Find coefficient of static friction

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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 tangential acceleration and its motion before skidding off the track.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the relationship between tangential and radial accelerations, with one participant noting the importance of considering both components to accurately determine the coefficient of static friction.

Discussion Status

There is an ongoing exploration of the problem, with participants providing insights into the calculations and acknowledging the complexities involved in real-world applications, such as weight distribution and wheel drive dynamics.

Contextual Notes

Participants mention the challenges of applying theoretical concepts to practical scenarios, particularly regarding the behavior of a car on a circular track and the implications of acceleration on frictional forces.

Sefrez
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The problem:
A car traveling on a flat (unbanked) circular track accelerates uniformly from rest with a tangential acceleration of 1.60 m/s2. The car makes it one quarter of the way around the circle before it skids off the track. Determine the coefficient of static friction between the car and track from these data.

I keep ending up with acceleration_radial = PI*acceleration_tangential, and thus u = PI*acceleration_tangential/g = ~0.513. My assignment says that it is wrong.
 
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Sefrez said:
The problem:
A car traveling on a flat (unbanked) circular track accelerates uniformly from rest with a tangential acceleration of 1.60 m/s2. The car makes it one quarter of the way around the circle before it skids off the track. Determine the coefficient of static friction between the car and track from these data.

I keep ending up with acceleration_radial = PI*acceleration_tangential, and thus u = PI*acceleration_tangential/g = ~0.513. My assignment says that it is wrong.

Your answer has accounted for the centripetal acceleration developed [and done it most elegantly], but has ignored the forward acceleration of 1.60, which after you apply Pythagoras gives a slightly higher μ.

[If course a real car presents a great problem here as not all wheels necessarily drive, and the weight distribution affects the reaction force for each tyre.]
 
PeterO said:
Your answer has accounted for the centripetal acceleration developed [and done it most elegantly], but has ignored the forward acceleration of 1.60, which after you apply Pythagoras gives a slightly higher μ.

[If course a real car presents a great problem here as not all wheels necessarily drive, and the weight distribution affects the reaction force for each tyre.]

Ah. Thank you for that. That makes sense other than as you said that with a real car not all wheels necessarily drive. I am surprised I did not see this after the time put into thinking about it. I only had a single component of the acceleration! *bangs head*

Thanks again!
 
Sefrez said:
Ah. Thank you for that. That makes sense other than as you said that with a real car not all wheels necessarily drive. I am surprised I did not see this after the time put into thinking about it. I only had a single component of the acceleration! *bangs head*

Thanks again!

Don't feel so bad - my post had begun with "lloks great to me I can't understand why that is not the answer .." when I suddenly realized the effect of acceleration.

A practical application here is that if you are rounding a corner on a loose surface, and are on the point of slipping/spinning out, then it is not possible to accelerate or brake without dire consequences! - a real problem when you come across an obstacle.
That explains many of the crashes in NASCAR.
 

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