Question

Kyle boards a Ferris wheel at the 3-o'clock position and rides the Ferris wheel for multiple revolutions. The Ferris wheel rotates at a constant angular speed of 5 radians per minute and has a radius of 40 feet. The center of the Ferris wheel is 47 feet above the ground. Let t represent the number of minutes since the Ferris wheel started rotating.
Required:
Write an expression (in terms of t) to represent the varying number of radians θ Ryan has swept out since the ride started.

Answer 1

Answer:

a) The expression for the change in angular position is $\theta = 5\cdot t$.

b) The expression for the height of Ryan regarding the center of the Ferris wheel is $H(t) = 40\cdot \sin (5\cdot t)$.

c) The expression for the height of Ryan above the ground is $H(t) = 40\cdot \sin (5\cdot t) + 47$.

Step-by-step explanation:

The statement is incomplete. Complete form is introduced below:

Kyle boards a Ferris wheel at the 3-o'clock position and rides the Ferris wheel for multiple revolutions. The Ferris wheel rotates at a constant angular speed of 5 radians per minute and has a radius of 40 feet. The center of the Ferris wheel is 47 feet above the ground. Let t represent the number of minutes since the Ferris wheel stated rotating.

a) Write an expression (in terms of t) to represent the varying number of radians $\theta$ Ryan has swept out since the ride started.

b) Write an expression (in terms of t) to represent Ryan's height (in feet) above the center of the Ferris wheel.

c) Write an expression (in terms of t) to represent Ryan's height (in feet) above the ground.

a) Let suppose that Ferris wheel rotates counterclockwise. As the Ferris wheel rotates at constant rate, this kinematic expression can be used to determine the change in angular position ($\theta$), in radians:

$\theta = \omega \cdot t$ (1)

Where:

$\omega$ - Angular velocity, in radians per minute.

$t$ - Time, in second.

If we know that $\omega = 5\,\frac{rad}{min}$, then the expression for the change in angular position is $\theta = 5\cdot t$.

b) Geometrically speaking, Ryan's height with respect to the center of the Ferris wheel is described by the following formula:

$H(t) = r \cdot \sin (\omega\cdot t)$ (2)

Where:

$r$ - Radius of the Ferris wheel, in feet.

$H(t)$ - Height of Ryan with respect to the center of the Ferris wheel, in feet.

If we know that $\omega = 5\,\frac{rad}{min}$ and $r = 40\,ft$, then the expression for the height of Ryan regarding the center of the Ferris wheel is $H(t) = 40\cdot \sin (5\cdot t)$.

c) We use the following geometric expression to model Ryan's height above the ground:

$H(t) = r\cdot \sin (\omega\cdot t) +h_{o}$ (3)

Where $h_{o}$ is the height of the center of the Ferris wheel above the ground, in feet.

If we know that $h_{o} = 47\,ft$, $\omega = 5\,\frac{rad}{min}$ and $r = 40\,ft$, then the expression for the height of Ryan above the ground is $H(t) = 40\cdot \sin (5\cdot t) + 47$.