Designing a speed bump given the maximum vertical wheel acceleration

[Bling Fizikst]

Homework Statement

below

Relevant Equations

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With the given information , we can find $y=h\cos\frac{\pi x}{L}$
Since , the wheel has constant velocity $v$ , we can write the $$\vec{a}=0\hat{e_t}+\frac{v^2}{\rho+r}\hat{e_n}$$
where $$\rho=\frac{\left[1+\frac{h^2 \pi^2}{L^2}\sin\frac{\pi x}{L}^2\right]^{\frac{3}{2}}}{\mid \frac{h\pi^2}{L^2}\cos\frac{\pi x}{L}\mid}$$

So , we are given : $$\frac{v^2}{\rho+r}<ng$$ but i am stuck here . How do i simplify? Morever it seems like i might get a lower bound on $h$ rather than an upper bound

 

[kuruman]

Morever it seems like i might get a lower bound on h rather than an upper bound

Please show how you arrived at this conclusion.

What would happen if the acceleration exceeds its maximum value? Where on the bump is that most likely to happen?

 

[Bling Fizikst]

Please show how you arrived at this conclusion.

What would happen if the acceleration exceeds its maximum value? Where on the bump is that most likely to happen?

I am sorry my previous take was completely idiotic . please ignore that .

What i am thinking is that , as $v,r$ are constant , so the only remaining parameter is $\rho$. To get max out of it , it should be minimum . $$\rho\geq \frac{L^2}{\pi^2 h} $$ equality occurring at $$x=0$$ . So that means : $$a_{max} <ng$$ Plugging the value for $\rho$ , we get: $$h<\frac{ngL^2}{\pi^2[v^2-ngr]}$$

But the answer doesn't seem to match

 

[Charles Link]

I have not been able to work out yet where the $r$ in the denominator comes from, but suggestion is to use $y=(h/2)+(h/2)(\cos(2 \pi x/L )$ for the cosine curve. Then my answer agrees with theirs if you set their $r=0$.

Edit: I don't know if I agree with their $r$ term in the denominator. For $ng$ very large, it still places a restriction that $h$ must be small if the wheel radius $r$ is large and that I think is incorrect. [Edit 2: See post 7. I now agree with the book's answer.]

 

[TSny]

suggestion is to use $y=(h/2)+(h/2)(\cos(2 \pi x/L )$ for the cosine curve.

Ahh. Yes! I believe this will yield the given answer.

@Bling Fizikst : Use this form of the bump so that there is no discontinuous slope of the bump at $x = \pm L/2$.

Consider the acceleration as the wheel arrives at the bump. Compare this to the acceleration when the wheel is at the top. Are they the same?

 

[kuruman]

I have not been able to work out yet where the $r$ in the denominator comes from

It comes from the FBD at the top of the bump $~N-mg=-\dfrac{mv^2}{\rho+r}$. The assumption is that the mass of the wheel is concentrated at its center which is at distance $\rho+r$ from the center of curvature. Solving for the normal force gives $~N=m\left(g-\dfrac{v^2}{\rho+r}\right)$. For the wheel to stay in contact with the surface, the normal force must be positive. With a factor of $n$ of order 1 thrown in, the condition for staying on the surface is $ng>\dfrac{v^2}{\rho+r}.$

. . . $y=(h/2)+(h/2)(\cos(2 \pi x/L )$ for the cosine curve.

It is simpler to write it as $y=h\cos^2(\pi x/L).~~$I think it still qualifies as a "cosine curve."

Ahh. Yes! I believe this will yield the given answer.

I have not been able to achieve that goal . . . yet.

 

[Charles Link]

I think I just figured out where the $r$ term in the denominator comes from, and this really then is much more than a beginner physics problem: For finite $r$, in order for the first derivative of the motion of the center of the wheel to remain continuous, there must not be an abrupt contact made between the wheel and the bump. Basically the condition that ensures this is that the radius of curvature of the cosine function , which is approximately and very nearly $|1/(d^2 y / dx^2)|$ must be greater than the radius $r$ of the wheel. When I worked through the calculation, I did indeed get the $r$ term as a secondary condition, i.e. $h<L^2 /(2 \pi^2 r)$.

