Question

a 40cm diameter wheel accelerates uniformly from 240 rpm to 360 rpm in 6.5 seconds. how far will a point on the edge of the wheel have traveled in this time

Answer 1

S = V1 t + 1/2 a t^2 distance traveled for linear acceleration

θ = ω1 t + 1/2 α t^2 equivalent angle for rotational acceleration

ω1 = 240 / min * 2 π * 240 / (60 s/min) = 8 π / s

ω2 = 360 rpm = 12 π / s

α = 4 π/ 6.5 s^2 = .62 π / s^2 angular acceleration

θ = 8 π / s 6.5 s + .62 π / 2 * 6.5^2

θ = 52 π + 13 π = 65 π = 65 * π / (2 π / rev) = 32.5 rev

1 rev = 2 * π * R = 2 * π * .4 m = 2.51 m / rev

S = 32.5 rev * 2.5 m/rev = 81 m