1
$\begingroup$

My reasoning as to why this could be used to approximate the maximum speed of a car around a corner is because;

$\frac{mv^2}{r}$ is the centripetal force of the car.

The force to overcome this centripetal force, is the lateral friction of the car's tyres. Therefore, the maximum of this can be given by $\mu R$, where $\mu$ is the lateral coefficient of friction, and $R$ is the force applied between the car's tyres and the ground (in this case, weight added to downforce).

Therefore, surely the maximum speed of the car around a corner can be calculated by using $\mu R = \frac{mv^2}{r}$?

Is this valid for an approximation?

$\endgroup$
  • $\begingroup$ Is the Stig driving? Rear wheels spinning and opposite lock applied? $\endgroup$ – Solar Mike May 18 at 10:55
  • $\begingroup$ To make @SolarMike's comment a bit more explicit, it is only a "valid" approximation if the road is a horizontal surface (not banked around the curve) and the wheels are not skidding. You can go round corners faster by deliberately making the wheels spin, so the engine is providing additional centripetal force. $\endgroup$ – alephzero May 18 at 11:24
  • $\begingroup$ @PhysicsGuy123: I would rather say that it is the frictional force (caused by limited slip) between the tire and the ground that provides the centripetal force. $\endgroup$ – Winter Soldier May 18 at 12:32
  • $\begingroup$ @alephzero. Please can you explain how wheelspin causes you to go faster around the corner? I compete in racing, and generally the fastest way around the corner is to have the car on the limit of grip, but with little or no wheelspin, as this generally causes oversteer which eventually means you go wider (hence making you go slower). I admit, you do use the throttle pedal to control the car through the corner, but from my experience, when you get wheelspin it tends to negatively impact lap time, not make you faster. $\endgroup$ – PhysicsGuy123 May 18 at 12:59
0
$\begingroup$

Yes your reasoning is correct. The centripetal force acts towards the centre of curvature of the bend and the friction acts directly opposite this. The dependence on the mass of the system cancels. A different case arises with the popular fair ground ride in which a rotating wall supports your weight. It is interesting to compare the direction of the forces in each case.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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