# Math Help - wheel problem

1. ## wheel problem

okay so this is an end of the unit problem and im completly stuck, i just need to know sort of where to start.

question;
Analyze the motion of a pebble stuck in a tire
write an equation for the motion at 50km/h, and 100km/h for two differnt wheels(radius)

So, so far if the radius was 10 cm,
then the info i can get is

F(x)=10Cos___(x)+10

but i cant figure out how to figure out the _____ part with my info.
Please any help just to start it would be apprciated !

2. Originally Posted by ohman
Analyze the motion of a pebble stuck in a tire
write an equation for the motion at 50km/h, and 100km/h for two differnt wheels(radius)
...
Have a look here: Cycloid -- from Wolfram MathWorld

With your problem I think it is the easiest way to use parametric equation.

3. but in class we are learning sinusoidal functions.

4. Originally Posted by ohman
but in class we are learning sinusoidal functions.
Note that the parametric form for a cycloid does involve sinusoidal functions.

-Dan

5. i see that.
just we have done the whole unit and never did anything like that.
but it makes sence, but one more question if i changed the radius then the only thing that would change would be the amplitude and vertical translation then ? or would the period also change.

6. Here's a diagram of your tire with P(x,y) denoting the pebble.

$x=h-rsin(t), \;\ y=k-rcos(t)$

Since the height of the wheel's center is the radius we have k=r.

The distance moved by the center is the same as the circular arc length subtended by t. We have h=rt.

So, from the above equations we get:

$x=rt-rsin(t), \;\ y=r-rcos(t)$

You can use whatever radii you wish.

These are the parametric equations of a cycloid.

7. either im really dumb, or since your giving me things i have never learned its making it harder then it really is.

i'll try to clairify :S

Okay so what i dont understand is how you get a period. In the question the teacher did not give a certain distance. So do i use 50km as a distance and then one hour is the time. but when we ussauly do the graphs time is on the bottom and hight on the side. So basicly im just stuck on how to get the rotations of the wheel. like how many rotations at what time. Its not really clear to me on what goes on the bottom of my graph and how to get it. Otherwise i get the rest, such as amplitude, horizontal translation... etc

8. Originally Posted by ohman
either im really dumb, or since your giving me things i have never learned its making it harder then it really is.

i'll try to clairify :S

Okay so what i dont understand is how you get a period. In the question the teacher did not give a certain distance. So do i use 50km as a distance and then one hour is the time. but when we ussauly do the graphs time is on the bottom and hight on the side. So basicly im just stuck on how to get the rotations of the wheel. like how many rotations at what time. Its not really clear to me on what goes on the bottom of my graph and how to get it. Otherwise i get the rest, such as amplitude, horizontal translation... etc
You are given a speed for the wheel. This is the speed the center of the wheel will move at. The rolling-without-slipping condition is
$v = r \omega$
where $\omega$ is the angular speed of the wheel. You can use that to derive the period.

-Dan

9. Originally Posted by ohman
...

Okay so what i dont understand is how you get a period. In the question the teacher did not give a certain distance. So do i use 50km as a distance and then one hour is the time. but when we ussauly do the graphs time is on the bottom and hight [ * ] on the side.
...
Your considereations are OK:

Let v denote the speed then you have:

$v = 50\ \frac{km}{h}=\frac{50000}{3600} \frac ms=\frac{125}{9} \frac ms$

Let r denote the radius of the wheel and k the number of revolutions per second. Then the circumference of the wheel is:

$c = 2 \pi r$ The distance the wheel travels in one second is:

$d = k \cdot 2 \pi r = \frac{125}{9} \frac ms$

The number of revolutions per second is:

$k = \frac{125}{9\cdot 2 \pi r} \frac1s$ (the meter cancel out!)

And therefore the time per one revolution is:

$\frac1k = \frac{18 \pi r}{125} s$

[ * ] Could it be that you should only determine the vertical displacement of the pebble from the ground?