# Finding static friction from circular motion

1. Aug 23, 2015

### Zynoakib

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. 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

F = ma = [PLAIN]https://upload.wikimedia.org/math/9/3/9/939974a71dda1b83cce5ab82a2d2cec1.pngn [Broken]
= 0.545

what's wrong? Thanks!

Last edited by a moderator: May 7, 2017
2. Aug 23, 2015

### PeroK

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

Last edited by a moderator: May 7, 2017
3. Aug 23, 2015

### Zynoakib

Ok, I get it. Thanks!

4. Aug 23, 2015

### haruspex

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.

5. Oct 14, 2016

### Erik coolguy

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

You see, this is an interesting problem since it looks for static friction. im 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.

6. Oct 14, 2016

### haruspex

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.

7. Oct 15, 2016

### Erik coolguy

Ok i understand! thank you for your help haruspex!