Chapter 4.PS, Problem 1PS

Angle of Rotation The restaurant at the top of the Space Needle in Seattle, Washington, is circular and has a radius of 47.25 feet. The dining part of the restaurant revolves, making about one complete revolution every 48 minutes. A dinner party, seated at the edge of the revolving restaurant at 6:45 P.M., finishes at 8:57 P.M.

(a) Find the angle through which the dinner party rotated.

(b) Find the distance the party traveled during dinner.

To determine

a)

To find:

The angle by which the dinner party is rotated from 6-45 P.M to 8-57 P.M.

Answer to Problem 1PS

Solution:

$990°$.

Explanation of Solution

Given,

The radius of the circular restaurant $=47.25\mathrm{ }$ feet.

Time of one complete revolution of the dining part $=48$ minutes.

Duration of the dinner party is from 6-45 P.M to 8-57 P.M.

The party between 6-45 P.M to 8-57 P.M corresponds to the duration of 2 hours and 12 minutes or 132 minutes.

First let us calculate the angle by which the dinner party is rotated in a minute.

Angle by which the dinner party rotated in $48$ Minute $=360°$ (Full rotation).

Thus, the angle by which the dinner party rotated in. $1$ Minute is $\frac{360°}{48}$ which is $=7.5°$.

Now, the angle by which the dinner party rotated in 132 minutes is,

$7.5°\times 132=990°$.

Therefore, the dinner party was rotated through the angle of $990°$.

To determine

b)

To find:

The distance by which the party travelled during dinner.

Answer to Problem 1PS

Solution:

816.01feet

Explanation of Solution

Given,

The radius of the circular restaurant $=47.25\mathrm{ }$ feet.

Time of one complete revolution of the dining part $=48$ minutes

Duration of the dinner party from 6-45 P.M to 8-57 P.M.

The dinner party was rotated by the angle of $990°$ from 6-45 P.M to 8-57 P.M.

Thus, the dinner party completes two full rotations $(360°+360°)$ and a rotation of $270°$.

Thus, two full rotation corresponds to twice the circumference of the circular dinner party and a $270°$ corresponds to the length (Distance) of the arc of the circle (Dinner Party) with central angle $270°$.

Thus, total distance is given by,

$2\left(2\pi r\right)+\frac{\theta }{360}\left(2\pi r\right)$

$=2\pi r\left[2+\frac{\theta }{360}\right]$

$=\left(2\right)\left(3.14\right)\left(47.25\right)\left[2+\frac{270}{360}\right]$

$=\left(296.73\right)\left(2.75\right)$

$=816.01$ feet.

Therefore, the dinner party was rotated by the distance of 816.01feet.