Chapter 8, Problem 84P

To determine

The speed of merry goes round after child gets into it. How much the rotational kinetic energy change.

Answer to Problem 84P

The speed of merry goes round after child gets into it is $\underset{\_}{{\omega }_{\text{f}}=0.61\text{rev/s}}$, and change in kinetic energy is $\underset{\_}{\Delta K_{\text{r}}=-660\text{J}}$.

Explanation of Solution

Write the expression for initial rotational inertia before child enters to the merry go round

$I_{\text{i}}=\frac{1}{2}M{R}^2$ (I)

Here, $M$ is the mass of merry go round, $R$ is the radius of the merry go round

After the child enters into merry go round, the rotational inertia changes. It is required to take account of the weight of the child.

Write the expression for the rotational inertia after the child enters the merry go round

$I_{\text{f}}=\frac{1}{2}M{R}^2+\frac{F_g{R}^2}{g}$ (II)

Here, $F_g$ is the weight of the child, $g$ is the acceleration due to gravity

Write the expression of initial angular momentum

$L_{\text{i}}=I_{\text{i}}{\omega }_{\text{i}}$ (III)

Here, $L_{\text{i}}$ is the initial angular momentum, ${\omega }_{\text{i}}$ is the initial angular speed

Substitute equation (I) in (III)

$L_{\text{i}}=\frac{1}{2}M{R}^2{\omega }_{\text{i}}$ (IV)

Write the expression of final angular momentum

$L_{\text{f}}=I_{\text{f}}{\omega }_{\text{f}}$ (V)

Here, $L_{\text{f}}$ is the final angular momentum, ${\omega }_{\text{f}}$ is the final angular speed

Substitute equation (II) in (V)

$L_{\text{f}}=\left(\frac{1}{2}M{R}^2+\frac{F_g{R}^2}{g}\right){\omega }_{\text{f}}$ (VI)

According to law of conservation of angular momentum, $L_{\text{i}}$, and $L_{\text{f}}$ are same. Equate equation (VI) and (IV)

$\begin{array}{c}{L}_{\text{i}}=L_{\text{f}}\\ \frac{1}{2}M{R}^2{\omega }_{\text{i}}=\left(\frac{1}{2}M{R}^2+\frac{F_g{R}^2}{g}\right){\omega }_{\text{f}}\end{array}$ (VII)

Rearrange equation (VII) to obtain an expression for ${\omega }_{\text{f}}$

$\begin{array}{c}\frac{1}{2}M{R}^2{\omega }_{\text{i}}=\left(\frac{1}{2}M{R}^2+\frac{F_g{R}^2}{g}\right){\omega }_{\text{f}}\\ {\omega }_{\text{f}}=\frac{\frac{1}{2}M{R}^2{\omega }_{\text{i}}}{\frac{1}{2}M{R}^2+\frac{F_g{R}^2}{g}}\\ ={\left(1+\frac{2F_g}{gM}\right)}^{-1}{\omega }_{\text{i}}\end{array}$ (VIII)

Write the expression for change rotational kinetic energy

$\Delta K_{\text{r}}=\frac{1}{2}{I}_{\text{f}}{\omega }_{\text{f}}{}^2-\frac{1}{2}{I}_{\text{i}}{\omega }_{\text{i}}{}^2$ (IX)

Substitute equation (I) and (II) in (IX)

$\Delta K_{\text{r}}=\frac{1}{2}\left(\frac{1}{2}M{R}^2+\frac{F_g{R}^2}{g}\right){\omega }_{\text{f}}{}^2-\frac{1}{2}\left(\frac{1}{2}M{R}^2\right){\omega }_{\text{i}}{}^2$ (X)

Conclusion:

Substitute $180\text{N}$ for $F_g$, $9.8{\text{m/s}}^2$ for $g$, $160\text{kg}$ for $M$, and $0.75\text{rev/s}$ ${\omega }_{\text{i}}$ in equation (VIII)

$\begin{array}{c}{\omega }_{\text{f}}={\left(1+\frac{2\left(180\text{N}\right)}{9.8{\text{m/s}}^2\times 160\text{kg}}\right)}^{-1}0.75\text{rev/s}\\ =0.6099\text{rev/s}\\ \approx \text{0}\text{.61}\text{rev/s}\end{array}$

Substitute $180\text{N}$ for $F_g$, $9.8{\text{m/s}}^2$ for $g$, $160\text{kg}$ for $M$, $2.0\text{m}$ for $R$, $0.75\text{rev/s}$ ${\omega }_{\text{i}}$, and $0.61\text{rev/s}$ for ${\omega }_{\text{f}}$ in equation (X)

$\begin{array}{c}\Delta K_{\text{r}}=\frac{1}{2}\left\{\left(\frac{1}{2}\left(160\text{kg}\right){\left(2.0\text{m}\right)}^2+\frac{\left(180\text{N}\right){\left(2.0\text{m}\right)}^2}{9.8{\text{m/s}}^2}\right){\left(0.61\text{rev/s}\right)}^2-\frac{1}{2}\left(\frac{1}{2}\left(160\text{kg}\right){\left(2.0\text{m}\right)}^2\right){\left(0.75\text{rev/s}\right)}^2\right\}{\left(\frac{2\pi \text{rad}}{\text{rev}}\right)}^2\\ =-662\text{J}\end{array}$

Therefore, the speed of merry goes round after child gets into it is $\underset{\_}{{\omega }_{\text{f}}=0.61\text{rev/s}}$, and change in kinetic energy is $\underset{\_}{\Delta K_{\text{r}}=-660\text{J}}$.