# Friction as a Centripetal Force

When an object is moving in a circle, won't the frictional force oppose it's motion? And if the velocity is always tangential to the path, there is no component of it towards the center, so there is no component of friction towards the center (since the frictional force is antiparallel). So how can friction be a centripetal force, like in the case of a car rounding a corner?

• That would be true if tires where sliding, but they are rolling. As a result the relative motion of the contact patch is not tangential to the circle but radially out (see slip angles). Nov 10 '14 at 12:26

Frictional force opposes sliding motion, basically. Car tires produce centripetal force by changing their angle relative to the rest of the car's orientation. The tires do not slide in the direction of the tires' orientation: they roll. Friction in this direction rotates the tires, or if the engine is applying force to the wheels during the turn, friction prevents the tires from "burning rubber", and pushes the car in this direction.

Meanwhile, motion in the direction of the rest of the car's orientation is opposed by friction only to the extent that it is not motion in the direction of the tires' orientation. The velocity vector corresponding to the rest of the car's orientation can be understood in terms of these two orthogonal components. The component corresponding to the tires' orientation is basically not subject to friction for our purposes (ignoring whether one's foot is on the gas pedal). The component that does not correspond to that other component is orthogonal and opposed by centripetal friction.

Based on GIF by Droidmakr.

If tires' orientation $=$ rest of the car's orientation, basically no centripetal friction results.