# Motorcycle Accident - Coefficient of Friction issues

1. Jan 5, 2013

### pmoon.pt

Hi Everyone,

I'm trying to find the initial speed of a motorcycle that was involved in an accident. The driver lost control and the bike dragged along the asphalt for some distance. I know several variables but my physics is not a strong suit to say the least. Any help would be much appreciated!

Here is what I know:

- bike travelled on its side (independently of the driver) for 200 meters
- nothing but the friction of the road brought the bike to a stop
- grade of the road - uphill approx 10 degrees
- motorbike weight - approx 145 kg
- dry road
- coefficient of friction of asphalt?

I'd like to know several things:
- the initial speed
- the formula used to find the initial speed
- if the type (i.e. rubber vs metal vs plastic) of contact points and actual surface area of the contact points of the bike on the asphalt make a difference for this calculation and if so how would that factor into the formula? Negligible? Can reasonable assumptions be made for a reasonably accurate initial speed?

Thanks in advance!

2. Jan 5, 2013

### AlephZero

You could get an estimate of the minimum bike speed by assuming no friction at all.

The bike climbs 200 x sin 10 degrees = about 35 meters

The starting speed to "free wheel" up the hill is given by v = sqrt(2gh) = 26 m/s or about 60 mph minimum.

I don't think you can do much more than that, without a lot more information.

Note: when you say "grade: uphill about 10 degrees" are you sure about that? That's a very steep hill. If you mean the grade is "10:1" that would reduce the speed estimate to about 45 mph minimum.

Last edited: Jan 6, 2013
3. Jan 6, 2013

### pmoon.pt

Thanks Aleph, thats very helpful!

Yes I meant 10:1

4. Jan 6, 2013

### haruspex

The frictional aspect is likely to be the more important. Based on various other metal/nonmetal combinations, if the road was clean and dry I would expect a coefficient of around .4 to .5. Let's say it was at least .3. With a gradient of 0.1, that makes it three times as significant. v = sqrt(2gd(0.1+0.3)) = sqrt(2*10*200*0.4) m/s = 40 m/s = 90mph. Any chance of getting a better estimate for mu?

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