Angular Acceleration: finding the coefficient of friction

[vipertongn]

Homework Statement

A car traveling on a flat (unbanked) circular track accelerates uniformly from rest with a tangential acceleration of 1.80 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.

Homework Equations

[tex]\Delta[/tex][tex]\theta[/tex]=1.57 rad
Fc = mv2/r
f= Fc
f=uN

The Attempt at a Solution

I really don't understand how to go at this problem. I was told to find Vf, but how do you do that? Knowing only the tangential acceleration, how would I use that to find Vf. And not knowing what the radius is how do I find the centripetal force and connecting that with the normal force? Without the mass being given?

 

[tiny-tim]

A car traveling on a flat (unbanked) circular track accelerates uniformly from rest with a tangential acceleration of 1.80 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.
…
Knowing only the tangential acceleration, how would I use that to find Vf. And not knowing what the radius is how do I find the centripetal force and connecting that with the normal force? Without the mass being given?

Hi vipertongn!

You don't need the mass because it cancels.

You can forget the tangential acceleration once you've calculated V_f (for which you use the usual constant acceleration equations … call the radius r, it'll cancel out later).

Then just use good ol' Newton's second law in the radial direction.

 

[vipertongn]

I was wondering which one though? I'm not sure if it starts from rest or not and i believe Vf=vi+at might be one equation i could use...but i don't know the time! so how would i calculate for Vf?

 

[tiny-tim]

I was wondering which one though? I'm not sure if it starts from rest or not and i believe Vf=vi+at might be one equation i could use...but i don't know the time! so how would i calculate for Vf?

Yes, V_i = 0 … it "accelerates uniformly from rest".

You have V_i V_f a and s …

so the equation to use is … ?

 

[vipertongn]

OHHHH i would use change in Vf^2=Vi^2-2as

However, where would s come from? I only know the angular distance thingy. i do know that theta is equal to s/r ,but like... i don't know what r is.

 

[tiny-tim]

OHHHH i would use change in Vf^2=Vi^2-2as

However, where would s come from? I only know the angular distance thingy. i do know that theta is equal to s/r ,but like... i don't know what r is.

The constant acceleration equations work for any constant double derivative (rate of rate of change) …

we normally use it for distance speed and acceleration, but we can also use it for angle angular speed and angular acceleration (θ θ' and θ'').

 

[vipertongn]

so does that mean it's interchangable w/ S?

Vf^2=Vi^2-2a[tex]\Delta[/tex][tex]\theta[/tex]

 

[tiny-tim]

so does that mean it's interchangable w/ S?

Vf^2=Vi^2-2a[tex]\Delta[/tex][tex]\theta[/tex]

Sorry …

I was making it too complicated …

I should just have said that you can treat distance round the track as if it was straight …

even though s doesn't represent a straight distance, it's still a variable with constant second derivative, so Vf²=Vi²-2as is still valid

 

[vipertongn]

but what would s be then?

sould i assume its like 1/4 2pi then?

 

[tiny-tim]

just got up … :zzz:

I think I've confused you

No, s is just ordinary distance, as is shown on the car's "mileometer" … it just isn't in a straight line.

(you could use combinations of r θ θ' and θ', but in this case it isn't necessary)

 

[vipertongn]

yes ! i got hte correct answer! i just had to use variables nad plugi n the equations to cancel them out! haha yea you did confuse me abit ther but its ok you helped me ^^