# Calculating max. speed of car over humpback bridge before it leaves the ground?

## Homework Statement

A car travels over a humpback bridge of radius of curvature 45m, what is the max speed it can reach before the wheel lose contact with the road?

## Homework Equations

mg - R = (mv^2) / r

## The Attempt at a Solution

When the car is about to lose contact with the road R tends to zero therefore:

mg = (mv^2) / r

v = (rg)^0.5

v = (450)^0.5 = 21 m/s

The problem I have with this in my head is that this answer seems to only have been resolved when the car is a the peak of the bridge.

For example if the car is somewhere else other than the top, the reaction force would be perpendicular to the tangent, and the component of weight would be opposite this, so:

mgcos(x) - Rcos(x) = (mv^2) / r

where x is the angle between the vertical and the raius at that point,

I just don't understand, could somebody explain.

Hootenanny
Staff Emeritus
Gold Member

## Homework Statement

A car travels over a humpback bridge of radius of curvature 45m, what is the max speed it can reach before the wheel lose contact with the road?

## Homework Equations

mg - R = (mv^2) / r

## The Attempt at a Solution

When the car is about to lose contact with the road R tends to zero therefore:

mg = (mv^2) / r

v = (rg)^0.5

v = (450)^0.5 = 21 m/s

The problem I have with this in my head is that this answer seems to only have been resolved when the car is a the peak of the bridge.

For example if the car is somewhere else other than the top, the reaction force would be perpendicular to the tangent, and the component of weight would be opposite this, so:

mgcos(x) - Rcos(x) = (mv^2) / r

where x is the angle between the vertical and the raius at that point,

I just don't understand, could somebody explain.
You're half right. The quantity on the RHS of your final equation is the centripetal force, yes? This acts perpendicular to the tangent, toward the centre of the circle. Therefore, you need only consider the components of force parallel and anti-parallel to the centripetal force. The normal reaction force always acts perpendicular to the surface, but as you correctly note, the weight of the car always acts vertically down, so you need to resolve this into the component parallel to the centripetal force.

Does that make sense?

You're half right. The quantity on the RHS of your final equation is the centripetal force, yes? This acts perpendicular to the tangent, toward the centre of the circle. Therefore, you need only consider the components of force parallel and anti-parallel to the centripetal force. The normal reaction force always acts perpendicular to the surface, but as you correctly note, the weight of the car always acts vertically down, so you need to resolve this into the component parallel to the centripetal force.

Does that make sense?

So Rcos(x) would just be R? the thing is with this I think is that the further from the peak of the bridge you get, the smaller the component of mg is acting towards the centre, therefore a smaller speed (v) is needed, which to me doesn't make sense (mainly because there is no mention of this in my worked example I've got).

Hootenanny
Staff Emeritus
Gold Member
So Rcos(x) would just be R? the thing is with this I think is that the further from the peak of the bridge you get, the smaller the component of mg is acting towards the centre, therefore a smaller speed (v) is needed, which to me doesn't make sense (mainly because there is no mention of this in my worked example I've got).
That is indeed correct. However, you are only interested in the velocity at the top of the bridge since this is when your car will leave the bridge. That is why there is no need to consider any point other than the apex. Besides, it would be physically impossible for a car to drive over a truly semi-circular bridge

I think my problem is assuming that the car could take off at any point as it was going up the humpback, as long as it had enough speed.

For example if you take a point before the apex and instantly accelerate the car up to a large speed along the tangent - it would leave the track, you could do this with smaller and smaller speeds until it stays on the track and that would be your velocity for that point.

Why would it only leave the track at the apex?

Hootenanny
Staff Emeritus
Gold Member
I think my problem is assuming that the car could take off at any point as it was going up the humpback, as long as it had enough speed.

For example if you take a point before the apex and instantly accelerate the car up to a large speed along the tangent - it would leave the track, you could do this with smaller and smaller speeds until it stays on the track and that would be your velocity for that point.

Why would it only leave the track at the apex?
Technically, the maximum speed for the car to be undergoing circular motion in $v_\text{max} = \sqrt{rg\cos\theta}$, where $\theta$ is the angular position of the car. However, as you are given no information regarding the initial starting point of the car, it is safe to assume that the question is asking for the maximum speed at the apex. Technically, the question should have stated this.

Technically, the maximum speed for the car to be undergoing circular motion in $v_\text{max} = \sqrt{rg\cos\theta}$, where $\theta$ is the angular position of the car. However, as you are given no information regarding the initial starting point of the car, it is safe to assume that the question is asking for the maximum speed at the apex. Technically, the question should have stated this.

Thanks