(If the first derivative is discontinuous, (corresponding to direction of travel of the center of the wheel=$dy_c /dx$), mathematically that means the second derivative goes infinite for that instant).

I need to add some detail to the mathematics above:
Given $\vec{v}=(ds/dt) \hat{T}$, we have $\vec{a}=d \vec{v} /dt=(d^2 s/ dt^2) \hat{T}+(ds/dt)^2 ( d \hat{T} /d \phi) (d \phi /ds)$.

$d \phi /ds=1/r_o$ where $r_o$ is the radius of curvature at that point. For the case at hand, $\phi \approx \tan{\phi} =dy / dx$ , so that $d \phi / ds \approx d(dy/dx)/dx =d^2 y / d x^2$.

Note: $\hat{T}=\cos{\phi} \hat{i}+\sin{\phi} \hat{j}$, and $d \hat{T} / d \phi=-\sin{\phi} \hat{i} +\cos{\phi} \hat{j}= \hat{N}$. Note also $ds \approx dx$.

Using $y=(h/2)+(h/2)( \cos(2 \pi x/L)$, you just need to compute $d^2 y /d x^2$ to get this result. The $r_o$ has a minimum when $|\cos(2 \pi x / L )|$ has a maximum. The major point of interest is at $x=-L/2$. We must have $r_o > r$.

Thereby I agree with the book's answer. The finite $r$ can be significant, even for very slow velocity $v$. We must ensure that $d y_c / dx$ is not discontinuous, and the $r$ term in the denominator makes sure of that.

 

[TSny]

The acceleration of the center of the wheel at an arbitrary point $p$ of the bump is $a = \dfrac {v^2}{R}$ where $R$ is the radius of curvature of the trajectory of the center of the wheel when the wheel is at point $p$. Thus, the maximum acceleration of the center of the wheel occurs at the point of the bump where $R$ is minimum.

Since , the wheel has constant velocity $v$ , we can write the $$\vec{a}=0\hat{e_t}+\frac{v^2}{\rho+r}\hat{e_n}$$

Yes, $R = \rho + r$ is correct if the wheel is on the convex side of the cosine curve (such as when the wheel is near the top of the bump). For Charles Link's bump, however, the wheel is on the concave side of the cosine curve as it starts rolling up the bump. At the inflection point of the cosine curve, the wheel transitions to the convex side.

When the wheel is on the concave side, $R \neq \rho +r$. What is the correct relation?

At what point(s) of the ramp does $R$ have its minimum value?

 

[Charles Link]

For a couple of comments and additional hints, it seems to be necessary to make the approximations that $ds \approx dx$ and that $v \approx dx/dt$. Otherwise the problem as stated really becomes tremendously difficult, and really would be too difficult. Even though it isn't exact, suggestion for the OP is to assume to a first order or better approximation that the height of the center of the wheel $y_c$ at position $x$ is $y_c=y+r$.

also suggest that the OP first consider just the acceleration in the y direction instead of normal to the curve, especially when trying to compute the acceleration at an arbitrary point.

This is IMO a rather advanced kinematics problem, rather than introductory physics. I thought the OP did rather well in the first post trying to separate the acceleration into tangential and normal components, but it is still necessary to make certain approximations. See also my post 7 above. If I did it correctly, and I think I did, to me it is asking a lot of the student to come up with the $r$ term in the denominator of the expression for the answer of what $h$ needs to be, as I finally did in post 7. The $d \phi / ds$ concept I learned as an undergraduate student from a calculus book by Purcell, but I don't know if that currently is part of the standard calculus or physics curriculum.

 

[Bling Fizikst]

Well , I thought i could simply assume that the curve is of the form $A\cos kx$ .

