In this article we are going to discuss 5 facts related to frictional force and centripetal acceleration.

**Before starting with the detailed facts related to frictional force and centripetal acceleration we need to have the basic idea of frictional force and centripetal acceleration. Frictional force is a force that resists the motion of a sliding body or surface over the other.**

If a body travels along a circular path with a change in the direction of its velocity,then the acceleration of it is known as centripetal acceleration. Centripetal force is acted on that body so that is why centripetal acceleration is generated.

**How is frictional force related to centripetal acceleration?**

**To describe the relation between frictional force and centripetal acceleration we have to take an example of a car that is moving on a flat curve. The curve is an unbanked curve. As the car is moving on the road there should be a tendency for it to get slipped.**

But it does not slip,why? The answer is static friction. Static frictional force acts in the opposite direction of the motion of the car on the road that is why it does not slip. The only force that acts on the tires of the car in the horizontal direction is the static frictional force.

We all know that centripetal force is needed by a car to move on an unbanked curve. So here the static frictional force must be the centripetal force. The mathematical expression for the static frictional force is equal to F_{s}= μ.N where N denotes the normal force and μ denotes the coefficient of static friction.

The normal N is equivalent to the weight of the car that is acting in downward direction. Therefore , F_{s}= μ.N = μ.m.g. Now we all know that the mathematical expression for the centripetal force is F_{c} = mv^{2}/r where v and r denote the velocity of the car and radius of the curve respectively.

Therefore , F_{s}= μ.N = μ.m.g = F_{c} = m.v^{2}/r

v^{2}/r = μ.g

Here the mathematical expression for the centripetal acceleration is v2/r. Hence we can say that we can calculate the value of centripetal acceleration from the frictional force. The relation between frictional force and centripetal acceleration is v2/r = μ.g.

**Can frictional forces produce centripetal accelerations?**

**A very common example of banking of roads can help us to understand if frictional forces can be able to produce centripetal accelerations. At first we need to know what is banking of roads? Sometimes we see tires of cars skid while taking an U turn on a road. **

To remove this possibility of skidding centripetal force is provided to the car. The phenomenon of maintaining an angle to reduce the probability of slipping of a car is known as **banking of the road.**

There are 3 types of banking. Here only the case of how frictional forces can produce centripetal accelerations is going to be illustrated. If the banking angle is zero then only frictional forces can produce centripetal accelerations.

When the banking angle remains zero then the car has to take turns on the flat curves. In this case the upward directed normal force(N) balances the weight(m.g) of the car that acts downwards as it is vertical in nature. Therefore , N =m.g………(1)

If the road is smooth enough then the car cannot take the turns as it will slip in the absence of frictional forces. But in case of a rough road the centripetal force will be provided by the frictional force(f) that acts upon the road.

Therefore , f = μ.N = μ.m.g …………….(2) [from equation (1)]

As the centripetal force ( F_{c}= m.v^{2}/r) is provided by the frictional force (f) hence

f = μ.N = μ.m.g = F_{c}= m.v^{2}/r [from equation (2)]

Therefore , μ.g = v^{2}/r

v = √μ.r.g …………….. (3)

Here v= v _{max} = the maximum velocity that can be obtained by the car on a flat road

**When frictional forces produce centripetal accelerations?**

**Here we can take an example of a merry go round. Say a person is standing on a merry go round and the merry go round is moving but not in a very fast way. Then the person will find himself standing on the same position where he was standing when the merry go round was at rest. **

Now the question arises how it is possible as the merry go round is moving the person should also change his position. The answer is due to frictional force. In this case the acceleration that is gained by the person is centripetal acceleration and it is produced by the frictional force.

**How frictional forces produce centripetal accelerations?**

**Here again we will take the example of a moving car that wants to take a turn on an unbanked road. In the previous portion of this concept is already discussed in a detailed way. So if a car wants to take a turn on a flat road it will be slipped. If it is not getting slipped then it is sure that there is another force that has reduced its chance of skidding. **

This force is static frictional force. It provides centripetal force to the tires of the car. The acceleration gained by the car which is the centripetal acceleration is also produced from this static frictional force.

**Examples of frictional forces that produce centripetal accelerations**

**Here we will take a mathematical example to show that frictional forces produce centripetal accelerations.**

**Numerical problem :**

**A car is moving on a flat road. What is the value of the maximum velocity at which the car should move to remove the probability of skidding? The value of coefficient of static fraction is 0.4 and the mass of the car is 100 kg and the radius of the road is 4 m. (g= 10 m/s ^{2})**

**Answer :**

It is mentioned in the question that the road is flat. It means that the angle of banking is zero.

Hence the centripetal force that is needed by the car to take a turn is provided by the static friction.

Here μ=0.4,m=100 kg and g= 10 m/s^{2},r = 4 m

F_{s}= μ.m.g

μ.m.g = m.v^{2}/r

v _{max} = v = √μ.r.g

v _{max} = √0.4 x 4 x 10

v _{max} = 4 m/s where v _{max} is the maximum velocity at which the car should move to remove the probability of skidding.

**Conclusion**

in this article the relation between frictional force and centripetal acceleration along with a numerical example is described in an elaborate manner.