Angular velocity from rod data

Data note. The printed θ at t = 0.6 is 0.123, which breaks the smooth trend (0, 0.122, 0.493, …, 2.022; the second differences are almost constant only if θ(0.6) = 1.123). It is taken as a misprint of 1.123.

The time steps are equal (h = 0.2) and t = 0.2 is the second tabulated point, so Newton's forward formula with t₀ = 0 and p = 1 is used. Angular velocity ω = dθ/dt and angular acceleration α = d²θ/dt².

Difference table

Take t₀ = 0, h = 0.2, so p = (0.2 − 0)/0.2 = 1.00. Differentiating Newton's forward formula with respect to t gives

dθ/dt  = (1/h)[Δθ₀ + (2p−1)/2·Δ²θ₀ + (3p²−6p+2)/6·Δ³θ₀ + (4p³−18p²+22p−6)/24·Δ⁴θ₀]
d²θ/dt² = (1/h²)[Δ²θ₀ + (p−1)Δ³θ₀ + (6p²−18p+11)/12·Δ⁴θ₀]

First derivative

Sum = 0.2448, so dθ/dt = 0.2448/0.2 = 1.2242

Second derivative

Sum = 0.2490, so d²θ/dt² = 0.2490/0.2² = 6.2250

Answer: ω(0.2) ≈ 1.224 rad/s, α(0.2) ≈ 6.225 rad/s²

(With the printed 0.123 taken literally the same method gives ω ≈ 3.724 rad/s and α ≈ -2.108 rad/s², which is not physically consistent with the rest of the table.)