Analytical Mechanics - rolling wheel (accelerating)

[Ecoi]

Homework Statement

A wheel of radius b rolls along the ground with constant forward acceleration a0. Show that at any given instant, the magnitude of the acceleration of any point on the wheel is

(a0^2 + (v^4 / b^2))^(1/2)

relative to the center of the wheel.

Here v is the instantaneous forward speed.

(I left out part of the question because I think I could get it after I get started on this.)

Homework Equations

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The Attempt at a Solution

I'm having difficulty getting this problem started really. I can go through my thought process on this problem. By looking at what they wish for me to solve, it seems those two accelerations are perpendicular to each other (a0 and v^2 / b). I think that v^2 / b is the centripetal acceleration actually, so this implies that a0 is actually the tangential component of the acceleration.

However, that is more of backwards reasoning in order to get the acceleration they want. I do not know why a0 would be the tangential component without knowing the answer in advanced. (Maybe I should? I don't think I've done a rotating wheel problem with acceleration in it before.) It's been a while since I've taken Physics 1 xD...

I tried to write down the position vector relative to the wheel, but I do not think I wrote it down correctly. I would think it would be of the form:

r(t) = bsin(wt)i + bcos(wt)j

But I realize w is not a constant here... So I'm not entirely sure.

 

[diazona]

You're thinking along the right lines. See if you can figure out an expression for w as a function of time, then plug that in and see where it gets you.

By the way, the expression you have for r(t) only applies to points on the edge of the wheel.

 

[Ecoi]

You're thinking along the right lines. See if you can figure out an expression for w as a function of time, then plug that in and see where it gets you.

By the way, the expression you have for r(t) only applies to points on the edge of the wheel.

I'm not getting the correct answer with what I did; I got

sqrt(4a0^2 + (2v)^4 / r^2)

It's in a similar format, but those coefficients there are pesky.

What I did:

r(t) = rsin(a0*t^2 / r)i + rcos(a0*t^2 / r)j

Since w = v(tangential) / r, I realized that w = a0 * t / r and plugged that in and took two derivatives and got the result I just put up there earlier. Is this correct or no? I don't see anything wrong in my calculations (I did it 3 times now).

If you want, I can post my steps here of what I did exactly.

I still am not sure why the a0 is part of the tangential acceleration. Is there a way to reason this better?

Also, does the question want every point on the wheel, including r < b? I don't see how I would get the expression they want for every point (the formula they asked to prove has a b in it).

Thank you for helping by the way =).

 

[diazona]

Ah, there is something sneaky going on, which I missed at first. You actually do have an error in the expression for \vec{r}(t). In your chosen coordinates, the correct expression is
\vec{r}(t) = r\sin(\phi)\hat{i} + r\cos(\phi)\hat{j}
where \phi is the angular position (which is a function of time). Hopefully it makes sense why that's the case. Now, under certain circumstances, \phi = \omega t (exercise for the reader: figure out which circumstances those are), but in this case \phi \neq \omega t which is why you're getting the wrong results.

I still am not sure why the a0 is part of the tangential acceleration. Is there a way to reason this better?

Here's one way to think about it: imagine that, instead of a wheel rolling along the ground with acceleration a_0, you have a wheel suspended in space on a stationary axle, but with the same angular speed \omega(t) as the wheel in the problem. Effectively, this is just translating your problem to a reference frame that moves along with the center of the wheel. Think about what kind of acceleration corresponds to a change in the angular speed.

Also, does the question want every point on the wheel, including r < b? I don't see how I would get the expression they want for every point (the formula they asked to prove has a b in it).

That's right, so I'd guess that they're only asking about the edge. Though the problem is not well worded in that case.

 

[Ecoi]

Ah, there is something sneaky going on, which I missed at first. You actually do have an error in the expression for \vec{r}(t). In your chosen coordinates, the correct expression is
\vec{r}(t) = r\sin(\phi)\hat{i} + r\cos(\phi)\hat{j}
where \phi is the angular position (which is a function of time). Hopefully it makes sense why that's the case. Now, under certain circumstances, \phi = \omega t (exercise for the reader: figure out which circumstances those are), but in this case \phi \neq \omega t which is why you're getting the wrong results.


Here's one way to think about it: imagine that, instead of a wheel rolling along the ground with acceleration a_0, you have a wheel suspended in space on a stationary axle, but with the same angular speed \omega(t) as the wheel in the problem. Effectively, this is just translating your problem to a reference frame that moves along with the center of the wheel. Think about what kind of acceleration corresponds to a change in the angular speed.


That's right, so I'd guess that they're only asking about the edge. Though the problem is not well worded in that case.

Oh okay, I think I get this problem now. I can't believe this problem caused me so much confusion. All the other problems I worked on I understood perfectly, but this one stumped me for some reason. Okay, I did:

phi = (1/2)a0/b * t^2

because for angular position:

theta = theta0 + w0 *t + (1/2)alpha*t^2

So if theta0 and w0 is zero, then it becomes:

r(t) = b sin((1/2)a0/b * t^2)i + b cos ((1/2)a0/b * t^2)j

If I take the derivatives, I get their desired answer =)!

Thank you x_x. I worked on the other part of this question and I got the other desired formula (well the sines and cosines were different than the book, but I might have defined my system differently, so I will check that). Thank you for your help.