UY1: Uniform Circular Motion & Non-uniform Circular Motion

This lesson covers the mechanics of circular motion (uniform and non-uniform) and how to think about “centrifugal force” correctly via frames of reference.

At a glance

• Prerequisites: Basics & Kinematics (UY1), Newton’s 2nd law
• You will learn: how to decompose acceleration into radial/tangential parts; how to apply Newton’s 2nd law component-wise; how tension varies in a vertical circle; what “centrifugal force” means and when it is (and isn’t) used
• Key results: aᵣ = v²/r = rω², ∑ Fᵣ = mv²/r
• Common trap: using “centrifugal force” in an inertial frame, or mixing up “radial inward” vs “radial outward” sign conventions

Setup (coordinates and sign conventions)

At any point on a circular path, define two perpendicular directions:

• Radial direction r hat: toward the center (this is the “centripetal” direction).
• Tangential direction t hat: along the instantaneous direction of motion.

Then the acceleration decomposes as: vector a = vector aᵣ + vector aₜ, where: aᵣ = v²/r = rω², aₜ = dv/dt.

1) Uniform circular motion (constant speed)

Uniform circular motion means v is constant in time, so aₜ = 0. The velocity changes direction, so there is still a non-zero radial acceleration.

Result (centripetal acceleration)

Magnitude: aᵣ = v²/r.

Direction: toward the center (radially inward). In uniform circular motion, vector aᵣ is perpendicular to vector v.

Newton’s 2nd law in the radial direction

Apply Newton’s 2nd law along the radial direction: ∑ Fᵣ = m aᵣ = mv²/r.

This does not mean “there is a special extra force called centripetal force”. It means: whatever real forces act on the object, their radial inward resultant must equal m v²/r.

Checks

• Units: v²/r has units (m²s⁻²)/m = m s⁻².
• Scaling: if you double v at fixed r, then aᵣ increases by a factor of 4.

2) Non-uniform circular motion (changing speed)

If the speed changes, there is a tangential component of acceleration: vector a = vector aᵣ + vector aₜ.

Correspondingly, Newton’s 2nd law can be applied in components: vector F = vector Fᵣ + vector Fₜ, ∑ Fᵣ = maᵣ, ∑ Fₜ = maₜ.

• vector Fᵣ is the radial inward resultant that produces the direction change.
• vector Fₜ is the tangential resultant that changes the speed.

Worked example

Example: mass around a vertical circle (tension)

Scroll diagram horizontally to read all labels.

Problem

• Given: a mass m on a light cord of length R moving in a vertical circle, speed v at an instant when the cord makes angle θ with the vertical
• Find: the tension T in the cord at that instant

Approach

• Resolve forces into radial (toward the center) and tangential components.
• Use Newton’s 2nd law in the radial direction: ∑ Fᵣ = m v²/R.

Working

1. Choose the radial inward direction (toward O) as positive.
2. Resolve forces in the radial direction:
   • tension T is inward,
   • the radial component of weight is outward with magnitude mg cos θ.
3. Apply Newton’s 2nd law radially: T - mg cos θ = mv²/R.
4. Solve for tension: T = m(v²/R + g cos θ).
5. Useful special cases:
   • bottom ( θ = 0°): T_bottom = m(v_bottom²/R + g).
   • top ( θ = 180°): Tₜₒₚ = m(vₜₒₚ²/R-g).

Answer + checks

• T = m(v²/R + g cos θ)
• Direction/sign: at the top, if vₜₒₚ² < gR then Tₜₒₚ would be negative (meaning the string would go slack; the model “string under tension” breaks)

Mini-example: combine energy + radial dynamics (vertical circle)

Often you are given the speed at one point and asked about tension at another point. A fast workflow is:

1. Use energy to relate the speeds (gravity is conservative): K + U = constant.
2. Use radial Newton’s 2nd law at the point of interest to get tension/normal force.

Example: a mass m moves in a vertical circle of radius R. If its speed at the bottom is v_b, what is the tension at the top?

