Answer: 0.975 m^2
Step-by-step explanation:
θ = 0.5αt^2
8.20 revolutions = 8.20 x 2π = 51.52 rad
51.52 = 0.5α(12)^2
α = 0.716 rad/s^2
ω = αt = 0.716(12) = 8.592 rad/s
KE = 1/2Iω^2
36 = 0.5I(8.592)^2
I = 0.975 m^2
Therefore, the moment of inertia of the wheel is approximately 0.975m^2