teaching values; the checklist is PhysicsUK’s own marking guidance.

Related examiner evidence: AQA 7408/1, June 2018 — Q05.2; report p. 6 · AQA 7408/1, June 2025 — Q07.4; report pp. 5–6 · AQA 7408/1, June 2025 — Q07.1; report p. 5

Answer: r = 1.53 m; v = 2.65 m/s; P = 3.64 s; rope tension = 325 N. At the same angle and rate, tension is proportional to total mass.

1. Horizontal reach from pivot to centre of mass = h tan θ = 2.00 tan 25.0° = 0.933 m;
   r = a + h tan θ = 1.53 m.
2. The circle is centred on the rotation axis. Its radius is not the rope length and not just the 0.933 m reach from the pivot.
3. S cos θ = mg; S sin θ = mv²/r;
   v = √(rg tan θ) = 2.65 m s⁻¹;
   P = 2πr/v = 3.64 s.
4. S = mg/cos θ = (30.0 × 9.81)/cos 25.0° = 325 N.
5. Mass cancels in the relation between angle, radius and speed, but S is proportional to mass. Increasing the mass at the same θ increases weight and tension proportionally.

Check: Rope length is h/cos θ = 2.21 m. Neither this length nor h is the path radius. Tension’s inward component is approximately 137 N.

Mark checklist · 6 marks

• 1 mark — Uses h tan θ to find horizontal reach, approximately 0.933 m.
• 1 mark — Adds a, obtaining approximately 1.53 m from the rotation axis.
• 1 mark — Uses resolved forces/v² = rg tan θ to obtain approximately 2.65 m/s; carry radius forward.
• 1 mark — Uses 2πr/v to find approximately 3.64 s; carry previous values forward.
• 1 mark — Uses mg/cos θ, approximately 325 N.
• 1 mark — Tension increases in proportion to combined mass at fixed angle; speed/geometry need not change.

Common mistake: Using h sin θ treats a vertical drop as a rope length; omitting a measures from the wrong centre. Draw the right triangle, then locate the rotation axis.

Exam technique: Identify the rotation axis and real forces first. These are original teaching values; the checklist is PhysicsUK’s own marking guidance.

Related examiner evidence: AQA 7408/1, June 2018 — Q05.1; report p. 6 · AQA 7408/1, June 2018 — Q05.3; report p. 6

Answer: α = 23.5°; N = 8.56 × 10³ N. Weight is vertical; the contact force is perpendicular to the bank and tilts inward. Design speed is independent of mass.

1. Real forces are weight mg vertically downward and the combined normal contact force perpendicular to the road, tilted towards the centre. Do not add a separate centripetal force. In an extended-car diagram, the individual contact forces act through the tyres; this question explicitly combines them into a particle-model resultant.
2. Vertically: N cos α = mg.
   Horizontally inward: N sin α = mv²/r.
   Therefore tan α = v²/(rg) = 16.0²/(60.0 × 9.81) = 0.435;
   α = 23.5°.
3. N = mg/cos α = 8.56 × 10³ N.
4. The mass cancels when the component equations are divided. For fixed r and α, v = √(rg tan α) is independent of m under the stated no-sideways-friction model.

Check: N exceeds mg = 7.85 kN; N sin α = 3.41 kN independently equals mv²/r.

Mark checklist · 6 marks

• 1 mark — Weight vertical down, not perpendicular to road.
• 1 mark — Normal reaction perpendicular to the road and tilted inward; no extra centripetal force.
• 1 mark — Correct vertical/radial component equations and division.
• 1 mark — Obtains approximately 23.5° from tan α = v²/(rg).
• 1 mark — Uses mg/cos α to find approximately 8.56 kN; carry their angle forward.
• 1 mark — Explains mass cancellation/independence in the stated model.

Common mistake: Weight does not acquire a horizontal component merely because the road is sloped. Resolve in vertical and horizontal directions; it is the tilted contact force that has an inward horizontal component.

Exam technique: Identify the rotation axis and real forces first. These are original teaching values; the checklist is PhysicsUK’s own marking guidance.

Related examiner evidence: AQA 7408/1, June 2025 — Q07.4; report pp. 5–6 · AQA 7408/1, June 2025 — Q07.5; report p. 6

Answer: Ncrest = 2.69 × 10³ N; Ndip = 9.09 × 10³ N; limiting crest speed = 10.8 m/s. Faster than the limit, contact is lost.

1. Crest, inward downward:
   mg−N = mv²/r;
   N = 600(9.81)−600(8.00²)/12.0 = 2.69 × 10³ N.
2. Dip, inward upward:
   N−mg = mv²/r;
   N = 600(9.81)+600(8.00²)/12.0 = 9.09 × 10³ N.
3. The minimum allowed normal force is zero.
   At the limiting crest speed, mg = mv²/r;
   vlimit = √(gr) = √(9.81 × 12.0) = 10.8 m s⁻¹.
4. Above the limit, the circular-path equation would demand a downward contact pull (negative N). This track cannot provide that. The cart leaves the surface and follows unsupported motion under gravity rather than that prescribed circle.

