Chapter 6.2, Problem 37E

a.

To determine

To find: the linear velocity in feet per second.

Answer to Problem 37E

  $7.1\text{ feet per second}$

Explanation of Solution

Given information:

Angular velocity $\left(\omega \right)=\text{ 3 revolutions per minute}$

Distance of the rider from the centre $\left(r\right)=22\frac{1}{2}\text{ feet}$

Calculation:

The linear velocity can be evaluated as follows:

  $\begin{array}{c}v=r\omega \\ =22\frac{1}{2}\text{feet}\times \frac{3\text{ revolutions}}{1\text{ minute}}\left[\text{Put the value of }r\text{ and }\omega \right]\\ =\frac{45}{2}\text{feet}\times \frac{3\text{ revolutions}}{1\text{ minute}}\times \frac{2\pi \text{ radians}}{\text{1 revolution}}\left[\text{Convert revolutions into radians}\right]\\ =\frac{135\pi \text{ feet}}{1\text{ minute}}\times \frac{1\text{ minute}}{60\text{ seconds}}\left[\text{Convert minute into second}\right]\\ =7.1\text{ feet per second}\end{array}$

Thus the linear velocity is $7.1\text{ feet per second}$ .

b.

To determine

To find: the distance between the person from the center of the carousel.

Answer to Problem 37E

9.9 feet

Explanation of Solution

Given information:

Angular velocity $\left(\omega \right)=\text{ 3 revolutions per minute}$

Distance of the rider from the centre $\left(r\right)=22\frac{1}{2}\text{ feet}$

Liner velocity of person inside the carousel $\left({\nu }_{inside}\right)=$ 3.1 feet per second

Calculation:

The angular velocity of both the persons i.e. sitting inside and outside of the carousel will be same. So first evaluate the angular velocity in radian per second as shown below.

  $\begin{array}{c}\omega =\frac{3\text{ revolutions}}{1\text{ minute}}\\ =\frac{3\text{ revolutions}}{1\text{ minute}}\times \frac{2\pi \text{ radians}}{\text{1 revolution}}\left[\text{Convert revolutions into radians}\right]\\ =\frac{6\pi \text{ feet}}{1\text{ minute}}\times \frac{1\text{ minute}}{60\text{ seconds}}\left[\text{Convert minute into second}\right]\\ =0.314\text{ radians per second}\end{array}$

Thus the angular velocity is $0.314\text{ radians per second}$ .

The distance from the center of the carousel to the person $\left(r\right)$ is evaluated as follows:

  $\begin{array}{c}v=r\omega \\ 3.1\text{ feet per second}=r\left(0.314\text{ radians per second}\right)\left[\text{Put the value of }v\text{ and }\omega \right]\\ r=\frac{3.1\text{ feet per second}}{0.314\text{ radians per second}}\\ =9.9\text{ feet}\end{array}$

Thus the distance is 9.9 feet.

c.

To determine

To find: the rate by which the rider on the outside going is faster than the rider on the inside.

Answer to Problem 37E

  $4\text{ feet per second}$

Explanation of Solution

Given information:

Liner velocity of person inside the carousel $\left({\nu }_{inside}\right)=$ 3.1 feet per second

Linear velocity of carousel = $7.1\text{ feet per second}$

Calculation:

The linear velocity of carousel and that of person outside the carousel $\left({\nu }_{outside}\right)$ is same. The difference between the rates of linear velocity of both the persons is evaluated as follows:

  $\begin{array}{c}\Delta \nu ={\nu }_{outside}-{\nu }_{inside}\\ =7.1\text{ feet per second}-3.1\text{ feet per second}\\ =4\text{ feet per second}\end{array}$

Thus rate difference is $4\text{ feet per second}$ .