Chapter 8, Problem 3SP

In the park, several children (having a total mass of 90 kg) are riding on a merry-go-round that has a rotational inertia of 1100 kg·m² and a radius of 2.4 m. The average distance of the children from the axle of the merry-go-round is 2.2 m initially, because they are all riding near the edge.

1. a. What is the rotational inertia of the children about the axle of the merry-go-round? What is the total rotational inertia of the children and the merry-go-round?
2. b. The children now move inward toward the center of the merry-go-round so that their average distance from the axle is 0.8 m. What is the new rotational inertia for the system?
3. c. If the initial rotational velocity of the merry-go-round was 1.3 rad/s, what is the rotational velocity after the children move in toward the center, assuming that the frictional torque can be ignored? (Use conservation of angular momentum.)
4. d. Is the merry-go-round rotationally accelerated during this process? If so, where does the accelerating torque come from?

To determine

The rotational inertia of the children about the axle of the merry-go-round and the total rotational inertia of the children and the merry-go-round.

Answer to Problem 3SP

The rotational inertia of the children is $435.6\text{ kg}\cdot {\text{m}}^2$ and the total rotational inertia is $1535.6\text{ kg}\cdot {\text{m}}^2$.

Explanation of Solution

Given info: Mass is $90\text{ kg}$, rotational inertia is $\text{1100kg}\cdot {\text{m}}^2$, radius is $2.4\text{ m}$ and the average distance of the children from the axle is $2.2\text{ m}$.

Write the expression for the rotational inertia.

$I=m{r}^2$

Here,

$I$ is the inertia

$m$ is the mass

$r$ is the distance

Substitute $90\text{ kg}$ for $m$ and $\text{2}\text{.2 m}$ for $r$ to find $I$.

$\begin{array}{c}I=90\text{ kg}\times {\left(2.2\right)}^2\\ =435.6\text{ kg}\cdot {\text{m}}^2\end{array}$

The total rotational inertia acting on the merry-go—round is given by adding with the rotational inertia of the merry-go-round.

$I_{\text{total}}=I_{\text{merry}}+I$

$\begin{array}{c}{I}_{\text{total}}=\left(1100+435.6\right)\text{ kg}\cdot {\text{m}}^2\\ =1535.6\text{ kg}\cdot {\text{m}}^2\end{array}$

Conclusion:

Therefore, the rotational inertia of the children is $435.6\text{ kg}\cdot {\text{m}}^2$ and the total rotational inertia is $1535.6\text{ kg}\cdot {\text{m}}^2$.

To determine

The new rotational inertia of the merry-go-round.

Answer to Problem 3SP

The new rotational inertia of the merry-go-round is $1157.6\text{ kg}\cdot {\text{m}}^2$.

Explanation of Solution

Write the expression for the rotational inertia.

$I=m{r}^2$

Substitute $90$ for $m$ and $0.8$ for $r$ to find $I$.

$\begin{array}{c}I=90\times {\left(0.8\right)}^2\\ =57.6\text{ kg}\cdot {\text{m}}^2\end{array}$

The total rotational inertia acting on the merry-go—round is given by adding with the rotational inertia of the merry-go-round.

$I_{\text{total}}=I_{\text{merry}}+I$

$\begin{array}{c}{I}_{\text{total}}=\left(1100+57.6\right)\text{ kg}\cdot {\text{m}}^2\\ =1157.6\text{ kg}\cdot {\text{m}}^2\end{array}$

Conclusion:

Therefore, the new rotational inertia of the merry-go-round is $1157.6\text{ kg}\cdot {\text{m}}^2$.

To determine

The rotational velocity of the merry-go-round after the children move in towards the center.

Answer to Problem 3SP

The rotational velocity of the merry-go-round after the children move towards the center is $1.72\text{rad}/\text{s}$

Explanation of Solution

Write the expression for the conservation of angular momentum.

$I_1{\omega }_1=I_2{\omega }_2$

Here,

$I$ is the inertia

$\omega$ is the angular velocity

Substitute $1535.6\text{ kg}\cdot {\text{m}}^2$ for $I_1$, $1157.6\text{ kg}\cdot {\text{m}}^2$ for $I_2$ to find ${\omega }_2$.

$1535.6\text{ kg}\cdot {\text{m}}^2\times 1.3\text{rad}/\text{s}=1157\text{ kg}\cdot {\text{m}}^2\times {\omega }_2$

$\begin{array}{c}{\omega }_2=\frac{1535.6\text{ kg}\cdot {\text{m}}^2\times 1.3\text{rad}/\text{s}}{1157\text{ kg}\cdot {\text{m}}^2}\\ =1.72\text{rad}/\text{s}\end{array}$

Conclusion:

Therefore, the rotational velocity of the merry-go-round will be $\text{1}\text{.72}\text{rad}/\text{s}$.

To determine

Whether the merry-go-round rotationally accelerated during the process and where does the accelerating torque come from.

Answer to Problem 3SP

Yes, the merry-go-round rotationally accelerated during the process

Explanation of Solution

Write the expression for the rotational acceleration.

$\alpha =\frac{\tau }{I}$

When the children moving, at that time the friction between the feet of the children and the merry-go-round produces an accelerating torque.

Conclusion:

Therefore, the merry-go-round rotationally accelerated during the process