# Pebble in a rolling tire - finding velocity and acceleration

1. Oct 28, 2007

### gills

Pebble in a rolling tire -- finding velocity and acceleration

1. The problem statement, all variables and given/known data

ou are to find the coordinates of a pebble stuck in the tread of a rolling tire that is rotating counterclockwise (i.e., in the positive sense) with angular velocity w. The tire rolls without slipping on the ground (which is at y=0). The outer radius of the tire is R. At time t=0, the pebble is at the top of the tire, as shown.

(a)Find the velocity of the axle of the tire relative to a fixed point on the ground, $$\vec{V}$$ag. Note the order of the subscripts: velocity of axle measured relative to the ground.

Express your answer in terms of R, w, and $$\hat{x}$$ and/or $$\hat{y}$$.

(b)Find the position vector of the pebble relative to the initial point of contact between the wheel and ground at a time t, $$\vec{r}$$pg(t).
Express the position vector of the pebble in terms of R, w, t, and the unit vectors $$\hat{x}$$ and/or $$\hat{y}$$ of the xy coordinate system shown.

(c)Find $$\vec{V}$$pg(t), the velocity vector of the pebble with respect to a fixed point on the ground, in terms of the unit vectors $$\hat{x}$$ and $$\hat{y}$$ of the xy coordinate system shown.
Express the velocity vector in terms of R, w, t, and $$\hat{x}$$ and/or $$\hat{y}$$.

(d)Now find $$\vec{a}$$pg(t), the acceleration vector of the pebble with respect to a fixed point on the ground.
Express your answer in terms of R, w, t and $$\hat{x}$$ and/or $$\hat{y}$$ of the xy coordinate system shown.

2. Relevant equations

v = wR
atan = $$\alpha$$/R

3. The attempt at a solution

Well part (a) i got which =

-(wR)$$\hat{x}$$

The rest i need a little guidance on. Am i using the analogous equations of motion for rotational mototion? Or something different?

Any help would be great

Last edited: Oct 28, 2007
2. Oct 28, 2007

### saket

It's analogous.
It's something like, you can superimpose purely translatory motion and purely rotational motion.
The tyre is rolling away. So, it can be considered as sum of translation at a speed vCM and rotation with an angular speed w.
Thus, velocity of any particle on the tyre w.r.t. ground, vP, can be given as: (bold letters indicate vector quantites, 'X' denotes vector cross product.)
vP = vCM + w X r
where, vCM is velocity of centre (or, axle) of the tyre w.r.t. the ground;
w is the angular velocity of the tyre;
and, r is the position vector of any point on the tyre w.r.t. the centre of the tyre at any instant.
I hope the rest can be done.

3. Oct 28, 2007

### saket

Basically, the above formula can be obtained using simple geometry for distances, and then differentiating it once. (It's no big deal.)

4. Oct 28, 2007

### gills

I'm completely lost in this problem.

5. Oct 28, 2007

### saket

hullo.. where are you lost?
jokes apart.. try doing part (c) first, using the formula, i gave.

6. Oct 28, 2007

### saket

For position part you may proceed like this:
Initial co-ordinate of the pebble is (0, 2R). Let the wheel translate by a distance 'x' in time 't'. Also, let the wheel rotate by an angle Θ (in radians) in that time. Obviously, x = R*Θ. Now, try to locate the co-ordinate of the pebble. Note that Θ is the angle from the vertical, in the same sense as rotation of the tyre.

7. Oct 28, 2007

### saket

Basically, what is being said is, at any instant position vector of the pebble (w.r.t. ground) can be found as vector sum of position of the centre of the tyre (w.r.t. ground) and position vector of the pebble (w.r.t. the centre of the tyre).

8. Oct 28, 2007

### gills

i ended up using a hint from my web homework program. I got the idea now, but it was difficult for me to see it

9. Oct 28, 2007

### saket

Problem posted by "Hellz Angel" with heading "Pebble in a rolling tire", is also same! You may cross-refer!!
You added, "Pebble in a rolling tire -- finding velocity and acceleration" -- therefore I concentrated on velocity.

10. Oct 28, 2007

### saket

hmm.. i wasn't able to see how much you were able to do, you didn't post your attempt. anyways, good to know that you have got the idea now.

11. Oct 28, 2007

### gills

I didn't post my attempt, because it wasn't worth posting!:rofl:

12. Oct 28, 2007

### saket

but that is something we have to decide, isn't it? that (showing your attempt) is our only hope to understand the measure of grip of the student on the problem.. which lets us decide wha amount of help is to be provided. (personally, i do not believe in posting the whole solution if it can be handled otherwise.) show, plz do show your attempts, however bad they seem to you.