Aflywheel completes 42.3 rev as it slows from an angular speed of 1.44 rad/s to a complete stop. assuming uniform acceleration, what si the time required for it to come to rest? what is the angular acceleration?

t = 6.15 minutes

α = - 3.9 x 10^-3 rad/s^2

Explanation:

n = 42.3 rev

angle turns in one revolution = 2 π radians

Angle turns in 42.3 rev = 2 x 3.14 x 42.3 = 265.644 rad

ωo = 1.44 rad/s, ω = 0

Let t be the time taken to stop and the angular acceleration is α.

Use third equation of motion for rotational motion

ω^2 = ωo^2 + 2 α θ

0 = 1.44 x 1.44 + 2 α x 265.644

α = - 3.9 x 10^-3 rad/s^2

Use first equation of motion for rotational motion

ω = ωo + α t

0 = 1.44 - 3.9 x 10^-3 x t

t = 368.95 second = 6.15 minutes

if you answer my question i will answer yours

1/0.14nm = 10^10/1.4 a lot