Finding static friction from circular motion

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Homework Help Overview

The problem involves a car accelerating uniformly on a flat circular track and seeks to determine the coefficient of static friction based on the car's motion before it skids off the track.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between tangential and radial accelerations and their implications for static friction. Some express confusion about the relevance of tangential acceleration after overcoming static friction, while others suggest working symbolically to clarify the dynamics involved.

Discussion Status

Participants are actively engaging with the problem, questioning assumptions about the forces at play, and exploring the implications of different types of acceleration. Some guidance has been offered regarding the importance of considering both tangential and radial components of acceleration.

Contextual Notes

There is a noted confusion regarding the treatment of tangential acceleration in the context of static friction, as well as the need for clarity on the forces acting on the car during its motion.

Zynoakib
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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!
 
Last edited by a moderator:
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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.
 
Last edited by a moderator:
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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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