Check: At zero speed both expressions give mg. At the stated speed the crest force is below weight and the dip force above weight.

Mark checklist · 6 marks

• 1 mark — Crest radial equation mg−N = mv²/r.
• 1 mark — Finds approximately 2.69 kN with units.
• 1 mark — Dip radial equation N−mg = mv²/r.
• 1 mark — Finds approximately 9.09 kN with units.
• 1 mark — Sets N = 0 and finds approximately 10.8 m/s.
• 1 mark — Explains loss of contact when the required N would be negative.

Common mistake: Writing N = mv²/r forgets gravity. A negative calculated N is not a physical pulling reaction from this track; it signals that the assumed constrained path fails.

Exam technique: Identify the rotation axis and real forces first. These are original teaching values; the checklist is PhysicsUK’s own marking guidance.

Related examiner evidence: AQA 7408/1, Autumn 2021 — Q04.1; report p. 4

Answer: vB = 4.20 m/s; top tension = 2.44 N; limiting minimum bottom speed = 6.26 m/s.

1. The rise from A to B is 2r.
   ½mv_A² = ½mv_B² + mg(2r);
   v_B² = 7.00²−4(9.81)(0.800) = 17.608;
   v_B = 4.20 m s⁻¹.
2. At B, weight and string tension both act inward/downward:
   S + mg = mv_B²/r;
   S = 0.200(17.608)/0.800−0.200(9.81) = 2.44 N.
3. In the limiting just-taut case S_B = 0, so v_B,min² = gr.
   Energy then gives v_A,min² = gr+4gr = 5gr;
   v_A,min = √(5gr) = 6.26 m s⁻¹.
4. This is a limiting boundary case: a small extra bottom speed gives positive top tension. Radial acceleration is v²/r at each point; gravity also causes tangential acceleration away from the top and bottom, so the full motion is not uniform circular motion.

Check: At the stated bottom speed tension is 14.2 N. The top tension is positive and the bottom speed exceeds 6.26 m/s, consistent with the taut-circle assumption.

Mark checklist · 7 marks

• 1 mark — Uses energy with height difference 2r, obtaining vB² = vA²−4gr.
• 1 mark — Finds approximately 4.20 m/s; retain unrounded squared speed for later calculation.
• 1 mark — Correct top radial equation S+mg = mvB²/r.
• 1 mark — Finds approximately 2.44 N; carry their speed squared forward.
• 1 mark — Sets top tension to zero, obtaining vB,min² = gr.
• 1 mark — Combines energy to show vA,min² = 5gr.
• 1 mark — Finds approximately 6.26 m/s with units.

Common mistake: Using only energy to reach the top ignores the nonnegative-tension condition. The top speed cannot be zero for a complete string circle: gravity still needs to supply its required inward acceleration.

Exam technique: Identify the rotation axis and real forces first. These are original teaching values; the checklist is PhysicsUK’s own marking guidance.

Answer: Gradient = 1.00 N Hz⁻²; inferred radius = 0.507 m. Bounds are 0.489–0.524 m, overlapping the direct interval 0.505–0.515 m.

1. F = mv²/r and v = 2πfr;
   F = 4π²mr f².
   Plot F vertically against f² horizontally: at fixed m and r the ideal line passes through the origin, with gradient 4π²mr.
2. Using the endpoints: gradient = (1.00−0.160)/(1.00²−0.400²) = 1.00 N Hz⁻².
   m = 0.0500 kg;
   r = 1.00/[4π²(0.0500)] = 0.507 m.
3. r_min = 0.970/[4π²(0.0502)] = 0.489 m;
   r_max = 1.030/[4π²(0.0498)] = 0.524 m.
4. The direct interval is 0.505–0.515 m. It overlaps the graph interval, so the results are consistent within the stated bounds. This does not prove the model or identify the exact true radius.
5. Time many complete revolutions and repeat timings to reduce fractional timing scatter; do not time only a short arc. Measure radius from the rotation axis to the puck’s centre and monitor tether stretch so r stays fixed. Keep each trial at a stable speed, repeat force readings, and zero/check the force sensor. Any two justified contextual improvements earn credit.

Check: All four supplied rows have F/f² = 1.00 N Hz⁻². A frequency-squared graph is linear; a frequency graph is quadratic.

