Wheel Trig Problem

**stones44** · November 19th 2007, 08:39 AM

A pebble is wedged in the bottom of a tire. The horizontal distance of the pebble from the left-most edge of the tire varies sinusoidally with the distance you have traveled. Diameter is 5pi inches

L=distance of the pebble from the left-most edge of the tire after you have traveled x inches. Write en equation relating L and X….work in degrees and radians? Degrees first I guess than convert…I dno.

Then…..determine the first positive x value that makes L = 1

please use diagrams

**Soroban** · November 19th 2007, 10:28 AM

Hello, stones44!

A pebble is wedged in the bottom of a tire.
The horizontal distance of the pebble from the left-most edge of the tire varies
sinusoidally with the distance the wheel has traveled. The diameter is $5\pi$ inches.

$L$ = distance of the pebble from the leftmost edge of the tire.
$x$ = horizontal displacement of the wheel.

(a) Write an equation relating $L$ and $x.$

(b) Determine the first positive $x$ value that makes $L = 1$

Code:

              * * *
          *           *
        *               *
       *                 *

      * E       O         *
    D *-+-------*         *
      * :     / |         *
        :   / θ |
       *: /     |        *
       C* - - - +A      *
          *     |     *
    - - - - - * * * - - - - - - 
                B

Let the radius be $r = OB = OC = OD$
Let $L = DE$ .

When the wheel has turned $\theta$ radians,
. . it has moved a distance $x$ equal to arc(BC): . $x \:=\:r\theta\quad\Rightarrow\quad\theta \:=\:\frac{x}{r}$ .[1]

We see that: . $AC \:=\: OE\:=\:r\sin\theta$

. . Hence: . $L \:=\:r - r\sin\theta \:=\:r(1 - \sin\theta)$ .[2]

Substitute [1] into [2]: . $L \;=\;r\left(1 - \sin\frac{x}{r}\right)$

Since $r = \frac{5\pi}{2}$ . we have: . $\boxed{L \;=\;\frac{5\pi}{2}\left(1 - \sin\frac{2x}{5\pi}\right)}$ .(a)

If $L = 1$ , we have: . $1 \;=\;\frac{5\pi}{2}\left(1 - \sin\frac{2x}{5\pi}\right) \quad\Rightarrow\quad \frac{2}{5\pi} \;=\;1 - \sin\frac{2x}{5\pi}$

. . $\sin\frac{2x}{5\pi} \;=\;1 - \frac{2}{5\pi}\;=\;\frac{5\pi-2}{5\pi} \quad\Rightarrow\quad \frac{2x}{5\pi} \;=\;\sin^{-1}\!\left(\frac{5\pi-2}{5\pi}\right)$

Therefore: . $x \;=\;\frac{5\pi}{2}\sin^{-1}\!\left(\frac{5\pi-2}{5\pi}\right) \;\approx\;\boxed{8.33\text{ inches}}$ .(b)

**stones44** · November 19th 2007, 02:48 PM

yeah the distance L.....its DC not DE....the pebble is on the circumference somewhere....

oh wait..but it said horizontal distance...OK!

heres what i have so far...

**stones44** · November 19th 2007, 03:52 PM

wait..what if the pebble is in the first quadrant..than it would be r+rsintheta

**stones44** · November 19th 2007, 05:45 PM

Since

. we have: .

.(a)
why is it 2x/5pi

shouldnt it be x/2.5pi...are they the same thing?

Thread: Wheel Trig Problem

LinkBack

Thread Tools

Display

Wheel Trig Problem

Similar Math Help Forum Discussions

Ratio of Left Wheel to Right Wheel Based on Coordinate Data

Double Ferris wheel problem

Ferris wheel problem calculus

wheel problem

wheel trig problem

Search Tags