ω = 0 + 2(20) = 40 (according to rotational equation ω = ω_{0} + αt)

and θ = 0 + 1/2(2)(20)^{2} = 400 radian

in middle uniform part angle rotated = 40*10 = 400 radian

and since acceleration in first part and retardation in last part are equal so angle rotated in last part = 400 radian

total = 1200 radian