When the wheel is on the concave side then , then $R=\rho-r$ , this would give max centripetal accn .
If $y=h\cos \frac{\pi x}{L} ^2$ then : $$\rho \geq \frac{L^2}{2\pi^2 h}$$ equality occurring at $x=0\in\left[\frac{-L}{2},\frac{L}{2}\right]$

We are given that : $$\frac{v^2}{R}<ng\implies \frac{v^2}{R}\mid _{\text{max}}<ng$$

We get the maximum at $R_{\text{min}}=\rho_{\text{min}}-r=\frac{L^2}{2\pi^2 h}-r$

Plugging this in the given inequality and simplifying yields:
$$h<\frac{L^2}{2\pi^2 \left[r+\frac{v^2}{ng}\right]}$$

 

[Charles Link]

What you have may be correct, but you are not showing or explaining the steps you are taking thoroughly enough, so it looks like you are doing some handwaving to get to the answer. Please try to add a little detail with a couple more steps. I tried to follow your steps including $R=\rho-r$ and they just are not computing completely.

Edit: I see I need to go up to your original post for how you defined $\rho$. It's starting to look a little more sensible. and I see better now by carefully studying post 8 by @TSny . It's basically the problem of a wheel of radius $r$ traveling inside a ring of radius $\rho$ so that the center of the wheel goes in a circle of radius $R =\rho-r$ for anywhere near the point $x=-L/2$. The denominator you have with $\frac{v^2}{\rho-r }$ ensures that the radius of curvature of the cosine curve, (what I called $r_o$ in post 7 that you call $\rho$) is greater than $r$, so that it isn't necessary to make it a separate requirement, like I did in post 7. I like your solution. :)

and I thought you might need to make the approximations that I suggested in post 9, but I see you got the answer without needing to approximate things. I went the route of $y=y(t)$, and $a \approx d^2 y / dt^2$ and $dy/dt=(dy/dx)( dx/dt)$ with $v \approx dx/dt$, etc.

 

[Charles Link]

@Bling Fizikst I'm looking over the expression for $\rho$ of the OP. Using $y=(h/2)+(h/2) \cos(2 \pi x/L)$, it then needs to have a 2 in a couple of places, but the expression for $\rho=ds /d \phi$ is correct otherwise, as I was able to verify by taking derivatives, with e.g. $\phi=\arctan(dy/dx)$. also $ds=(1+(dy/dx)^2)^{1/2} dx$, and $d \phi /dx=\frac{1}{1+(dy/dx)^2} d^2y/dx^2$.

Thereby I worked out the result of the OP for $\rho$ from first principles and I agree with it. It does help to show some of the steps though, because others reading through it can more easily verify it. :)

and I do like your solution. :)

 

[Charles Link]

I thought of something else on this problem that is I think worth asking. In the case of the concave upward part of the bump, we have $R=\rho-r$ so that we can say instantaneously the axle is moving in a circle of radius $R$, and the whole thing is operating on the inside of a ring of radius $\rho$.

Let's consider the case of $r$ almost as large as $\rho$. In that case, we really have what I think is a dilemma. We have a wheel moving on a ring of radius $\rho$, and the wheel must go around the ring fast enough, to keep the axle moving at a speed of $v$. In the limit $R \rightarrow 0$, we have that the wheel must be moving along the pavement at a speed that approaches infinity. This case also has an acceleration that approaches infinity for the axle with $a=v^2/R$.

It is odd that a small wheel, (small $r$), has no trouble running this racecourse, but that if we make the wheel too large, the bump appears to grow in size. What is even more puzzling is that the bump $h$ needs to be smaller as $r$ gets larger.

I think this may be worth taking a second look at it, because presently it appears there might be some logic that is not completely correct in how our wheel makes it over the bump.

Something with a small $h$ and a small $L$ , with $h$ even as large as $L$, is usually very easy for a truck with a large wheel radius $r$ to get over. I think what we designed here is a roller coaster, (to keep the $g$ within reasonable limits), rather than a model for how to design a speed bump for cars.

 

[Charles Link]

@Bling Fizikst , @TSny I really thought the mathematics of this problem was very neat, but when I took another look at it last night, I think the case of $r>> h$ is really much different than what the above mathematics gives. Perhaps it needs some flexibility of the tire or something. In any case, the formula we obtained for large $r$ appears to be in error. See post 13 above.

Edit: also a discontinuous first derivative causes what may be an infinite acceleration, but that happens for zero length of time, so I don't think that is really at all problematic.

