Question

Three children are riding on the edge of a merry-go-round that is a disk of mass 94 kg, radius 1.1 m, and is spinning at 25 rpm. The children have masses of 21.2 kg, 28 kg, and 34.2 kg. IF THE CHILD WHO HAS THE MASS OF 28 kg moves to the centre of merry go ground what is the new magnitude of angular velocity in rpm?

Answer 1

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.

`L_i=L_f`

The momentum is the product of the moment of inertia and the angular velocity, so the equation becomes

`\begin{gathered} I_i\cdot\omega_i=I_f\cdot\omega_f \\ \omega_f=(I_i\cdot\omega_i)/(I_f) \end{gathered}`

Where the initial moment of inertia is the sum of the initial moment of inertia of merry-go-round and all the children.

`\begin{gathered} I_i=I_m+I_1+I_2+I_3 \\ I_i=(m_mr^2)/(2)+m_1r^2+m_2r^2_{}+m_3r^2_{} \end{gathered}`

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 .

`\begin{gathered} I_f=I_m+I_1+I_3 \\ I_f=(m_mr^2)/(2)+m_1r^2+m_3r^2_{} \end{gathered}`

First, convert the angular velocity of merry-go-round from rpm to rad/s

`25\; (r)/(\min)*(2\pi)/(\sec)*(1\min)/(60\sec)=(5)/(6)\pi\; \; (rad)/(s)`

So, the new magnitude of angular velocity is

`\begin{gathered} \omega_f=(I_i\cdot\omega_i)/(I_f) \\ \omega_f=\frac{((m_mr^2)/(2)+m_1r^2+m_2r^2_{}+m_3r^2_{})\cdot w_i}{(m_mr^2)/(2)+m_1r^2+m_3r^2_{}} \\ \omega_f=(((94\cdot1.1^2)/(2)+21.2\cdot1.1^2+28\cdot1.1^2+34.2\cdot1.1^2)\cdot(5)/(6)\pi)/((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}`

Finally, convert the final angular velocity to rpm

`3.33\; \frac{\text{rad}}{\text{sec}}*\frac{1\; \text{rev}}{2\pi\; \text{rad}}*(60\sec)/(1\min)=31.80\; \text{rpm}`

Therefore, the new magnitude of angular velocity is 31.80 rpm