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Proof 01: Speed in Circular and Spiral Motion

We want to prove that the speed of a particle whose radius r and angle θ may both change with time is:

v = √[(dr/dt)² + (rω)²]

1. Express position in Cartesian coordinates

For a particle at polar coordinates (r, θ):

x = r cos θ
y = r sin θ

Both r and θ may depend on time.

2. Differentiate with respect to time

Use the product rule and chain rule:

v_x = dx/dt = (dr/dt) cos θ − r sin θ (dθ/dt)
v_y = dy/dt = (dr/dt) sin θ + r cos θ (dθ/dt)

Since angular velocity ω = dθ/dt, and writing ṙ = dr/dt:

v_x = ṙ cos θ − rω sin θ
v_y = ṙ sin θ + rω cos θ

3. Find the magnitude of velocity

By Pythagoras' theorem, v² = v_x² + v_y²:

v² = (ṙ cos θ − rω sin θ)²
+ (ṙ sin θ + rω cos θ)²

Expand and collect terms:

v² = ṙ²(cos² θ + sin² θ)
+ r²ω²(sin² θ + cos² θ)
− 2ṙrω sin θ cos θ + 2ṙrω sin θ cos θ

The cross terms cancel. Since sin² θ + cos² θ = 1:

v² = ṙ² + r²ω²

Therefore: v = √[(dr/dt)² + (rω)²] ∎

4. Apply the result to the experiment

Fixed-radius circular motion: r = r₀, so dr/dt = 0 and v = r₀|ω|.

Inward spiral: r(t) = r₀ − kt, so dr/dt = −k and v = √[k² + (rω)²], until the particle reaches the centre.

Physical interpretation: Radial velocity (dr/dt) points along the radius, while tangential velocity (rω) points perpendicular to it. Their perpendicular components combine by Pythagoras. A spiral can have inward radial motion even when ω = 0.

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