Chapter 8, Problem 3SP

(a)

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 $960\text{ kg}\cdot {\text{m}}^2$ and the total rotational inertia is $2460\text{ kg}\cdot {\text{m}}^2$.

Explanation of Solution

Given info: The total mass of the children is $240\text{ kg}$ and the rotational inertia of the merry-go-round is $\text{1500 kg}\cdot {\text{m}}^2$. The radius of the merry-go-round is $2.2\text{ m}$ and the average distance of the children from the axle is $2.0\text{ m}$.

Write the expression for the rotational inertia of the children.

$I=m{r}^2$

Here,

$I$ is the rotational inertia of the children

$m$ is the mass of the children

$r$ is the distance

Substitute $240\text{ kg}$ for $m$ and $2.0\text{ m}$ for $r$ to find $I$.

$\begin{array}{c}I=\left(240\text{ kg}\right){\left(2.0\text{ m}\right)}^2\\ =960\text{ kg}\cdot {\text{m}}^2\end{array}$

Write the expression for total rotational inertia acting on the merry-go-round by adding the new rotational inertia of the children with the rotational inertia of the merry-go-round.

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

Here,

$I_{\text{merry}}$ is the rotational inertia of the merry-go-round

Substitute $\text{1500 kg}\cdot {\text{m}}^2$ for $I_{\text{merry}}$ and $960\text{ kg}\cdot {\text{m}}^2$ for $I$ in the above equation.

$\begin{array}{c}{I}_{\text{total}}=\text{1500 kg}\cdot {\text{m}}^2+960\text{ kg}\cdot {\text{m}}^2\\ =2460\text{ kg}\cdot {\text{m}}^2\end{array}$

Conclusion:

Therefore, the rotational inertia of the children is $960\text{ kg}\cdot {\text{m}}^2$ and the total rotational inertia is $2460\text{ kg}\cdot {\text{m}}^2$.

(b)

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 $1560\text{ kg}\cdot {\text{m}}^2$.

Explanation of Solution

Write the expression for the rotational inertia.

$I=m{r}^2$

Substitute $240\text{ kg}$ for $m$ and $0.5\text{ m}$ for $r$ to find $I$.

$\begin{array}{c}I=\left(240\text{ kg}\right){\left(0.5\text{ m}\right)}^2\\ =60\text{ kg}\cdot {\text{m}}^2\end{array}$

Write the expression for total rotational inertia acting on the merry-go-round by adding the new rotational inertia of the children with the rotational inertia of the merry-go-round.

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

Here,

$I_{\text{merry}}$ is the rotational inertia of the merry-go-round

Substitute $\text{1500 kg}\cdot {\text{m}}^2$ for $I_{\text{merry}}$ and $60\text{ kg}\cdot {\text{m}}^2$ for $I$ in the above equation.

$\begin{array}{c}{I}_{\text{total}}=\text{1500 kg}\cdot {\text{m}}^2+60\text{ kg}\cdot {\text{m}}^2\\ =1560\text{ kg}\cdot {\text{m}}^2\end{array}$

Conclusion:

Therefore, the new rotational inertia of the merry-go-round is $1560\text{ kg}\cdot {\text{m}}^2$.

(c)

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 $1.89\text{rad}/\text{s}$

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 $2460\text{ kg}\cdot {\text{m}}^2$ for $I_1$, $1560\text{ kg}\cdot {\text{m}}^2$ for $I_2$ and $1.2\text{rad}/\text{s}$ for ${\omega }_1$ to find ${\omega }_2$.

$\begin{array}{c}{\omega }_2=\frac{\left(2460\text{ kg}\cdot {\text{m}}^2\right)\left(1.2\text{rad}/\text{s}\right)}{1560\text{ kg}\cdot {\text{m}}^2}\\ =1.89\text{rad}/\text{s}\end{array}$

Conclusion:

Therefore, The rotational velocity of the merry-go-round after the children move towards the center is $1.89\text{rad}/\text{s}$

(d)

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 is rotationally accelerated during the process

Explanation of Solution

The rotational acceleration is defined as the ratio of torque and rotational inertia. The main causes of rotational acceleration is friction.

Write the expression for the rotational acceleration.

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

Here,

$\alpha$ is the rotational acceleration

$\tau$ is the torque acting on the merry-go-round

$I$ is the rotational inertia of the merry-go-round

When the children are 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 is rotationally accelerated during the process