Angular Acceleration: finding the coefficient of friction

In summary, the car skids off the track after making it one-quarter of the way around the circle. The coefficient of static friction between the car and track is 1.57.
  • #1
vipertongn
98
0

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?
 
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  • #2
vipertongn said:
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! :smile:

You don't need the mass because it cancels.

You can forget the tangential acceleration once you've calculated Vf (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. :wink:
 
  • #3
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?
 
  • #4
vipertongn said:
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, Vi = 0 … it "accelerates uniformly from rest".

You have Vi Vf a and s …

so the equation to use is … ? :smile:
 
  • #5
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.
 
  • #6
vipertongn said:
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''). :smile:
 
  • #7
so does that mean it's interchangable w/ S?

Vf^2=Vi^2-2a[tex]\Delta[/tex][tex]\theta[/tex]
 
  • #8
vipertongn said:
so does that mean it's interchangable w/ S?

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

Sorry :redface:

I was making it too complicated :rolleyes:

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 :smile:
 
  • #9
but what would s be then?

sould i assume its like 1/4 2pi then?
 
  • #10
just got up … :zzz:

I think I've confused you :redface:

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

(you could use combinations of r θ θ' and θ', but in this case it isn't necessary)
 
  • #11
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 ^^
 

Related to Angular Acceleration: finding the coefficient of friction

1. What is angular acceleration?

Angular acceleration is the rate at which an object's angular velocity changes over a given amount of time. It is measured in radians per second squared (rad/s²) and is a measure of how quickly an object's rotational speed is changing.

2. How is angular acceleration related to coefficient of friction?

The coefficient of friction is a measure of the amount of resistance between two surfaces when in contact with each other. In the context of angular acceleration, it is the ratio of the frictional torque to the normal force acting on an object. This coefficient is important in determining the amount of angular acceleration an object experiences.

3. How do you calculate angular acceleration?

Angular acceleration can be calculated using the formula α = τ / I, where α is the angular acceleration, τ is the net torque acting on the object, and I is the moment of inertia of the object. Alternatively, it can also be calculated by taking the second derivative of an object's angular displacement over time.

4. What factors affect the coefficient of friction?

The coefficient of friction is affected by several factors, including the roughness of the surfaces in contact, the type of material, the normal force acting between the surfaces, and the presence of any lubricants. It is important to consider these factors when trying to determine the coefficient of friction in order to accurately calculate angular acceleration.

5. How can the coefficient of friction be experimentally determined?

The coefficient of friction can be experimentally determined by conducting a series of tests where the normal force and frictional force are measured for different surfaces in contact. The coefficient can then be calculated by dividing the frictional force by the normal force. This process may need to be repeated multiple times in order to obtain an accurate average value.

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