Question

A child of mass 50 kg is standing on the edge of a merry go round of mass 250 kg and
radius 3.0 m which is rotating with an angular velocity of 3.0 rad s−1
. The child then
starts walking towards the centre of the merry go round. What will be the final angular
velocity of the merry go round when the child reaches the centre?

Answer 1

Answer on Question 73308, Physics, Mechanics, Relativity

Question:

A child of mass $50\,kg$ is standing on the edge of a merry-go-round of mass $250\,kg$ and radius $3.0\,m$ which is rotating with an angular velocity of $3.0\,rad/s^{-1}$. The child then starts walking towards the centre of the merry-go-round. What will be the final angular velocity of the merry-go-round when the child reaches the centre?

Solution:

We can find the final angular velocity of the merry-go-round from the law of conservation of angular momentum:

$$L_i = L_f,$$ $$I_i \omega_i = I_f \omega_f,$$

here, $I_i$ is the initial rotational inertia of the system, $I_f$ is the final rotational inertia of the system, $\omega_i$ is the initial angular velocity of the merry-go-round, $\omega_f$ is the final angular velocity of the merry-go-round.

We can find the initial rotational inertia of the system as follows:

$$I_i = \left(I_{disk,i} + I_{child,i}\right) = \left(\frac{1}{2} m_{disk} r_{disk,i}^2 + m_{child} r_{child,i}^2\right),$$

here, $I_{disk,i} = \frac{1}{2} m_{disk} r_{disk,i}^2$ is the initial rotational inertia of the merry-go-round, $I_{child,i} = m_{child} r_{child,i}^2$ is the initial rotational inertia of the child, $m_{disk}$ is the mass of the merry-go-round, $m_{child}$ is the mass of the child, $r_{disk}$ is the radius of the merry-go-round, $r_{child}$ is the distance from the centre of the merry-go-round to the child.

Then, we can calculate $I_i$:

$$I_i = \left(\frac{1}{2} m_{disk} r_{disk,i}^2 + m_{child} r_{child,i}^2\right) = \left(\frac{1}{2} \cdot 250\,kg \cdot (3.0\,m)^2 + 50\,kg \cdot (3.0\,m)^2\right) = 1575\,kg \cdot m^2.$$

Similarly, we can find the final rotational inertia of the system:

$$I_f = \left(I_{disk,f} + I_{child,f}\right) = \left(\frac{1}{2} m_{disk} r_{disk,f}^2 + m_{child} r_{child,f}^2\right),$$

here, $I_{disk,f} = \frac{1}{2} m_{disk} r_{disk,f}^2$ is the final rotational inertia of the merry-go-round, $I_{child,f} = m_{child} r_{child,f}^2$ is the final rotational inertia of the child, $m_{disk}$ is the mass of the merry-go-round, $m_{child}$ is the mass of the child, $r_{disk}$ is the radius of the merry-go-round, $r_{child}$ is the distance from the centre of the merry-go-round to the child.

Then, we can calculate $I_{f}$:

$$\begin{array}{l} I_{f} = \left(\frac{1}{2} m_{disk} r_{disk,f}^2 + m_{child} r_{child,f}^2\right) = \\ = \left(\frac{1}{2} \cdot 250 \, kg \cdot (3.0 \, m)^2 + 50 \, kg \cdot (0.0 \, m)^2\right) = 1125 \, kg \cdot m^2. \end{array}$$

Finally, we can calculate the final angular velocity of the merry-go-round from the law of conservation of angular momentum:

$$I_{i} \omega_{i} = I_{f} \omega_{f},$$ $$\omega_{f} = \omega_{i} \frac{I_{i}}{I_{f}} = 3.0 \, \frac{rad}{s} \cdot \frac{1575 \, kg \cdot m^2}{1125 \, kg \cdot m^2} = 4.2 \, \frac{rad}{s}.$$

Answer:

$$\omega_{f} = 4.2 \, \frac{rad}{s}.$$

Answer provided by https://www.AssignmentExpert.com