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Pebble in a rolling tire  finding velocity and acceleration 
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#1
Oct2807, 06:40 PM

P: 124

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, [tex]\vec{V}[/tex]ag. Note the order of the subscripts: velocity of axle measured relative to the ground. Express your answer in terms of R, w, and [tex]\hat{x}[/tex] and/or [tex]\hat{y}[/tex]. (b)Find the position vector of the pebble relative to the initial point of contact between the wheel and ground at a time t, [tex]\vec{r}[/tex]pg(t). Express the position vector of the pebble in terms of R, w, t, and the unit vectors [tex]\hat{x}[/tex] and/or [tex]\hat{y}[/tex] of the xy coordinate system shown. (c)Find [tex]\vec{V}[/tex]pg(t), the velocity vector of the pebble with respect to a fixed point on the ground, in terms of the unit vectors [tex]\hat{x}[/tex] and [tex]\hat{y}[/tex] of the xy coordinate system shown. Express the velocity vector in terms of R, w, t, and [tex]\hat{x}[/tex] and/or [tex]\hat{y}[/tex]. (d)Now find [tex]\vec{a}[/tex]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 [tex]\hat{x}[/tex] and/or [tex]\hat{y}[/tex] of the xy coordinate system shown. 2. Relevant equations v = wR atan = [tex]\alpha[/tex]/R 3. The attempt at a solution Well part (a) i got which = (wR)[tex]\hat{x}[/tex] 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 


#2
Oct2807, 07:33 PM

P: 169

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 v_{CM} and rotation with an angular speed w. Thus, velocity of any particle on the tyre w.r.t. ground, v_{P}, can be given as: (bold letters indicate vector quantites, 'X' denotes vector cross product.) v_{P} = v_{CM} + w X r where, v_{CM} 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
Oct2807, 07:37 PM

P: 169

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



#4
Oct2807, 08:18 PM

P: 124

Pebble in a rolling tire  finding velocity and acceleration
I'm completely lost in this problem.



#5
Oct2807, 08:25 PM

P: 169

jokes apart.. try doing part (c) first, using the formula, i gave. 


#6
Oct2807, 08:40 PM

P: 169

For position part you may proceed like this:
Initial coordinate 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 coordinate of the pebble. Note that Θ is the angle from the vertical, in the same sense as rotation of the tyre. 


#7
Oct2807, 08:44 PM

P: 169

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
Oct2807, 08:55 PM

P: 124

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
Oct2807, 08:57 PM

P: 169

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


#10
Oct2807, 09:00 PM

P: 169

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
Oct2807, 09:02 PM

P: 124




#12
Oct2807, 09:08 PM

P: 169

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.



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