Lesson 7: Centripetal Force and ApplicationsAP

Lesson 4 established the requirement: circular motion needs a net inward force of mv²/r. Lesson 7 (CED Unit 3: Circular Motion and Gravitation) asks the follow-up question on every problem: which real forces supply it? The answer is never a new force called "centripetal" — it is friction, tension, the normal force, gravity, or a combination, and the FBD is how you find out.

The method: FBD, then ΣF toward the centre

• Draw the free-body diagram with real forces only.
• Choose the positive direction toward the centre.
• Write ΣF_centre = mv²/r, using components of the real forces along that direction.

This single equation, plus ΣF_perpendicular = 0 when appropriate, solves every case below.

Flat curves, banked curves and the conical pendulum

On a flat curve, static friction alone points toward the centre: f_s = mv²/r, so the maximum safe speed is v_max = √(μ_s g r). On a banked curve at angle θ, the normal force tilts; its horizontal component N sinθ points inward while N cosθ balances weight. With no friction needed, dividing the two equations gives the design condition tanθ = v²/(rg). A conical pendulum (mass swinging in a horizontal circle on a string) is the same physics with tension playing the normal force's role: tanθ = v²/(rg) where θ is the string's angle from vertical and r = L sinθ.

Vertical circles: energy meets dynamics

For a car cresting a hill of radius r, gravity points toward the centre and the normal force points away: mg − N = mv²/r, so N = m(g − v²/r) — you feel lighter, and at v = √(gr) the normal force vanishes (the car would leave the road). At the top of a loop, both gravity and the normal force (or tension) point inward: mg + N = mv²/r, so the minimum speed to stay on the track is v = √(gr) (where N → 0). At the bottom, the inward direction is up: N − mg = mv²/r, giving the largest normal force of the circuit — the heavy feeling in a roller-coaster dip. Because speed varies around a vertical circle, these problems pair dynamics at a point with energy conservation between points (Units 4/5 ideas arriving early — a deliberate College Board habit).

Worked example 1 — banked curve design speed

A curve of radius 100 m is banked so that a car can round it at 20 m/s with no friction (g = 10 m/s²). Find the banking angle. Given: ΣF_horizontal = N sinθ = mv²/r; ΣF_vertical = N cosθ = mg. Method: divide.

tanθ = v²/(rg) = 400/(100 × 10) = 0.40
θ = tan⁻¹(0.40) ≈ 22°

Check: faster design speeds or tighter radii need steeper banking; as v → 0, θ → 0 (a stationary car needs no bank) — the formula's limits behave correctly.

Worked example 2 — car over a hill and loop minimum

(a) A 60 kg passenger rides over a hill of radius 30 m at 15 m/s. Find the normal force (scale reading) at the crest. (b) A roller-coaster loop has radius 10 m; find the minimum speed at the top for the car to stay on the track. Given: g = 10 m/s².

(a) N = m(g − v²/r) = 60 × (10 − 225/30) = 60 × (10 − 7.5) = 150 N
(b) At the top with N = 0: mg = mv²/r → v = √(gr) = √(10 × 10) = 10 m/s

Check: in (a) the passenger's weight is 600 N, so the scale reads a quarter of that — the familiar stomach-lifting lightness. In (b) the mass cancels: the minimum top speed depends only on g and r.

Real-world application: designing for the centre-seeking force

Velodrome tracks are banked up to 47° because at racing speeds v²/(rg) demands it; highway off-ramps add banking plus friction margin for wet days (lower μ means the normal-force component must do more of the work). Pilots pulling out of a dive experience the bottom-of-circle case: at v = 200 m/s in a radius of 500 m, v²/r = 80 m/s², so the seat must exert N = m(g + v²/r) ≈ m × 90 N/kg — about nine times the pilot's ordinary weight, beyond untrained tolerance. That is precisely why g-suits and pilot training exist.

Experimental design note

Test the conical-pendulum relation tanθ = v²/(rg) by swinging a mass on a string of length L, measuring the radius of its circle and its period (time 20 swings). The model's hidden trap: students measure the string angle with a protractor while it swings, blurring θ; better, compute r and θ from a photograph with a scale in frame. Plotting tanθ against v² should give a straight line of slope 1/(rg) — a linearisation the exam loves.

Common mistakes (from College Board scoring commentary)

• Adding a separate "centripetal force" arrow to the FBD alongside tension/friction/gravity — double counting.
• At the top of a loop, writing N − mg = mv²/r (signs wrong: both mg and N point toward the centre at the top).
• For a hill crest, using the same equation as a dip — the centre is below the car on a crest, above it in a dip; direction first, formula second.
• Assuming the speed around a vertical circle is constant and applying one v everywhere.

