Find coefficient of static friction

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The discussion focuses on calculating the coefficient of static friction for a car accelerating on a flat circular track. A participant initially misapplied the centripetal acceleration, leading to an incorrect value for the coefficient. Another contributor pointed out that the forward tangential acceleration must also be considered, which would yield a higher coefficient when combined using Pythagorean principles. The conversation highlights the complexities of real-world driving dynamics, such as weight distribution and wheel drive, which can affect friction and handling. Ultimately, understanding these factors is crucial for accurately predicting vehicle behavior during cornering.
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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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