Answer:
1.64 rad/s
Explanation:
Given,
radius of merry-go-round, r = 1.80 m
mass of merry-go-round, M = 120 Kg
angular speed, ω = 0.350 rev/s
Initial angular speed = 2π x 0.350 = 2.12 rad/s
Mass of child, m = 36.5 Kg
Moment of inertia of the merry-go-round
[tex]I = \dfrac{1}{2}Mr^2[/tex]
[tex]I = \dfrac{1}{2}\times 120\times 1.8^2[/tex]
[tex]I = 349.92\ kg.m^2[/tex]
L= I ω = 349.92 x 2.12 = 769.41 kgm²/s
Child moment of inertia
I₂ = m r^2 = 36.5 x 1.8² = 118.26 kg.m²
Final moment of inertia,
I_f = 349.92+118.26 = 468.18 kg.m²
Final speed = [tex]\dfrac{769.41}{468.14}[/tex]
[tex]\omega_f = 1.64 \rad/s[/tex]
angular speed of the child is equal to 1.64 rad/s