Three children are riding on the edge of a merry-go-round that is 100 kg, has a 1.60-m radius, and is spinning at 20.0 rpm. The children have masses of 22.0, 28.0, and 33.0 kg. If the child who has a mass of 28.0 kg moves to the center of the merry-go-round, what is the new angular velocity in rpm?

According to the law of conservation of angular momentum, when no external torque is applied, then the angular momentum of the system remains constant.

Initial angular momentum = final angular momentum

(1/2 x M x r^2 + m1 x r^2 + m2 x r^2 + m3 x r^2) x ω =

(1/2 x M x r^2 + m1 x r^2 + m3 x r^2 ) x ω'

(1/2 M + m1 + m2 + m3) x 2 x π x f = (1/2 M + m1 + m3) x 2 x π x f'

( 1/2 x 100 + 22 + 28 + 33) x 20 = (1/2 x 100 + 22 + 33) x f'

2660 = 105 x f'

f' = 25.33 rpm

onyebuchinnaji onyebuchinnaji · Answer 2 · 2021-11-04T22:00:55+01:00

The new angular velocity of the merry-go-round is 25.34 rpm.

The given parameters;

mass of the merry go round, M = 100 kg
mass of the children, m₁, m₂, m₃ = 22 kg, 28 kg, 33 kg
initial angular velocity, ω₁ = 20 rpm
radius of the merry - go - round, r = 1.6 m

Apply the principle of conservation of angular momentum;

[tex]I_1\omega _1 = I_2 \omega _2\\\\(\frac{1}{2} Mr^2 + m_1r^2 + m_2r^2 + m_3r^2 )\omega_1 = (\frac{1}{2} Mr^2 + m_1r^2 + m_3r^2 )\omega_2\\\\\(\frac{1}{2} M + m_1 + m_2 + m_3) r^2 \omega_1 = (\frac{1}{2} M + m_1 + m_3 ) r^2 \omega_2\\\\(\frac{1}{2} M + m_1 + m_2 + m_3) \omega_1 = (\frac{1}{2} M + m_1 + m_3 ) \omega_2\\\\(0.5M + m_1 + m_2 + m_3)\omega _1 = (0.5M + m_1 + m_3) \omega _2\\\\\omega_2 = \frac{(0.5M + m_1 + m_2 + m_3)\omega _1}{0.5M + m_1 + m_3 } \\\\[/tex]

[tex]\omega_2 = \frac{(0.5\times 100\ +\ 22 \ +\ 28 \ +\ 33) 20}{0.5\times 100 \ + \ 22 \ + \ 33 }\\\\\omega_2 = 25.34 \ rpm[/tex]

Thus, the new angular velocity of the merry-go-round is 25.34 rpm.

Learn more here:https://brainly.com/question/4126751