Question

Three children are riding on the edge of a merry-go-round that is 182 kg, has a 1.60 m radius, and is spinning at 15.3 rpm. The children have masses of 17.4, 28.5, and 32.8 kg. If the child who has a mass of 28.5 kg moves to the center of the merry-go-round, what is the new angular velocity in rpm

Answer 1

Answer:

The new angular velocity of the merry-go-round is 18.388 revolutions per minute.

Explanation:

The merry-go-round can be represented by a solid disk, whereas the three children can be considered as particles. Since there is no external force acting on the system, we can apply the principle of angular momentum conservation:

$\left(\frac{1}{2}\cdot M + m_{1}+m_{2} + m_{3} \right)\cdot R^{2}\cdot \dot n_{o} = \left(\frac{1}{2}\cdot M + m_{1} + m_{3})\cdot R^{2}\cdot \dot n_{f}$ (1)

Where:

$M$ - Mass of the merry-go-round, in kilograms.

$m_{1}$, $m_{2}$, $m_{3}$ - Masses of the three children, in kilograms.

$R$ - Radius of the merry-go-round/Distance of the children with respect to the center of the merry-go-round, in meters.

$\dot n_{o}$, $\dot n_{f}$ - Initial and final angular speed, in revolutions per minute.

If we know that $M = 182\,kg$, $m_{1} = 17.4\,kg$, $m_{2} = 28.5\,kg$, $m_{3} = 32.8\,kg$, $R = 1.60\,m$ and $\dot n_{o} = 15.3\,\frac{rev}{min}$, then the final angular speed of the system is:

$\dot n_{f} = \dot n_{o}\cdot \left(\frac{\frac{1}{2}\cdot M + m_{1} + m_{2} + m_{3} }{\frac{1}{2}\cdot M + m_{1} + m_{3} } \right)$

$\dot n_{f} = \left(15.3\,\frac{rev}{min} \right)\cdot \left[\frac{\frac{1}{2}\cdot (182\,kg) + 17.4\,kg +28.5\,kg + 32.8\,kg }{\frac{1}{2}\cdot (182\,kg) + 17.4\,kg + 32.8\,kg } \right]$

$\dot n_{f} = 18.388\,\frac{rev}{min}$

The new angular velocity of the merry-go-round is 18.388 revolutions per minute.