Find the angular velocity of a body rotating with an acceleration of $2 r e v / s^{2}$ as it completes the 4th revolution after the start.

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Solution

$θ = 4 r e v$

initial angular speed= $ω_{o} = 0$

$α = 2 r e v / s^{2}$

Comparing with linear equation of $v^{2} - u^{2} = 2 a s$

$ω^{2} - ω_{o}^{2} = 2 α θ$

$⟹ ω^{2} = 2 \times 2 \times 4 = 16$

$⟹ ω = 4 r e v / s$

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Q. Find the angular velocity of a body rotating with an acceleration of 2rev/s² as it completes the 5th revolution after the start.[Concepts of Physics]
BOOK ANSWER - 4.47 rev/s
MY SOLUTION
In one revolution the distance covered is 2π,
so, in 5 revolution the distance moved is 10π
Angular Acceleration = 2rev/s
Initial angular velocity = 0
Final angular velocity = ?
So using equation,
v²= u²+ 2as
= 0 + 2* 2*10π
= 40π
v = 11.2rev/sec
Why my Answer is different? Where and what I am missing.
NOTE : I can't write omega on my tablet also you understand it well that we can type only that which is given in keyboard.

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Find the angular velocity of a body rotating with an acceleration of 2rev/s2 as it completes the 4th revolution after the start.

θ=4revinitial angular speed=ωo=0α=2rev/s2Comparing with linear equation of v2−u2=2asω2−ω2o=2αθ⟹ω2=2×2×4=16⟹ω=4rev/s

Find the angular velocity of a body rotating with an acceleration of $2 r e v / s^{2}$ as it completes the 4th revolution after the start.

$θ = 4 r e v$

initial angular speed= $ω_{o} = 0$

$α = 2 r e v / s^{2}$

Comparing with linear equation of $v^{2} - u^{2} = 2 a s$

$ω^{2} - ω_{o}^{2} = 2 α θ$

$⟹ ω^{2} = 2 \times 2 \times 4 = 16$

$⟹ ω = 4 r e v / s$