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Hello everyone!

Just a little problem for you, if you have a little time!

"A hump-back bridge follows an arc of a circle of radius 30m. What is the maximum speed at which a car may be driven over this bridge if its wheels are to remain in contact with the road?"

Thank you Just a little lost as to where to start!

Just a little problem for you, if you have a little time!

"A hump-back bridge follows an arc of a circle of radius 30m. What is the maximum speed at which a car may be driven over this bridge if its wheels are to remain in contact with the road?"

Thank you Just a little lost as to where to start!

Model the car as a particle.

So start by finding an expression for the normal reaction of the car as a function of angle around the (circular arc) hump. Then make this expression for normal reaction equal to zero, and you will get an expression for the limiting speed as a function of angle. Find the maximum value of this speed, and that's the maximum speed that you can go over the hump without the normal reaction going to zero (i.e. you leave the road).

Given the only variables can be R, radius of hump and g, acceleration due to gravity, by dimensional analysis (k is a dimensionless constant):

[v] = k[R]^{a}[g]^{b}

LT^{-1} = L^{a}(LT^{-2})^{b}

L ---> 1 = a + b

T ---> -1 = -2b

So b = 1/2, a = 1/2 --> $v_{max} \propto \sqrt{Rg}$

...which makes sense because if you increase the radius of the hump, you'd expect to be able to go faster, and so too would you if gravity was stronger.

So start by finding an expression for the normal reaction of the car as a function of angle around the (circular arc) hump. Then make this expression for normal reaction equal to zero, and you will get an expression for the limiting speed as a function of angle. Find the maximum value of this speed, and that's the maximum speed that you can go over the hump without the normal reaction going to zero (i.e. you leave the road).

Given the only variables can be R, radius of hump and g, acceleration due to gravity, by dimensional analysis (k is a dimensionless constant):

[v] = k[R]

LT

L ---> 1 = a + b

T ---> -1 = -2b

So b = 1/2, a = 1/2 --> $v_{max} \propto \sqrt{Rg}$

...which makes sense because if you increase the radius of the hump, you'd expect to be able to go faster, and so too would you if gravity was stronger.

when a car is driving on a level road there is no resultant vertical force acting on it.

When it is going over the hump it is effectively moving in a circlr so there MUST be a vertical resultant force towards the centre of the circle, i.e. downwards. This will equal W - R where W is the weight of the car and R is the normal contact force with the road. From F = m a it will also equal m v^2 / r where v is the velocity of the car and r is the radius of the hump.

As W - R = m v^2 / r on can see that as the car travels faster R must get less. (the right hand side of the equation increases but W is constant so R must decrease). Eventually at a certain value of v, R = 0 and you are no longer in contact with the road. so W = mg = m v^2 / r

Cancelling the m's we get v = root (gr) so for a hump of radius 30m this works out as v = 17.1 m/s

When it is going over the hump it is effectively moving in a circlr so there MUST be a vertical resultant force towards the centre of the circle, i.e. downwards. This will equal W - R where W is the weight of the car and R is the normal contact force with the road. From F = m a it will also equal m v^2 / r where v is the velocity of the car and r is the radius of the hump.

As W - R = m v^2 / r on can see that as the car travels faster R must get less. (the right hand side of the equation increases but W is constant so R must decrease). Eventually at a certain value of v, R = 0 and you are no longer in contact with the road. so W = mg = m v^2 / r

Cancelling the m's we get v = root (gr) so for a hump of radius 30m this works out as v = 17.1 m/s

Drummy

when a car is driving on a level road there is no resultant vertical force acting on it.

When it is going over the hump it is effectively moving in a circlr so there MUST be a vertical resultant force towards the centre of the circle, i.e. downwards. This will equal W - R where W is the weight of the car and R is the normal contact force with the road. From F = m a it will also equal m v^2 / r where v is the velocity of the car and r is the radius of the hump.

As W - R = m v^2 / r on can see that as the car travels faster R must get less. (the right hand side of the equation increases but W is constant so R must decrease). Eventually at a certain value of v, R = 0 and you are no longer in contact with the road. so W = mg = m v^2 / r

Cancelling the m's we get v = root (gr) so for a hump of radius 30m this works out as v = 17.1 m/s

When it is going over the hump it is effectively moving in a circlr so there MUST be a vertical resultant force towards the centre of the circle, i.e. downwards. This will equal W - R where W is the weight of the car and R is the normal contact force with the road. From F = m a it will also equal m v^2 / r where v is the velocity of the car and r is the radius of the hump.

