What is the difference between thrust and applied force?
1 answer:
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Answer:
(a) the angular velocity at θ1 is 11.64 rad/s
(b) the angular acceleration is 0.12 rad/
(c) the angular position was the disk initially at rest is - 428.27 rad
Explanation:
Given information :
θ1 = 16 rad
θ2 = 76 rad
ω2 = 11 rad/s
t = 5.3 s
(a) The angular velocity at θ1
First, we use the angular motion equation for constant acceleration
Δθ = (ω1+ω2)t/2
θ2 - θ1 = (ω1+ω2)t/2
ω1 + ω2 = 2 (θ2 - θ1) / t
ω1 = (2 (θ2 - θ1) / t ) - ω2
= (2 (76-16) / 5.3) - 11
= 11.64 rad/s
(b) the angular acceleration
ω2 = ω1 + α t
α t = ω2 - ω1
α = (ω2 - ω1)/t
= (11.64 - 11) / 5.3
= 0.12 rad/
(c) the angular position was the disk initially at rest, θ0
at rest ω0 = 0
ω2^2 = ω01 t + 2 α Δθ
2 α Δθ = ω2^2
θ2 - θ0 = ω2^2 / 2 α
θ0 = θ2 - (ω2^2) / 2 α
= 76 - (/ 2 x 0.12
= 76 - 504.16
= - 428.27 rad