Respuesta :
We are given the following information
Mass of merry-go-round = mm = 94 kg
Radius of merry-go-round = rm = 1.1 m
Angular velocity of merry-go-round = 25 rpm
Masses of three children = 21.2 kg, 28 kg, and 34.2 kg
If the child who has a mass of 28 kg moves to the center of merry go ground what is the new magnitude of angular velocity in rpm?
The initial momentum of the merry-go-round must be equal to the final momentum of the merry-go-round.
[tex]L_i=L_f[/tex]The momentum is the product of the moment of inertia and the angular velocity, so the equation becomes
[tex]\begin{gathered} I_i\cdot\omega_i=I_f\cdot\omega_f \\ \omega_f=\frac{I_i\cdot\omega_i}{I_f} \end{gathered}[/tex]Where the initial moment of inertia is the sum of the initial moment of inertia of merry-go-round and all the children.
[tex]\begin{gathered} I_i=I_m+I_1+I_2+I_3 \\ I_i=\frac{m_mr^2}{2}+m_1r^2+m_2r^2_{}+m_3r^2_{} \end{gathered}[/tex]The final moment of inertia is the sum of the final moment of inertia of merry-go-round and all the children except the child with the mass of 28 kg .
[tex]\begin{gathered} I_f=I_m+I_1+I_3 \\ I_f=\frac{m_mr^2}{2}+m_1r^2+m_3r^2_{} \end{gathered}[/tex]First, convert the angular velocity of merry-go-round from rpm to rad/s
[tex]25\; \frac{r}{\min}\times\frac{2\pi}{\sec}\times\frac{1\min}{60\sec}=\frac{5}{6}\pi\; \; \frac{rad}{s}[/tex]So, the new magnitude of angular velocity is
[tex]\begin{gathered} \omega_f=\frac{I_i\cdot\omega_i}{I_f} \\ \omega_f=\frac{(\frac{m_mr^2}{2}+m_1r^2+m_2r^2_{}+m_3r^2_{})\cdot w_i}{\frac{m_mr^2}{2}+m_1r^2+m_3r^2_{}} \\ \omega_f=\frac{(\frac{94\cdot1.1^2}{2}+21.2\cdot1.1^2+28\cdot1.1^2+34.2\cdot1.1^2)\cdot\frac{5}{6}\pi}{\frac{94\cdot1.1^2}{2}+21.2\cdot1.1^2+34.2\cdot1.1^2} \\ w_f=3.33\; \; \frac{\text{rad}}{s} \end{gathered}[/tex]Finally, convert the final angular velocity to rpm
[tex]3.33\; \frac{\text{rad}}{\text{sec}}\times\frac{1\; \text{rev}}{2\pi\; \text{rad}}\times\frac{60\sec}{1\min}=31.80\; \text{rpm}[/tex]Therefore, the new magnitude of angular velocity is 31.80 rpm