As W - R = m v^2 / r on can see that as the car travels faster R must get less. (the right hand side of the equation increases but W is constant so R must decrease). Eventually at a certain value of v, R = 0 and you are no longer in contact with the road. so W = mg = m v^2 / r

Cancelling the m's we get v = root (gr) so for a hump of radius 30m this works out as v = 17.1 m/s

This assumes that the car has zero normal reaction at the very top of the hump, where W and R are antiparallel. How have you shown that this is the case? Why would the car not have zero normal reaction at some other angle to the vertical apart from zero?

Hi Worzo,

What I have posted is the standard A level textbook answer to this classic problem. Your post about dimensional analysis looks OK but would be meaningless to the vast majority of A level students. I certainly didn't do DA until degree level.

The reaction force would be smaller he faster the car is travelling. In the derivation I conside the limiting case where R JUST reaches zero. This is therefore the minimum speed at which the car would leave the road. I suppose at faster speeds it may leave the road before the top of the road with the hump acting as some kind of ramp.

Do you know a way of embedding diagrmas done on paint into posts like these, or even a better way of doing formulae?

Best wishes.

What I have posted is the standard A level textbook answer to this classic problem. Your post about dimensional analysis looks OK but would be meaningless to the vast majority of A level students. I certainly didn't do DA until degree level.

The reaction force would be smaller he faster the car is travelling. In the derivation I conside the limiting case where R JUST reaches zero. This is therefore the minimum speed at which the car would leave the road. I suppose at faster speeds it may leave the road before the top of the road with the hump acting as some kind of ramp.

Do you know a way of embedding diagrmas done on paint into posts like these, or even a better way of doing formulae?

Best wishes.

Don't use the "tex" delimiters. Use the "latex" ones instead, as the image is rendered more clearly. Instead of a squiggly white image, you get a nice

$W-R=\frac{mv^2}{r}$ and $v=\omega^2r.$

EDIT: Worzo: As a matter of interest, how would you go about using your method? Also, why do you say that Drummy's method assumes that the weight and reaction force are antiparallel?

$W-R=\frac{mv^2}{r}$ and $v=\omega^2r.$

EDIT: Worzo: As a matter of interest, how would you go about using your method? Also, why do you say that Drummy's method assumes that the weight and reaction force are antiparallel?

Drummy

its been a steep learning curve but now I can say confidently that ...$v = \omega^2 r$

NO!

$v = r\omega$

Dharma

EDIT: Worzo: As a matter of interest, how would you go about using your method? Also, why do you say that Drummy's method assumes that the weight and reaction force are antiparallel?

What do you mean how would I go about using my method?

And Drummy's method does the calculation for the car when it is at the top of the hump. Weight always acts vertically downwards, but the normal reaction and centripetal force are perpendicular to the circular surface. The only time when the weight, normal reaction and centripetal force are all in a line (so you can add them) is when the car is at the top of the hump.

Worzo

What do you mean how would I go about using my method?

Worzo

And Drummy's method does the calculation for the car when it is at the top of the hump. Weight always acts vertically downwards, but the normal reaction and centripetal force are perpendicular to the circular surface. The only time when the weight, normal reaction and centripetal force are all in a line (so you can add them) is when the car is at the top of the hump.

I imagine this could be done generally by taking components. I'm just struggling with the bit about taking angles in a circle!

Dharma

My bad. What I meant was: how do you express the reaction force as a function of the angle and speed of the car?I see.

I imagine this could be done generally by taking components. I'm just struggling with the bit about taking angles in a circle!

I imagine this could be done generally by taking components. I'm just struggling with the bit about taking angles in a circle!

From the diagram you can see that

$\\R - mg\sin \theta = -\frac{mv^2}{r}[br]\\R = mg\sin \theta -\frac{mv^2}{r}$

Set $R=0$ for the limiting condition of staying on the road:

$\\mg\sin \theta = \frac{mv_{max}^2}{r}[br]\\v_{max}^2 = gr\sin \theta[br]\\v_{max} = \sqrt{gr\sin \theta}$

Now to maximise the speed, give $\sin \theta$ its maximum value of unity. This happens when $\theta = 90^{\mathrm{o}}$ i.e. at the top of the hump, so then you get:

$v_{max} = \sqrt{gr}$

Petty, I know, but you do have to show the angular dependence of the normal reaction.

Dharma

Thanks.

I also just realised that this shows you that you cannot have a speed of $\sqrt{gr}$ all the way over the hump. At angles of $\theta \ne 90^{\mathrm{o}}$, the normal reaction becomes zero at a lower speed.

You would expect this because at these non-vertical angles the component of the weight acting perpendicular to the surface is lower, hence the tangential speed must be lower to keep the normal reaction positve.

Original post by XseLeo

hence v=24.49

Just the tiniest bit late

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