A merry-go-round rotates from rest with an angular acceleration of 1.08 rad/s2. How long does it take to rotate through (a) the first 3.74 rev and (b) the next 3.74 rev

Let ω1 be the initial angular speed and ω2 the final angular speed:

α = (ω2- α1) / t      

corresponding to a = (v2 - v1) / t  

S (distance corresponds to theta)

1 rev = 2 pi      and 3.74 rev = 7.48 pi = 23.5 radians

S = 1/2 a t^2 linear     or S = 1/2 α t^2    angular acceleration

23.5 = 1/2 * 1.08 t^2     and t = 6.60 sec   for first 3.84 rev

b)  ω1 = 1.08 * 6.6 = 7.13 rad/sec    initial speed for second 3.74

23.5 = 7.13 t + .54 t^2        compare to S = v1 t + 1/2 a t^2

.54 t^2 + 7.13 t - 23.5 = 0    

t^2 + 13.2 t - 43.5 = 0

t = 2.7 sec for next 3.74

Check:

7.13 * 2.7 + .54 * 2.7^2 = 23.2 rad = 3.7 rad

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explanation:

true is the correct answer

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