AP Exam Experience

MCQ 1. A car rounds a flat, unbanked curve of radius 40 m. The coefficient of static friction is 0.50 (g = 10 m/s²). The maximum speed without skidding is:

• (A) 10 m/s
• (B) 14 m/s
• (C) 20 m/s
• (D) 28 m/s

Answer: (B). Friction supplies the centripetal force: μ_s mg = mv²/r gives v = √(0.50 × 10 × 40) = √200 ≈ 14 m/s.

MCQ 2. A ball on a string swings in a vertical circle. At the top of the circle, the forces on the ball (neglecting air resistance) are:

• (A) gravity only, if the speed is the minimum possible
• (B) gravity and tension, both toward the centre, always summing to mv²/r
• (C) gravity toward the centre and tension away from it
• (D) gravity, tension, and a centripetal force toward the centre

Answer: (B). Both real forces point toward the centre at the top and their sum is the required mv²/r. (A) is the special case T = 0 only; (D) invents a third force.

MCQ 3. A car of mass m crosses a bridge whose roadway is an arc of radius r. At the highest point, moving at speed v, the road exerts on the car a force of magnitude:

• (A) mg
• (B) mg + mv²/r
• (C) mg − mv²/r
• (D) mv²/r

Answer: (C). Net force toward the centre (downward here) is mg − N = mv²/r, so N = mg − mv²/r. If v² = gr the car loses contact entirely.

FRQ. A 0.40 kg ball attached to a 1.0 m string is swung in a vertical circle. At the bottom of the circle the ball moves at 8.0 m/s (g = 10 m/s²). (a) Calculate the tension in the string at the bottom. (b) Use energy conservation to calculate the ball's speed at the top of the circle. (c) Calculate the tension at the top, and state whether the string remains taut. (d) Explain, in terms of forces, why the tension is greatest at the bottom. Scoring guideline. (a) 1 point: at the bottom the centre is above the ball, so T − mg = mv²/r → T = 4.0 + 0.40 × 64/1.0 = 4.0 + 25.6 ≈ 30 N. (b) 2 points: ½mv²_bottom = ½mv²_top + mg(2r) → v²_top = 64 − 2 × 10 × 2.0 = 24, so v_top ≈ 4.9 m/s (1 point for energy set-up with height 2r, 1 for value). (c) 1 point: T + mg = mv²/r → T = 0.40 × 24 − 4.0 = 5.6 N > 0, so the string stays taut (also note v_top = 4.9 > √(gr) = 3.2 m/s, consistent). (d) 1 point: at the bottom, tension must both support the weight and supply the centripetal requirement (T = mg + mv²/r), and the speed is also greatest there, so both effects maximise T.

Worked example 3 — the conical pendulum, completely

A 0.30 kg mass swings on a 1.2 m string in a horizontal circle, the string making 30° with the vertical (g = 10 m/s²). Find (a) the radius of the circle, (b) the speed, (c) the period, (d) the tension.

(a) r = L sinθ = 1.2 × 0.50 = 0.60 m
(b) tanθ = v²/(rg) → v² = rg tanθ = 0.60 × 10 × 0.577 = 3.46 → v ≈ 1.9 m/s
(c) T_period = 2πr/v = 2π × 0.60/1.86 ≈ 2.0 s
(d) Tension from the vertical balance: T cosθ = mg → T = 3.0/0.866 ≈ 3.5 N

Check: verify (d) against the horizontal equation: T sinθ = 3.5 × 0.50 = 1.73 N, and mv²/r = 0.30 × 3.46/0.60 = 1.73 N ✓ — the same tension satisfies both components, as one string must. Also try the direct route: the period of a conical pendulum is T = 2π√(L cosθ/g) = 2π√(1.2 × 0.866/10) ≈ 2.0 s, agreeing again.

Ferris wheels: the vertical circle you have ridden

On a Ferris wheel of radius r turning at steady speed, the centre of your circle is beside you, horizontally. Wait — no: the centre of your circular path is the wheel's hub, which relative to your seat is sometimes above, sometimes level, sometimes below as you go around. At the top, the hub is below you: gravity points toward the centre and the seat's normal force points away, so N = m(g − v²/r) — you feel lightest. At the bottom the hub is above you: N = m(g + v²/r) — you feel heaviest. For a wheel with r = 10 m turning once a minute, v = 2π × 10/60 ≈ 1.05 m/s and v²/r ≈ 0.11 m/s² — a mere 1% modulation of your apparent weight, which is why Ferris wheels feel gentle while a loop-the-loop at v²/r = 2g does not.

Loose objects and dashboard physics

Why do loose objects slide across the dashboard on a bend? In the car's frame they seem flung outward; in the ground frame the truth is plainer: the object continues straight (Newton's first law) while the car turns beneath it, until the dashboard's edge or friction supplies an inward force. If the object's required mv²/r exceeds maximum static friction with the dash, no force can recruit it into the turn, and relative to the car it drifts outward. Same physics, two descriptions — but only the inertial description assigns forces correctly, which is why the AP exam insists on it.