Finding static friction from circular motion

AI Thread Summary
A car accelerates uniformly on a flat circular track with a tangential acceleration of 1.70 m/s² and skids after traveling one-quarter of the circle. To find the coefficient of static friction, the radial and tangential accelerations must be considered together, as friction acts to provide the necessary net acceleration. The radial acceleration after a quarter turn is expressed as π times the tangential acceleration. The discussion emphasizes the importance of considering both types of acceleration when calculating friction, as the frictional force is constrained by the normal force and the coefficient of friction. Understanding the interplay between tangential and radial forces is crucial for solving the problem accurately.
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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.

Homework Equations

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
5.3388m = 9.8[PLAIN]https://upload.wikimedia.org/math/9/3/9/939974a71dda1b83cce5ab82a2d2cec1.pngm
939974a71dda1b83cce5ab82a2d2cec1.png
= 0.545

what's wrong? Thanks!
 
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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.

Homework Equations

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
5.3388m = 9.8[PLAIN]https://upload.wikimedia.org/math/9/3/9/939974a71dda1b83cce5ab82a2d2cec1.pngm
939974a71dda1b83cce5ab82a2d2cec1.png
= 0.545

what's wrong? Thanks!

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

Could someone please clarify?
 
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.
 
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Ok i understand! thank you for your help haruspex!
 
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