What is the fastest speed around a curve?

[Natus Homonymus]

I understand that the centripetal force on an object of mass 'm' is (mv²)/r

However, isn't this for an object going around in a circle? Suppose I have a curve (0.0033x²+−1.0038x+98.2331). What would be the fastest speed around this curve on the bounds x ->

Please note that we would (obviously) have to take friction into account. Any help or methods/techniques to help me solve this would be greatly appreciated.

 

[lekh2003]

I don't think there's' a limit on the speed. A long as the curved path is strong enough to withstand the centripetal force and friction isn't infinite, you can theoretically get any torque you want on the wheels to go as fast as you want.

Why would there be a limiting factor on the speed?

 

[Natus Homonymus]

I don't think there's' a limit on the speed. A long as the curved path is strong enough to withstand the centripetal force and friction isn't infinite, you can theoretically get any torque you want on the wheels to go as fast as you want.

Why would there be a limiting factor on the speed?

In general, for an object to stay in orbit (in this case continue in circular/elliptical motion), the outward force on the object must be equivalent to the centripetal force (I may be mistaken here) - in the case of an ellipse, there are two focii - which one would you pick in this case.

Besides, the frictional force provides force to prevent the object from skidding, etc. Please correct me if I'm wrong.

 

[andrewkirk]

At each point on the curve there is a maximum speed at which it can be traversed without skidding sideways off the curve. That speed is determined by the radius of curvature and the formula you wrote above. It does not matter that the curve is not a circle because, at any point where there is nonzero curvature, it approximates a circle to second order.

You can use the formulas at the link to work out the radius of curvature at each point on the curve and hence work out the maximum non-skid speed (the 'slip speed') at each point using the coefficient of static friction.

If you plot slip speed as a function of distance along the curve you can identify the local maxima and minima of slip speed. What you need to do then is work out a curve that slows at each bend to just exceed its minimum slip speed, and accelerates out of bends until the next constraint needs to be prepared for. The acceleration out of a corner and braking into a corner is limited by the friction coefficient. If it is too high the wheels will skid or spin. To test this you need to vector-add the fore-aft acceleration from throttle/brake to the sideways acceleration from the track's curvature before comparing to the friction limit.

I would start by assuming the bike traveled the curve sitting on the slip speed at every point (ie ignoring the previous paragraph for a first pass), and then check whether the implied fore-aft acceleration/deceleration at any point would lead to a skid or wheel spin. If it does, some constraints will need to be made to fore-aft acceleration per the previous para. If it cannot be solved analytically it can be solved numerically by chopping the curve up into a series of small segments. That numeric approach to solving the problem is not far from what a racer actually does.

 

[A.T.]

In general, for an object to stay in orbit (in this case continue in circular/elliptical motion),

Orbits in space around a massive body are somewhat different from contact force constrained curving (car, train etc.). The common thing between them is the relation between the path curvature radius r, velocity v, and centripetal force F = (mv^2)/r. But note that "centripetal" here means the component "perpendicular to the path", or "towards the centre of path curvature".

 

[jbriggs444]

Suppose I have a curve $(0.0033x^2+−1.0038x+98.2331)$. What would be the fastest speed around this curve on the bounds x ->

That curve is a parabola. By inspection, the point at which the curvature is maximum would be the apex of the parabola. You can locate the apex for this parabola by taking the first derivative and solving for the x value that makes it zero.

 

[sophiecentaur]

I understand that the centripetal force on an object of mass 'm' is (mv²)/r

However, isn't this for an object going around in a circle? Suppose I have a curve (0.0033x²+−1.0038x+98.2331). What would be the fastest speed around this curve on the bounds x ->

Please note that we would (obviously) have to take friction into account. Any help or methods/techniques to help me solve this would be greatly appreciated.

Several of the answers here have assumed that the object is orbiting under gravity (central attractor). The general answer should include alternative paths and forces and the tightest part of the curve may not coincide with the highest speed (take a simple parabolic trajectory in a uniform g field where the speed is least at the top)
Speed relates to Kinetic Energy and in conservative situations, the sum of the Potential and Kinetic energies is constant so the maximum speed occurs where the Potential Energy is at a minimum. This sort of problem can often be approached easier in terms of Energy, rather than by trying to use Forces and Accelerations etc..

 

[CWatters]

I understand that the centripetal force on an object of mass 'm' is (mv²)/r

I prefer to say that the centripetal force required to make the object move at velocity v with a radius r is given by that equation.

However, isn't this for an object going around in a circle? Suppose I have a curve (0.0033x²+−1.0038x+98.2331). What would be the fastest speed around this curve on the bounds x ->

Any speed you like. You have specified a path so you can work out the raduis at any point. However you haven't specified what force is available. This could be a car moving on an infinitely strong roller coaster.

Please note that we would (obviously) have to take friction into account.

And possibly other forces acting on the object?
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[Natus Homonymus]

Any speed you like. You have specified a path so you can work out the raduis at any point. However you haven't specified what force is available. This could be a car moving on an infinitely strong roller coaster.

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My intended application for this is finding the top speed of a car around a particular curve governed by the equation above. The road would have friction with the tyres of the car. How would you suggest I find the radius at every point without doing so manually? Any mathematical techniques?

 

[CWatters]

http://mathworld.wolfram.com/RadiusofCurvature.html

Equation for the radius..

I haven't looked to see how easy/hard it would be to calculate for your equation.

The suggestion in #6 might make it easier.

 

[willem2]

I haven't looked to see how easy/hard it would be to calculate for your equation.

This is rather easy. If y = ax²+bx+c. The divisor will be a constant. The minimum of
(1 + (\frac{dy}{dx})^2) is obviously when dy/dx = 0. This is at the top of the parabola. The radius of curvature at this point will be 1/a