Chapter 12, Problem 36PQ

(a)

To determine

Angular speed of each child.

Answer to Problem 36PQ

Angular speed of each child is $\underset{\_}{4.17\text{rad}/\text{s}}$.

Explanation of Solution

Both the child in the question are situated in two different locations. But all points on a rigid rotator rotate with the same angular speed $\omega$. Let us calculate the angular speed of the child placed near the edge of merry go round.

Write the equation to find the angular speed of the child at the outer edge.

  $\omega =\frac{v_{\text{outer}}}{r}$                                                                                                              (I)

Here, $\omega$ is the angular speed, $v_{\text{outer}}$ is the velocity of the outer edge, and $r$ is the distance from the center of disc to the outer point.

Conclusion:

Substitute $12.5\text{m}/\text{s}$ for $v_{\text{outer}}$ and $3.00\text{m}$ for $r$ in equation (I) to get $\omega$.

  $\begin{array}{c}\omega =\frac{12.5\text{m}/\text{s}}{3.00\text{m}}\\ =4.17\text{rad}/\text{s}\end{array}$

Therefore, angular speed of each child is $\underset{\_}{4.17\text{rad}/\text{s}}$.

(b)

To determine

The angular distance travelled by each child in $5.0\text{s}$.

Answer to Problem 36PQ

The angular distance moved by each child in $5.0\text{s}$ is $\underset{\_}{21\text{rad}}$.

Explanation of Solution

The angular speed of both the children are same. Therefore the angular displacement is also equal.

Write the equation to find the angular velocity of the disc.

  $\omega =\frac{\Delta \theta }{\Delta t}$                                                                                                                   (II)

Here, $\Delta \theta$ is the angular displacement and $\Delta t$ is the time.

Rearrange equation (II) to get $\Delta \theta$.

  $\Delta \theta =\omega \Delta t$                                                                                                              (III)

Conclusion:

Substitute $4.17\text{rad}/\text{s}$ for $\omega$ and $5.0\text{s}$ for $\Delta t$ in equation (III) to get $\Delta \theta$.

  $\begin{array}{c}\Delta \theta =\left(4.17\text{rad}/\text{s}\right)\left(5.0\text{s}\right)\\ =21\text{rad}\end{array}$

Therefore, the angular distance moved by each child in $5.0\text{s}$ is $\underset{\_}{21\text{rad}}$.

(c)

To determine

The distance in meters moved by the child in $5.0\text{s}$.

Answer to Problem 36PQ

The distance moved by child on inner edge of disc is $\underset{\_}{16\text{m}}$ and the distance moved by the child on outer edge is $\underset{\_}{63\text{m}}$.

Explanation of Solution

The distance moved by each child is different since they are placed at different distance from the center of the disc.

Write the equation to find the distance travelled by the child on the outer edge.

  $d_{\text{outer}}=r_o\Delta \theta$                                                                                                            (IV)

Here, $d_{\text{outer}}$ is the distance moved by child on outer edge and $r_o$ is the distance between the center of disc and the child on outer edge.

Write the equation to find the distance travelled by the child on the inner edge.

  $d_{\text{inner}}=r_i\Delta \theta$                                                                                                            (V)

Here, $d_{\text{inner}}$ is the distance moved by the child on the inner edge and $r_i$ is the distance between the center of disc and the child on inner edge.

Conclusion:

Substitute $3.00\text{m}$ for $r_o$ and $21\text{rad}$ in equation (IV) to get $d_{\text{outer}}$.

  $\begin{array}{c}{d}_{\text{outer}}=\left(3.00\text{m}\right)\left(21\text{rad}\right)\\ =63\text{m}\end{array}$

Substitute $0.75\text{m}$ for $r_i$ and $21\text{rad}$ in equation (V) to get $d_{\text{inner}}$.

  $\begin{array}{c}{d}_{\text{inner}}=\left(0.75\text{m}\right)\left(21\text{rad}\right)\\ =16\text{m}\end{array}$

Therefore, the distance moved by child on inner edge of disc is $\underset{\_}{16\text{m}}$ and the distance moved by the child on outer edge is $\underset{\_}{63\text{m}}$.

(d)

To determine

The centripetal force of each child and find the one with more difficulty to hold on.

Answer to Problem 36PQ

The centripetal force on child on the outer edge is $\underset{\_}{1.30\times {10}^3\text{N}}$ and the centripetal force on child on inner edge is $\underset{\_}{326\text{N}}$.

Explanation of Solution

Write the equation to find the centripetal force on child on outer edge.

  $F_{\text{C,outer}}=m{\omega }^2{r}_o$                                                                                                       (VI)

Here, $F_{C,\text{outer}}$ is the centripetal force on child on the outer edge, $m$ is the mass of child.

Write the equation to find the centripetal force on child on inner edge.

  $F_{C,\text{inner}}=m{\omega }^2{r}_i$                                                                                                      (VII)

Here, $F_{C,\text{inner}}$ is the centripetal force on child on inner side.

Conclusion:

Substitute $25.0\text{kg}$ for $m$, $4.17\text{rad}/\text{s}$ for $\omega$ and $3.00\text{m}$ for $r_o$ in equation (VI) to get $F_{C,\text{outer}}$.

  $\begin{array}{c}{F}_{C,\text{outer}}=\left(25.0\text{kg}\right){\left(4.17\text{rad}/\text{s}\right)}^2\left(3.00\text{m}\right)\\ =1.30\times {10}^3\text{N}\end{array}$

Substitute $25.0\text{kg}$ for $m$, $4.17\text{rad}/\text{s}$ for $\omega$ and $0.75\text{m}$ for $r_i$ in equation (VII) to get $F_{C,\text{inner}}$.

  $\begin{array}{c}{F}_{C,\text{inner}}=\left(25.0\text{kg}\right){\left(4.17\text{rad}/\text{s}\right)}^2\left(0.75\text{m}\right)\\ =326\text{N}\end{array}$

Therefore, the centripetal force on child on the outer edge is $\underset{\_}{1.30\times {10}^3\text{N}}$ and the centripetal force on child on inner edge is $\underset{\_}{326\text{N}}$.

The centripetal force acting on the child on outer edge is more. As the magnitude of centripetal force increases, the child will experience strong outward force which makes holding on difficult.