 

[TSny]

@Bling Fizikst , @TSny I really thought the mathematics of this problem was very neat, but when I took another look at it last night, I think the case of $r>> h$ is really much different than what the above mathematics gives. Perhaps it needs some flexibility of the tire or something. In any case, the formula we obtained for large $r$ appears to be in error. See post 13 above.

Edit: also a discontinuous first derivative causes what may be an infinite acceleration, but that happens for zero length of time, so I don't think that is really at all problematic.

I think we can achieve a reasonable design of the bump without worrying about the case $r >> h$. Maybe I'm missing something. We do require $r$ to be less than the minimum radius of curvature of the bump.

Here's a particular example. I hope I didn't botch the calculations.

Suppose we take a cosine bump with $h$ = 0.1 m and $L$ = 1.0 m. Then, the bump's minimum radius of curvature turns out to be $\rho_{\rm min} \approx$ 0.51 m.

Let the car move over the bump with speed $v$= 1.5 m/s. [about 3.4 mph]
Take the tire radius to be $r =$ 0.25 m. I think this is reasonable. Note $r < \rho_{\rm min}$.

For these numbers, the maximum acceleration of the center of the wheel as it goes over the bump is about 8.8 m/s², or about 0.9g.

Here is a scale drawing for this case when the wheel is just starting up the bump.

The orange curve shows the curvature of the bump at x = -$L/2$.

 

[Charles Link]

@TSny Very good. Thank you. Just one little puzzle that I'm trying to iron out: What if the curvature of the road continues at a constant radius of curvature to send the wheel into motion around a ring of radius $\rho$ ? The axle will then be traveling in a circle of radius $R=\rho-r$. It never does that for this case though, but hypothetically, we could keep this same curvature at least for a finite distance. Even with a $\rho$ that is twice the radius of the wheel, to maintain the velocity/speed $v$ of the axle, the tread on the wheel needs to be moving over the road at twice this $v$. There must be a flaw in my logic, but I'm not sure where.

Maybe there is a clue in the actual path the axle covers when it goes over this sinusoidal hill. The position vs. time function is well behaved for the axle.

It seems though our model, and keeping the axle moving at speed $v$, seems to encounter difficulty if $r$ is a significant fraction of $\rho$.

 

[TSny]

Just one little puzzle that I'm trying to iron out: What if the curvature of the road continues at a constant radius of curvature to send the wheel into motion around a ring of radius $\rho$ ? The axle will then be traveling in a circle of radius $R=\rho-r$. . . . . Even with a $\rho$ that is twice the radius of the wheel, to maintain the velocity/speed $v$ of the axle, the tread on the wheel needs to be moving over the road at twice this $v$. There must be a flaw in my logic, but I'm not sure where.

Yes, the geometrical point of contact between the wheel and the ring moves twice as fast as the center of the wheel. I don't see any difficulty with this.

Of course, the physical point on the rim of the wheel that makes contact with the ring is instantaneously at rest for rolling without slipping.

 

[Charles Link]

Of course, the physical point on the rim of the wheel that makes contact with the ring is instantaneously at rest for rolling without slipping.

Glad you pointed this out. :)

 

[Charles Link]

I took another look at this, and I see as $r \rightarrow \rho$ that the point of contact moves around extremely rapidly, but the wheel (if I computed it correctly) is rotating with the rim of the wheel also moving along at speed $v$. One rotation of the axle in a circle does not go hand-in-hand with the wheel doing a rotation. That is not at all the case. Points on the rim do not in general (for arbitrary $r$ ) move at a constant speed, so I don't think this is a general result independent of $\rho$ and $r$, but for $r \rightarrow \rho$, it is interesting.

If the axle is doing a (small) counterclockwise circle, the outer part of the rim is rotating clockwise and very slowly=only a small fraction of a rotation for every time the axle does a circle. The two speeds are both $v$ if I computed it correctly.

If $\rho-r=\Delta$, in one circle of the axle, the axle moves a distance $2 \pi \Delta$, and a point on the rim moves along that very same distance.

(Note: This result surprised me a little when I looked at it carefully. It's not what I expected or what I had originally thought. I still need to take a second look at the general case).