Answer:
the final angular velocity of the platform with its load is 1.0356 rad/s
Explanation:
Given that;
mass of circular platform m = 97.1 kg
Initial angular velocity of platform ω₀ = 1.63 rad/s
mass of banana [tex]m_{b}[/tex] = 8.97 kg
at distance r = 4/5 { radius of platform }
mass of monkey [tex]m_{m}[/tex] = 22.1 kg
at edge = R
R = 1.73 m
now since there is No external Torque
Angular momentum will be conserved, so;
mR²/2 × ω₀ = [ mR²/2 + [tex]m_{b}[/tex] ([tex]\frac{4}{5}[/tex] R)² + [tex]m_{m}[/tex]R² ]w
m/2 × ω₀ = [ m/2 + [tex]m_{b}[/tex] ([tex]\frac{4}{5}[/tex] )² + [tex]m_{m}[/tex] ]w
we substitute
w = 97.1/2 × 1.63 / ( 97.1/2 + 8.97(16/25) + 22.1
w = 48.55 × [ 1.63 / ( 48.55 + 5.7408 + 22.1 )
w = 48.55 × [ 1.63 / ( 76.3908 ) ]
w = 48.55 × 0.02133
w = 1.0356 rad/s
Therefore; the final angular velocity of the platform with its load is 1.0356 rad/s
