Respuesta :
We are given the following information
Radius of merry-go-round = 1.60 m
Mass of merry-go-round = 120 kg
Angular velocity of merry-go-round = 0.58 rev/s
Mass of child = 18 kg
We are asked to find the angular velocity of the merry-go-round after the child gets onto it by grabbing its outer edge.
The angular momentum before and after the child gets onto the merry-go-round must be equal
[tex]L_b=L_a[/tex]The angular momentum before the child gets onto the merry-go-round is the product of the moment of inertia of the merry-go-round without the child and the angular velocity of the merry-go-round before the child gets onto it.
The angular momentum after the child gets onto the merry-go-round is the product of the moment of inertia of the merry-go-round with the child and the final angular velocity of the merry-go-round after the child gets onto it.
[tex]I_m\cdot\omega_b=I_{m+c}\cdot\omega_f[/tex]Arranging the equation for the final angular velocity of the merry-go-round,
[tex]\begin{gathered} \omega_f=\frac{I_m\cdot\omega_b}{I_{m+c}} \\ \omega_f=\frac{\frac{m_mr^2_m}{2}\cdot\omega_b}{\frac{m_mr^2_m}{2}+m_cr^2_c} \end{gathered}[/tex]Where
mm = Mass of merry-go-round = 120 kg
rm = Radius of merry-go-round = 1.60 m
mc = Mass of child = 18 kg
rc = Radius of child relative to merry-go-round = 1.60 m
wb = Angular velocity of merry-go-round = 0.58 rev/s
First, convert the angular velocity from rev/s to rad/s
[tex]0.58\times2\pi=3.64\; \; \frac{\text{rad}}{s}[/tex]So, the final angular velocity is
[tex]\begin{gathered} \omega_f=\frac{\frac{m_mr^2_m}{2}\cdot\omega_b}{\frac{m_mr^2_m}{2}+m_cr^2_c} \\ \omega_f=\frac{\frac{120\cdot1.60^2}{2}\cdot3.64}{\frac{120\cdot1.60^2}{2}+18\cdot_{}1.60^2_{}} \\ \omega_f=2.80\; \; \frac{rad}{s} \end{gathered}[/tex]Therefore, the angular velocity of the merry-go-round after the child gets onto it by grabbing its outer edge is 2.80 rad/s
