A flywheel turns through 24 rev as it slows from an angular speed of 1.2 rad/s to a stop.

(a) Assuming a constant angular acceleration, find the time for it to come to rest. s (b) What is its angular acceleration? rad/s² (c) How much time is required for it to complete the first 12 of the 24 revolutions? s

a) 251.36 s b) - 4.77 * 10⁻³ rad/s² c) 73.64 s

Explanation:

Using equation of angular motion

θ = (ω₀ + ω₁) t/2

2θ = (ω₁ + ω₀) t where ω₀ is the initial angular speed = 1.2 rad/s, ω₁ is the final angular speed = 0 and t is the time

2 * 24 * 2π = (0 + 1.2) t

t = 301.632 / 1.2 = 251.36 s

b) the angular acceleration can be calculated by:

ω₁ = ω₀ + αt

ω₁ - ω₀ = αt

0 - 1.2 = 251.36 α

α = - 1.2 / 251.36 = - 4.77 * 10⁻³ rad/s²

c) the amount of time required to complete the first 12 of the 24 revolutions

θ = ω₀t + 0.5αt²

12 * 2π = 1.2t - 0.00239t²

75.408 = 1.2t - 0.00239t²

- 0.00239t² + 1.2t - 75.408 = 0

multiply by - 1 both sides

0.00239t² - 1.2t + 75.408 = 0

using quadratic formula to solve for t

-b ±√ (b² - 4ac) / 2a

1.2 ±√ (1.44 - 0.721) / (2*0.00239)

1.2 ± √0.719 / 0.00478

1.2 - 0.848 / 0.00478 or 1.2 + 0.848 / 0.00478

0.352 / 0.00478 or 2.048 / 0.00478

t = 73.22 s or 428.5

t = 73.64 s