Mark checklist · 8 marks

• 1 mark — Substitutes v = 2πfr in F = mv²/r.
• 1 mark — Shows F = (4π²mr)f² and identifies fixed m/r.
• 1 mark — Central gradient 1.00 N Hz⁻² from separated points.
• 1 mark — Converts mass and obtains radius approximately 0.507 m.
• 1 mark — Uses opposite gradient/mass bounds to obtain approximately 0.489–0.524 m.
• 1 mark — Compares overlapping intervals and gives a consistency-within-uncertainty judgement.
• 2 marks — Two distinct contextual improvements with a reason each: long repeated timings, fixed/measured centre radius, stable speed/repeated forces, or force-sensor zero/calibration. One mark each.

Common mistake: Squaring the gradient, using 50.0 as kilograms, or treating an overlapping interval as proof loses the meaning of the analysis. Use the stated fit bounds; the idealised central table does not itself establish zero uncertainty.

Exam technique: Identify the rotation axis and real forces first. These are original teaching values; the checklist is PhysicsUK’s own marking guidance.

Answer: rA = 4.80 m; rB = 3.20 m; ω = 0.791 rad/s; P = 7.95 s; vA = 3.79 m/s and vB = 2.53 m/s. The floor’s contact force provides the passenger’s inward acceleration.

1. m_Aω²r_A = m_Bω²r_B, with r_B = L−r_A;
   r_A = m_BL/(m_A+m_B) = 4.80 m;
   r_B = 3.20 m.
2. ω = √(a_A/r_A) = √(3.00/4.80) = 0.791 rad s⁻¹;
   P = 2π/ω = 7.95 s.
3. v_A = ωr_A = 3.79 m s⁻¹;
   v_B = ωr_B = 2.53 m s⁻¹.
4. The floor exerts a normal contact force inward on the passenger, providing the required acceleration. The equal opposite force exerted by the passenger on the floor is on a different body. No separate outward force on the passenger is added in the inertial-frame analysis.

Check: Module A requires 2000 × 3.00 = 6000 N. B has acceleration 2.00 m/s² and requires 3000 × 2.00 = 6000 N, confirming equal link forces.

Mark checklist · 6 marks

• 1 mark — Equates mAω²rA to mBω²rB with a common ω and uses rA+rB = L.
• 1 mark — Obtains radii approximately 4.80 m and 3.20 m.
• 1 mark — Uses aA = ω²rA, approximately 0.791 rad/s.
• 1 mark — Finds approximately 7.95 s from 2π/ω.
• 1 mark — A has larger linear speed because v = ωr with common ω; gives a consistent comparison or values.
• 1 mark — Inward normal contact force from the floor supplies passenger acceleration/apparent weight.

Common mistake: Cancelling linear speeds between modules is wrong. The rigid link enforces common angular speed; v changes with radius.

Exam technique: Identify the rotation axis and real forces first. These are original teaching values; the checklist is PhysicsUK’s own marking guidance.

Related examiner evidence: AQA 7408/1, Autumn 2021 — Q04.2–04.3; report p. 4 · AQA 7408/1, Autumn 2021 — Q04.1; report p. 4

Answer: Initial horizontal velocity is north, 4.71 m/s. After release acceleration is vertically downward, 9.81 m/s²; t = 0.606 s and horizontal distance = 2.85 m.

1. Velocity is north along the tangent just before and immediately after the break. Velocity cannot jump merely because the string loses tension; the string’s inward force disappears.
2. The remaining force is gravity, so acceleration is vertically downward at g. Horizontal acceleration is zero in the stated model. Its horizontal projection follows the original tangent; its three-dimensional path curves downward as a projectile, not a horizontal straight line at fixed height.
3. v_horizontal = 2πfr = 2π(1.25)(0.600) = 4.71 m s⁻¹.
4. Initial vertical velocity is zero.
   h = ½gt²;
   t = √(2h/g) = √[(2 × 1.80)/9.81] = 0.606 s;
   d = vt = 2.85 m, north of the release point.

Check: The fall time depends on height, not the previous circular radius or frequency. Any inward curvature after the break would require an inward horizontal force that is absent.

Mark checklist · 6 marks

• 1 mark — North/tangent velocity immediately before and after release.
• 1 mark — Tension and its inward component vanish; no sudden velocity change.
• 1 mark — Acceleration vertically downward at g, with zero horizontal acceleration.
• 1 mark — Calculates horizontal speed approximately 4.71 m/s from 2πfr.
• 1 mark — Uses zero initial vertical speed to find approximately 0.606 s.
• 1 mark — Finds approximately 2.85 m horizontal distance; carry earlier speed/time forward.

Common mistake: “It flies outward” misses the initial tangent. “It continues at constant velocity” also misses gravity for this unsupported bob. Distinguish the straight horizontal projection from the downward-curving full path.

Exam technique: Identify the rotation axis and real forces first. These are original teaching values; the checklist is PhysicsUK’s own marking guidance.