Assuming a race car drives around in a circle of radius r, center (0,0), linear velocity v, and ignoring centrifugal forces and friction I can calculate the position at any time, t. Angular velocity

$$\omega = \frac{v}{r}$$

degrees turned through (assume at time t = 0 car is at position (r,0)):

$$\theta = \omega t$$

So the positional (x,y) coordinates, p, at time t, are

$$p(t) =(r\cos(\theta) , r\sin(\theta ))$$

Please correct me if I am wrong up to this point. Now I want to include slipping, so that the car will be thrust outwards by the centrifugal (?) force, with friction force countering it. Supposing the centrifugal force is stronger, then there will be a resultant force in the direction tangential to the linear velocity at time t.

Suppose this force is $F$, then I want to find the modified position of the car at time t, as it won't be travelling in a circle but, I think, will be travelling in an increasing spiral (this is theoretical, I haven't taken into account spinning around its axis or crashing into a barrier).

So this is where I am going to use very naive maths/physics to try and find the modified position.

Assume the car has mass $m$.

Then the acceleration outward from the centre of the circle will be

$$a = \frac{F}{m}$$

Now I'm going to assume acceleration is constant to make my life simpler, and I can fudge a value for outward velocity u,

$$u = \int_{0}^{t} a dt = at$$

and more dodgy physics, I can calculate the position vector by multiplying by t, assuming constant velocity outwards,

$$\text{position} = (u t \cos(\theta),u t \sin(\theta))$$

Remembering at time t, the angle is $\theta$

I can see that the new position, let's say $q(t)$, is given by

$$q(t) = p(t)+(u t\cos(\theta), u t\sin(\theta)) =p(t)+ (\frac{Ft}{m} t\cos(\theta),\frac{Ft}{m}t\sin(\theta))$$

One problem I have noted, just looking at this is that $q(t) \tilde{}t^{2}$ since to get the position from acceleration I multiply by t twice, which seems too much, the car will spiral too fast.

I am very skeptical this is correct. Please correct me where I have gone wrong.


Also I just pulled the value $F$ from thin air. Could you explain how I could calculate it based on the cars velocity, mass, radius of circle, coefficient of friction of surface or any other variables?


1 Answer 1


The centrifugal force is a fictitious force. A fictitious force appears in the equations of motion when you go from a inertial frame to an accelerating frame of reference (or rotating).

The centrifugal force appears in the equations of motion when you go to a rotating frame. As you're looking at the problem in the intertial frame, there is no centrifugal force.

Example to clarify this: Suppose you have some velocity, and there are no forces acting on you, then you will move in a straight line. If I (the observer), transform to a rotating frame, I will not see you moving in a straight line, but in some curve. As you are chaning direction, there appears to be a force acting on you: the centrifugal fictitious force.

  • $\begingroup$ I'm not sure I understand. I did put a question mark next to centrifugal force because I remember my high school physics teacher telling me something similar. So are you saying that the car would not spin outwards from the centre? Presumably in that case, if the driver hits the brake, the car will move in the direction outwards, slowing down if friction exists. Correct? $\endgroup$
    – JJG
    Commented Mar 24, 2013 at 10:18
  • $\begingroup$ Probably, but this has nothing to do wiht centrifugal force! If there is not enough force inwards (like friction), to keep it in the circular path, it will just continue in a straight line. $\endgroup$
    – Bernhard
    Commented Mar 24, 2013 at 10:21
  • $\begingroup$ Right, then I could possibly confusing real race cars motion outwards, when either the car spins inward or the brake is applied. $\endgroup$
    – JJG
    Commented Mar 24, 2013 at 10:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.