There is a crest in the road leading up to Red Gate HQ which has prompted a discussion on our forum about how fast someone would need to be traveling to ‘take off’ while passing over the crest. We decided that ‘taking off’ simply meant the tyres leaving the surface of the road for any amount of time.

A group of us went out and calculated that the crest has a radius of 30 meters. What is the minimum constant speed that a car would need to be travelling in order to ‘take off’ as it passes over the crest? Please post your answers and workings below in the comments section.

## 9 Comments

Jason Crease

06/06/2012

How many degress of arc does the bridge go through, or equivalently, what is the height of the bridge?

This is necssary to answer the question. For instance, if the bridge goes through 180 degrees of arc, the car will go directly upwards immediately on hitting the bridge.

All Things Oracle

06/06/2012

Hi Jason,

You don't need to know that :-) the image is not in anyway to scale - think of it as a crest in the road, rather than a bridge...

James

Jason Crease

06/06/2012

The car will leave the road when the rate of change of y-component is greatest. This occurs at the beginning (and end) of the bridge.

If the angle of arc is 180 degrees, the car will go directly upwards immediately at arrival at the bridge. A car going at 0.0001mph directly upwards will fly (albeit very briefly).

All Things Oracle

06/06/2012

Interesting, but let's assume that a car will drive smoothly over this part of the road, and any taking off will occur as it passes over the crest.

Phil

06/06/2012

I'm going to assume that the angle of the lump is small, and therefore that gravity and the centrifugal force due to travelling round the lump are parallel.

The acceleration is thus V^2/R, which must equal g, 9.8m/s/s (7.1e10 furlongs per fortnight squared for those who haven't gone metric yet)

V^2/R = 9.8

V = sqrt(9.8 * 30) = 17m/s = 38 mph or 10^5 furlongs/fortnight

Jason Crease

06/06/2012

This question can be partially answered by dimensional-analysis. The numbers involved are s (speed of car), r (radius of bridge) and g (9.8 m/s2). Hence the formula must be that it leaves the bridge when: s^2 > krg (for constant k).

Jason Crease

06/06/2012

AND since s = 0 when theta = 0, s = infinity when theta = 90, and s =0 when theta = 180, I bet k = tan theta.

So my answer (using NO physics) is s^2 > rg tan (theta)

All Things Oracle

08/06/2012

As Phil has described above:

In circular motion, acceleration due to change in direction = V^2/r

For the car to 'take off' this acceleration (upwards) must be equal to or greater than the downwards acceleration (gravity).

So…

V^2/r = g

V= square root of r x g, which = 17.2 m/s (62 km/h)

Lalit kumar

04/04/2013

This can be a question on projectile motion , where the object in the projectile trajectory is of course the car in motion .

The angle of projection can be 60 degree for the tyres to leave the ground .

which is governed by the equation

y = xtan(deg)- gx2/2u2cos2(deg), we can subsitute and work it out .

cheers :)

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