Answer:
(A) 180 kg·m²
(B) 0.111 rad/s²
(C) The number of revolutions the merry-go-round will complete until it finally stops is 3.142 or π rev
Explanation:
The equation of the moment of inertia of a solid cylinder is presented as follows;
Where:
I = Moment of inertia of the merry-go-round
M = Mass of the merry-go-round
R = Radius of the merry-go-round
Therefore, I = 1/2×90×2² = 180 kg·m²
(B) For the angular acceleration we have;
Therefore, since the force × radius = The torque, we have, angular acceleration is found as follows
F × R = τ
10.0 × 2.0 = 20 = I×α = 180×α
α = 20/180 = 0.111 rad/s².
angular acceleration = 0.111 rad/s².
(C) Here we have ω₀ = 20 rev/ min = 20×2×π rad/min = π·40/60 rad/s
2/3·π rad/s
ω = ω₀ - α×t
∴ t = ω₀/α = (2/3·π rad/s)/(0.111 rad/s²) = 18.85 s
Hence we have
θ = ω₀·t + 1/2·α·t², plugging in the values, we have;
θ = 2/3·π×18.85 - 1/2·0.111·18.85²
θ = 19.74 rad
Therefore, since 2·π radian = 1 revolution
The number of revolutions the merry-go-round will complete until it stops is 19.74/(2·π) = 3.142 or π revolutions.