• Height change bottom → top is 2R, so energy gives (1/2)mvₜ² + mg(2R) = (1/2)mv_b² ⇒ vₜ² = v_b²-4gR.
• At the top, radially inward is downward. The inward forces are T and mg, so T + mg = mvₜ²/R ⇒ T = m(vₜ²/R-g).

Checks: you need v_b² ≥ 4gR for vₜ² ≥ 0 (the object can reach the top). For the string to stay taut at the top, require T ≥ 0 ⇒ vₜ² ≥ gR.

3) Fictitious forces in non-inertial frames (optional but useful)

Newton’s laws apply directly in inertial frames. If you choose to work in a non-inertial frame (accelerating or rotating), you can keep the Newton’s-law form by adding fictitious (inertial) forces.

Case 1: linearly accelerating frame (pendulum in a car)

Consider a car accelerating with acceleration magnitude a (to the right). A pendulum bob of mass m hangs from the ceiling.

Scroll diagram horizontally to read all labels.

In an inertial frame, the bob accelerates with the car, so: ∑ Fₓ = T sin θ = ma, ∑ F_y = T cos θ - mg = 0. Divide the equations to eliminate T: tan θ = a/g ⇒ a = g tan θ.

In the non-inertial frame of the car, the bob is at rest, so you can add a fictitious force of magnitude F_fict = ma opposite the car’s acceleration, then set the net force to zero.

Case 2: rotating frame (centrifugal force)

Scroll diagram horizontally to read all labels.

In an inertial frame, uniform circular motion requires inward radial acceleration aᵣ = v²/r, so the inward resultant force must be mv²/r. If the only horizontal force is tension T, then: T = mv²/r.

In the rotating frame that co-rotates with the object, the object is at rest ( vector a = 0 in that frame). To keep Newton’s-law form, introduce a fictitious outward (centrifugal) force of magnitude: F_fict = mv²/r so it balances the inward tension.

Practice set (with hints + answers)

1. Show that aᵣ = rω² given v = rω.
2. A car goes around a flat curve of radius 50 m at 20 m s⁻¹. Find aᵣ and the required resultant radial force for m = 1200 kg.
3. In uniform circular motion, explain why the acceleration is perpendicular to velocity (1–2 sentences).
4. A mass on a string moves in a vertical circle. At the top, what condition on v ensures the string is just taut ( T = 0)?
5. A pendulum in a car is at steady angle θ = 15° to the vertical. Estimate the car’s acceleration.

Hints

2. Use aᵣ = v²/r and Fᵣ = maᵣ.
3. Use Tₜₒₚ = m(v²/R-g) and set Tₜₒₚ = 0.
4. Use tan θ = a/g.

Answers

1. aᵣ = v²/r = (r²ω²)/r = rω².
2. aᵣ = 20²/50 = 8.0 m s⁻²; Fᵣ = maᵣ = 1200(8.0) = 9.6 × 10³ N inward.
3. In uniform circular motion the speed (magnitude of velocity) is constant, so acceleration cannot have a component along velocity; it must point perpendicular to velocity to change direction only.
4. Tₜₒₚ = 0 ⇒ v²/R = g ⇒ v = square root of gR at the top.
5. a = g tan 15° ≈ 9.8(0.268) = 2.6 m s⁻².

Quiz (MCQ)

Quiz — Circular Motion (UY1)

Summary + next steps

• Decompose circular motion into radial (changes direction) and tangential (changes speed) parts.
• In uniform circular motion, aₜ = 0 but aᵣ = v²/r remains non-zero.
• “Centripetal force” means the inward resultant ∑ Fᵣ, not an extra force.
• Use radial/tangential axes locally; don’t confuse them with fixed x/y axes.
• Fictitious forces are frame-dependent tools; don’t mix inertial and non-inertial diagrams.

Next: Concept Of Work Previous: Resistive Forces Back To Mechanics (UY1)