The time interval needed for the disk to rev from leave to 25.0 m/s is 0.57 s.
Given:
vo = 0
v = 25.4 m/s
r = 0.95 m
n = 1.21 rev
A.
r*Ο = v
Ο = v/r
= 25.4/.95
= 26.74 rad/s
B.
ΞΈ = 2Οn
= 2Ο Γ (1.21)
= 7.6 m
Using equation of motion,
Οf^2 = Οi^2 + 2Ξ±ΞΈ
(26.74)^2 = 2 Γ Ξ± Γ (7.6)
Ξ± = 715.03/15.2
= 47.04 rad/s^2
C.
Using equation of motion,
ΞΈ = Οi*t + 1/2*Ξ±*t^2
ΞΈ = 0 + 1/2*Ξ±*t^2
7.6 = 47.04/2 Γ t^2
= sqrt(0.323)
= 0.57 s
Therefore the correct answer is 0.57 s
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