Motion of a particle

From the result of (a);

圖片參考：http://i388.photobucket.com/albums/oo325/loyitak1990/Nov10/Crazyapp1.jpg


Then from the given:

dθ/dt = ω and d2θ/dt2 = 0, also:


圖片參考：http://i388.photobucket.com/albums/oo325/loyitak1990/Nov10/Crazyapp2.jpg


So the acceleration vector a, in terms of tangential and radial component vectors eθ and er, can be rewritten as:


圖片參考：http://i388.photobucket.com/albums/oo325/loyitak1990/Nov10/Crazyapp3.jpg


So substituting r = 2 and θ = π/6, we have:

Radial acceleration = 12ω2 m/sTangential acceleration = 4√3 ω2 m/s

Also from θ = π/6 to θ = π/3, time elapsed = (π/3 - π/6)/ω = π/(6ω) s

And the particle only has horizontal displacement equal to:

2 cos π/6 - (2/√3) cos π/3 = 2/√3 m

So the average speed = (2/√3)/[π/(6ω)] = 4√3 ω/π m/s