Coefficient of static friction with radius, speed, etc. given

AI Thread Summary
A car accelerates on a flat circular track with a radius of 331 m until its tires skid at a speed of 27.8 m/s. The discussion focuses on calculating the coefficient of static friction, requiring the total acceleration, which includes both radial and tangential components. Initial attempts to find the coefficient using only radial acceleration were corrected to include the tangential acceleration as well. The correct approach involves summing the accelerations and applying the formula for friction. Understanding the role of friction in providing both tangential and centripetal forces is emphasized as crucial for solving the problem.
MissPenguins
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Homework Statement



To test the performance of its tires, a car travels along a perfectly flat (no banking) circular track of radius 331 m. The car increases its speed at uniform rate of at ≡ d|v|dt = 3.68 m/s^2 until the tires start to skid. If the tires start to skid when the car reaches a speed of 27.8 m/s, what is the coefficient of static friction between the tires and the road? The acceleration of gravity is 9.8 m/s^2.

Homework Equations


F = ma(subscript c) = m(v^2/r)
us = F/N

The Attempt at a Solution


I am Physics dummy, but I tried doing (27.8 m/s)^2 / 331 m. I tried to find the F of it. But I got all confused, I don't know what I am doing anymore. Can someone help? Please.
 
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Welcome to PF!

Hi MissPenguins! Welcome to PF! :smile:

(try using the X2 tag just above the Reply box :wink:)
MissPenguins said:
I am Physics dummy, but I tried doing (27.8 m/s)^2 / 331 m. I tried to find the F of it. But I got all confused, I don't know what I am doing anymore. Can someone help? Please.

Well, your approach looks ok …

that is the radial acceleration (but don't forget there's also a tangential acceleration in this case :wink:) …

where did you go from there? :smile:​
 


tiny-tim said:
Hi MissPenguins! Welcome to PF! :smile:

(try using the X2 tag just above the Reply box :wink:)


Well, your approach looks ok …

that is the radial acceleration (but don't forget there's also a tangential acceleration in this case :wink:) …

where did you go from there? :smile:​


I skipped to the next question because I got all confused and I don't know what else to do. :(
 
ok, next thing is to find the total acceleration, then find the friction force, then find the coefficient of friction. :smile:

(to put it another way: you know what acceleration you are trying to produce … so how much force does that need, and where does the force come from?)
 
tiny-tim said:
ok, next thing is to find the total acceleration, then find the friction force, then find the coefficient of friction. :smile:

(to put it another way: you know what acceleration you are trying to produce … so how much force does that need, and where does the force come from?)

But did I even start it right? I had ac = v2/r = (27.8)2/331 = 2.3348640...m/s2, and then I use u = a/g = 2.3348640.../9.8 = 0.238257...and I got it wrong again :(.
 
(have a mu: µ :wink:)

Yes, that method is correct (ma = Ffr, and |Ffr| = µmg), except that your a must be the total acceleration (ie the vector sum of the radial and tangential accelerations).
 
tiny-tim said:
(have a mu: µ :wink:)

Yes, that method is correct (ma = Ffr, and |Ffr| = µmg), except that your a must be the total acceleration (ie the vector sum of the radial and tangential accelerations).


Okie, I am confused when you use (ma = Ffr, and |Ffr| = µmg) since mine was ac = v2/r. But I think I got what u mean. So i redo it and have (2.3348640...+ 3.68 m/s2) / 9.8 = 0.61376163758555. Am I correct?
 
MissPenguins said:
(2.3348640...+ 3.68 m/s2)

Nope :redface: … you need the magnitude of the vector sum of aradial and atangential, which is … ? :smile:
 
tiny-tim said:
Nope :redface: … you need the magnitude of the vector sum of aradial and atangential, which is … ? :smile:

Okie, I did sqrt of [((2.334864048...)2 + (3.68)2) / 9.8 ] = 0.44471525601966. correct? thanks
 
  • #10
MissPenguins said:
Okie, I did sqrt of [((2.334864048...)2 + (3.68)2) / 9.8 ] = 0.44471525601966. correct? thanks

Should be! :smile:
 
  • #11
tiny-tim said:
Should be! :smile:

OMG, finally got it right. Thank you so much. :)
 
  • #12
I have the same exact problem and understand how to get to the end. But how do you continue the problem without a mass?
 
  • #13
Welcome to PF!

Hi ffejoriac! Welcome to PF! :smile:

Call the mass m …

you'll find it'll cancel in the end. :wink:​
 
  • #14
what would be the radial force? is it the friction? thanks I know that the tangential component of the total force is from the car's acceleration.. you only have those 2 forces in the horizontal direction , so it must be the friction?
 
  • #15
holezch said:
what would be the radial force? is it the friction? thanks I know that the tangential component of the total force is from the car's acceleration.. you only have those 2 forces in the horizontal direction , so it must be the friction?

Hi holezch! :smile:

Sorry, but you're putting the cart before the horse, and it's an imaginary cart.
:redface:

There aren't 2 forces in the horizontal direction

For the motion given in the question, the acceleration is completely fixed. So first find that acceleration.

Then find the force that will produce that acceleration …

that's the friction force … one force.
 
  • #16
I thought that the car produces the fixed acceleration itself( from the engine or something), and then that would be a tangential force.. and there should be a radial component to the force as well, since the car is moving in circularly.. and you'll have an opposite frictional force along with it right? and that's what makes the car turn? thanks
 
  • #17
You need the magnitude of acceleration which you find from the tangential acceleration and centripetal acceleration
 
  • #18
holezch said:
I thought that the car produces the fixed acceleration itself( from the engine or something)

Nooo :redface: …

that's an internal force …

only external forces can affect the movement of the car. :wink:

The engine exerts a force on (ultimately) the tyres, but all this is internal. Then the tyres exert a friction force on the Earth, which makes the Earth rotate very slightly (the exact direction depends mostly on the ways the four wheels are pointing).

The Earth in turn exerts an equal and opposite force on the tyres … that's the friction force that moves the car.​
 
  • #19
tiny-tim said:
Nooo :redface: …

that's an internal force …

only external forces can affect the movement of the car. :wink:

The engine exerts a force on (ultimately) the tyres, but all this is internal. Then the tyres exert a friction force on the Earth, which makes the Earth rotate very slightly (the exact direction depends mostly on the ways the four wheels are pointing).

The Earth in turn exerts an equal and opposite force on the tyres … that's the friction force that moves the car.​

oh wow :| I didn't even know that haha.. That's interesting.. thanks a lot! so the friction creates both tangential and centripetal accelerations
 

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