# Finding static friction from circular motion

• Zynoakib
In summary, a car traveling on a flat, circular track accelerates uniformly from rest with a tangential acceleration of 1.70 m/s2. After making one-quarter of a lap, the car skids off the track. To determine the coefficient of static friction between the car and the track, we can use the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the tangential acceleration, and s is the distance traveled. This leads to a coefficient of static friction of 0.545. The tangential and radial accelerations must be taken into account when considering the force of friction acting on the car.
Zynoakib

## Homework Statement

A car traveling on a flat (unbanked), circular track
accelerates uniformly from rest with a tangential acceleration
of 1.70 m/s2. The car makes it one-quarter of
the way around the circle before it skids off the track.
From these data, determine the coefficient of static
friction between the car and the track.

## The Attempt at a Solution

The distance traveled by the car:
2r(pi)(1/4) = 1.57r

s = 1.57r, v = ?, u = 0, a = 1.7

v^2 = u^2 + 2as
v = (5.338r)^(1/2)

ac = (v^2)/r = 5.338

= 0.545

what's wrong? Thanks!

Last edited by a moderator:
Zynoakib said:

## Homework Statement

A car traveling on a flat (unbanked), circular track
accelerates uniformly from rest with a tangential acceleration
of 1.70 m/s2. The car makes it one-quarter of
the way around the circle before it skids off the track.
From these data, determine the coefficient of static
friction between the car and the track.

## The Attempt at a Solution

The distance traveled by the car:
2r(pi)(1/4) = 1.57r

s = 1.57r, v = ?, u = 0, a = 1.7

v^2 = u^2 + 2as
v = (5.338r)^(1/2)

ac = (v^2)/r = 5.338

= 0.545

what's wrong? Thanks!

What's the acceleration of the car when it skids? Perhaps draw a diagram.

Last edited by a moderator:
Zynoakib
Ok, I get it. Thanks!

Further to PeroK's reply, I encourage you to work symbolically as far as possible, only plugging in numbers at the end. It has numerous advantages. In the present case, you would have found that the radial acceleration after a quarter turn was ##\pi a##, where a is the given acceleration.

Zynoakib and PeroK
I'm sorry for putting up a new question here but...

You see, this is an interesting problem since it looks for static friction. I am confused to why we should consider the tangential acceleration that is in the direction of the motion since this direction has already overcome the static friction. Why not just consider the direction where the car is static? Which is in the radial direction.

Erik coolguy said:
why we should consider the tangential acceleration that is in the direction of the motion since this direction has already overcome the static friction.
The force of friction makes no distinction between tangential and radial. To maintain rolling contact, under the given tangential and radial accelerations, the frictional force must act to provide both of those as a single net acceleration. Regardless of the direction of that, the magnitude is limited by the normal force and the coefficient of friction.

Erik coolguy
Ok i understand! thank you for your help haruspex!

## 1. What is static friction?

Static friction is a type of force that occurs when two surfaces are in contact and not moving relative to each other. It acts in the opposite direction of an impending motion and prevents objects from sliding against each other.

## 2. How is circular motion related to static friction?

Circular motion involves an object moving in a circular path at a constant speed. In order for this motion to occur, there must be a force acting towards the center of the circle, known as centripetal force. In the case of an object moving in a circular path on a surface, static friction provides the necessary centripetal force.

## 3. What factors affect the magnitude of static friction in circular motion?

The magnitude of static friction in circular motion is affected by the weight of the object, the coefficient of static friction between the object and the surface, and the radius of the circular path. As these factors change, the static friction force required to maintain circular motion also changes.

## 4. How can I calculate the static friction force in circular motion?

The static friction force in circular motion can be calculated using the formula F = mv²/r, where m is the mass of the object, v is the speed of the object, and r is the radius of the circular path. This formula is derived from the equation for centripetal force, F = mv²/r, where F represents the net force acting towards the center of the circle.

## 5. What happens if the static friction force is not enough to maintain circular motion?

If the static friction force is not enough to maintain circular motion, the object will start to slip or slide. This means that the kinetic friction force will take over and act in the opposite direction of the object's motion. The object will then continue to move in a straight line instead of a circular path.

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