A merry-go-round rotates from rest with an angular acceleration of 1.04 rad/s2. How long does it take to rotate through (a) the first 1.59 rev and (b) the next 1.59 rev

(a) 4.38 s.

(b) 1.817 s

Explanation:

(a)

Using

θ = ω₀t +1/2αt² Equation 1

Where θ  = number of revolution, t = time, α = angular acceleration,  ω₀ = angular velocity.

Given: θ  = 1.59 rev = 1.59×2π = 9.992 rad,  ω₀ = 0 rad/s, α = 1.04 rad/s².

Substitute into equation 1

9.992 = 0(t) + 1/2(1.04)(t²)

t² = (2×9.992)/1.04

t² = 19.984/1.04

t = √(19.215)

t =4.38 s.

(b)

also using

θ = ω₀t +1/2αt² Equation 1

Given: θ  =3.18 rev = 3.18×2π = 19.97 rad,  ω₀ = 0 rad/s, α = 1.04 rad/s².

Substitute into equation 1

19.97 = 0(t) + 1/2(1.04)(t²)

t² = 19.97×2/1.04

t = √(38.40)

t = 6.197 s

The time require = 6.197-4.38 = 1.817 s

find the sum of the two horizontal vectors and the sum of the two vertical vectors

isoceles right angled triangle