Chapter 6.2, Problem 38E

a.

To determine

To find: the angular velocity of each child.

Answer to Problem 38E

  $1.2\text{ radians per second}$

Explanation of Solution

Given information:

The angle travelled by the lighter child from ground to the highest point = $35°$

Time taken by lighter child to go from ground to the highest point = $\frac{1}{2}\text{ second}$

Calculation:

The angular velocity $\left(\omega \right)$ of lighter and heavier child is evaluated as follows:

  $\begin{array}{c}\omega =\frac{\theta }{t}\\ =\frac{35°}{\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}\left[\text{Put the values}\right]\\ =\frac{35°}{\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}\times \left(\frac{\pi \text{ radian}}{180°}\right)\left[\text{Changing degree into radian measure}\right]\\ =\frac{\left(35°\right)\left(2\right)}{1}\times \left(\frac{\pi }{180°}\right)\\ =\frac{7\pi }{18}\\ =1.2\text{ radians per second}\end{array}$

Thus the angular velocity of both the children is $1.2\text{ radians per second}$ .

b.

To determine

To find: the linear velocity of each child.

Answer to Problem 38E

Lighter child: $11\text{ feet per second}$

Heavier child: $7.3\text{ feet per second}$

Explanation of Solution

Given information:

Distance between lighter child from the fulcrum $\left(r_l\right)$ = 9 feet

Distance between heavier child from the fulcrum $\left(r_h\right)$ = 6 feet

The angle travelled by the lighter child from ground to the highest point = $35°$

Time taken by lighter child to go from ground to the highest point = $\frac{1}{2}\text{ second}$

  $\text{Angular velocity of both the child }\left(\omega \right)=1.2\text{ radians per second}$

Calculation:

The linear velocity $\left(\nu \right)$ of lighter child is evaluated as follows:

  $\begin{array}{c}v=r\omega \\ =\left(9\right)\left(1.2\right)\left[\text{Put the value of }r=r_l\right]\\ =11\text{ feet per second}\end{array}$

Thus the linear velocity of lighter child is $11\text{ feet per second}$ .

The linear velocity $\left(\nu \right)$ of heavier child is evaluated as follows:

  $\begin{array}{c}v=r\omega \\ =\left(6\right)\left(1.2\right)\left[\text{Put the value of }r=r_h\right]\\ =7.3\text{ feet per second}\end{array}$

Thus the linear velocity of heavier child is $7.3\text{ feet per second}$ .