CLASSICAL-MECHANICS · Interactive Practice | Unit 5 · Video 3

From Chord to Arc: Deriving v=rωv = r\omega with Pure Geometry

IKey Formulas

Formula Name Description
∣Δr∣=2rsin⁡ ⁣(Δθ2)\lvert\Delta\mathbf{r}\rvert = 2r\sin\!\left(\frac{\Delta\theta}{2}\right) Chord length Exact straight-line displacement (any angle)
sin⁡x≈x\sin x \approx x Small-angle approximation True when xx is small and in radians
∣Δr∣≈r Δθ\lvert\Delta\mathbf{r}\rvert \approx r\,\Delta\theta Chord →\to arc Chord length approaches arc length as Δθ→0\Delta\theta \to 0
v=rωv = r\omega Speed in circular motion Speed = radius ×\times angular velocity, with ω=dθdt\omega = \frac{d\theta}{dt}

IIVisualization 1 — Chord Becomes Arc

The straight chord is always shorter than the curved arc — but the two close in as the separation angle vanishes.

IIIVisualization 2 — Why sin⁡x≈x\sin x \approx x

The line y=xy = x hugs y=sin⁡xy = \sin x near the origin and peels away as xx grows.

💡 The gap x−sin⁡x≈x3/6x - \sin x \approx x^{3}/6 — the first dropped Taylor term — so the approximation is razor-sharp for small xx and degrades cubically.

IVVisualization 3 — Where Velocity Points

As the second position swings back toward PP, the displacement chord pivots onto the tangent.

💡 Because velocity points along Δr\Delta\mathbf{r}, in the limit it lies along the tangent — perpendicular to the radius, in the θ^\hat\theta direction.

VQuiz Questions

Question 1

Two position vectors of length rr are separated by a small angle Δθ\Delta\theta. Connecting their tips gives the displacement Δr\Delta\mathbf{r}. What is the exact length of this chord, valid for any angle?

✅ Correct! The half-angle comes from the bisected isosceles triangle.

❌ Not quite. That is the arc length (and the small-angle approximation of the chord), not the exact chord.

❌ Not quite. The right triangle uses the half-angle Δθ/2, and the chord is twice the opposite side.

Show solution

Solution:

The two radii and the chord form an isosceles triangle: two equal sides of length rr with the angle Δθ\Delta\theta between them. Drop the angle bisector to split it into two right triangles, each with hypotenuse rr, half-angle Δθ/2\Delta\theta/2, and opposite side ∣Δr∣/2\lvert\Delta\mathbf{r}\rvert / 2:

sin⁡ ⁣(Δθ2)=∣Δr∣/2r  ⟹    ∣Δr∣=2rsin⁡ ⁣(Δθ2)  \sin\!\left(\frac{\Delta\theta}{2}\right) = \frac{\lvert\Delta\mathbf{r}\rvert / 2}{r} \;\Longrightarrow\; \boxed{\;\lvert\Delta\mathbf{r}\rvert = 2r\sin\!\left(\frac{\Delta\theta}{2}\right)\;}

This is exact for any angle. The form r Δθr\,\Delta\theta is only the approximation that follows after applying the small-angle rule.

Question 2

The small-angle approximation sin⁡x≈x\sin x \approx x requires that the angle xx be measured in a particular unit. Which unit makes the approximation valid?

(Hint: think about the Taylor series sin⁡x=x−x33!+⋯\sin x = x - \frac{x^3}{3!} + \cdots.)

✅ Correct! The Taylor series — and arc length rΔθr\Delta\theta — both require radians.

❌ Not quite. In degrees, sin(30°) = 0.5 ≠ 30. The series only equals x when x is in radians.

Show solution

Solution:

The Taylor series sin⁡x=x−x33!+x55!−⋯\sin x = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \cdots holds only when xx is in radians. Keeping just the first term gives sin⁡x≈x\sin x \approx x.

If you used degrees, sin⁡x≈x\sin x \approx x would be wildly wrong — e.g. sin⁡(30∘)=0.5\sin(30^\circ) = 0.5, not 3030. The clean relationship ∣Δr∣≈r Δθ\lvert\Delta\mathbf{r}\rvert \approx r\,\Delta\theta (and ultimately v=rωv = r\omega) only works because Δθ\Delta\theta is in radians, which is also why arc length is simply r Δθr\,\Delta\theta.

Question 3

Start from the exact chord length ∣Δr∣=2rsin⁡(Δθ/2)\lvert\Delta\mathbf{r}\rvert = 2r\sin(\Delta\theta/2) and apply the small-angle approximation. Then form the speed v=lim⁡Δt→0∣Δr∣Δtv = \lim_{\Delta t \to 0} \dfrac{\lvert\Delta\mathbf{r}\rvert}{\Delta t}.

What is the resulting speed, given ω=dθdt\omega = \dfrac{d\theta}{dt}?

✅ Correct! Pull the constant r outside the limit; what's left is dθ/dt = ω.

❌ Not quite. After |Δr| ≈ rΔθ, dividing by Δt and taking the limit gives r·(dθ/dt) = rω — r appears to the first power, ω to the first power.

Show solution

Solution:

Apply sin⁡(Δθ/2)≈Δθ/2\sin(\Delta\theta/2) \approx \Delta\theta/2 to the exact chord length; the two factors of 22 cancel:

∣Δr∣≈2r⋅Δθ2=r Δθ\lvert\Delta\mathbf{r}\rvert \approx 2r \cdot \frac{\Delta\theta}{2} = r\,\Delta\theta

Now take the limit for the speed:

v=lim⁡Δt→0∣Δr∣Δt=lim⁡Δt→0r ΔθΔt=rlim⁡Δt→0ΔθΔt=r dθdt=rωv = \lim_{\Delta t \to 0} \frac{\lvert\Delta\mathbf{r}\rvert}{\Delta t} = \lim_{\Delta t \to 0} \frac{r\,\Delta\theta}{\Delta t} = r \lim_{\Delta t \to 0} \frac{\Delta\theta}{\Delta t} = r\,\frac{d\theta}{dt} = \boxed{r\omega}

The radius rr is constant, so it pulls outside the limit, and the remaining limit is the definition of the angular velocity ω\omega.

Question 4

True or False: In uniform circular motion, the velocity vector is always perpendicular to the position vector r\mathbf{r} (the radius) and points along the tangent to the circle.

✅ Correct! The tangent is perpendicular to the radius, so v ⊥ r and v points along θ̂.

❌ Not quite. In the limit the chord becomes the tangent, which is perpendicular to the radius — so the statement is true.

Show solution

Solution: True.

As Δt→0\Delta t \to 0, the chord Δr\Delta\mathbf{r} pivots until it lies along the tangent line at the particle's location. Velocity points along the displacement, so the velocity vector lines up with that tangent.

A tangent to a circle is always perpendicular to the radius at the point of contact. Therefore v⊥r\mathbf{v} \perp \mathbf{r}, and the velocity points in the θ^\hat\theta direction — along the direction of motion around the circle. (Its magnitude is the speed v=rωv = r\omega from Question 3.)

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