# Friction and centripetal force [closed]

Let us imagine that some object (A car possibly) is moving in a circular path. At some point of time let's say that the driver of car made a turn in order to follow the circular motion. Now my textbook claims that, The friction is but the centripetal force that is acting on the car.

Upon searching on the internet I found a plenty of answers but all of them were handwavy (At least for me).

My question

Is there any mathematical approach to this?

• Remember that centripetal force is the force required to make an object follow the circular path rather than continue straight as it would do without the force.
– Dan
Feb 16 at 20:13

TL;DR A particle that undergoes uniform circular motion has a net (resultant) force that points (always) towards the center of motion. In case of a level curved road, the friction force is the only force that can provide the radial force component.

You should start backwards with this problem, i.e. from a net (resultant) force. A particle in uniform circular motion has a net force that points towards the center of circular path, hence the name radial force or centripetal force:

$$\vec{F}_\text{net} = -m \frac{v^2}{R} \hat{r} \tag 1$$

where $$\hat{r}$$ is the position unit vector, $$R$$ is radius of the circular path, $$m$$ is mass of the particle, and $$v$$ is (tangential) velocity of the particle. Notice that $$-\hat{r}$$ is a unit vector that points towards the center of (circular) motion.

Now assume that the car travels on a level (horizontal) road. The three (significant) forces that act on the car are (i) gravitational force (weight) exerted by Earth, and (ii-iii) normal and friction forces exerted by the road surface. There is also air drag (resistance), bet let's neglect this force at the moment. In this particular case, weight $$\vec{w}$$ and normal force $$\vec{n}$$ are equal in magnitude and opposite in direction, i.e. $$\vec{w} = -\vec{n}$$.

Source: H. D. Young, R. A. Freedman, "University Physics with Modern Physics in SI Units", 15th. ed, 2019.

When the car enters a circular path, assuming it travels at constant (uniform) velocity, the net force on the car is given by

$$\vec{F}_\text{net} = -m \frac{v^2}{R} \hat{r} = \vec{w} + \vec{n} + \vec{f}_s$$

where $$\vec{f}_s$$ is the (static) friction force. Since weight and normal force cancel ($$\vec{w} + \vec{n} = 0$$), it turns out that the friction force provides the radial component which enables the car to follow a circular path. If there were no friction force ($$\vec{f}_s = 0$$) the car would not be able to follow circular path, as there is nothing to provide the radial force component!

Since the static friction force has a maximum value it can take, defined by the coefficient of static friction $$\mu_s$$, there is a maximum tangential speed $$v$$ which allows the car to follow circular path without slipping:

$$m \frac{v^2}{R} = m g \mu_s \qquad \rightarrow \qquad \boxed{v_\text{max} = \sqrt{g R \mu_s}}$$

## Banked curved road with no friction

A banked curved road enables car to follow a circular path without any friction. The trick is that normal force is no longer opposite in direction to the weight, but at some angle. In that case, vertical component of the normal force cancel the weight, and horizontal component of the normal force provide radial force component necessary for circular motion.

Source: H. D. Young, R. A. Freedman, "University Physics with Modern Physics in SI Units", 15th. ed, 2019.

Vertical component of the normal force cancels weight

$$mg = n \cos\beta$$

and the horizontal component of the normal force provides radial force component

$$n \sin\beta = m \frac{v^2}{R}$$

The car can move one a banked curve road at the following velocity

$$v = \sqrt{g R \tan\beta}$$

without any (static) friction force. This is of course idealization - in reality there is always some friction. I leave this analysis of banked curved road with friction to you.