Can I deduce the coefficient of static friction with Mass, Velocity and Radius?

AI Thread Summary
The discussion centers on calculating the coefficient of static friction for a car on a track using mass, velocity, and radius. Given a mass of 2000 kg, a velocity of 50 m/s, and a radius of 1000 m, the user calculates the centripetal force as 5000 N. They propose using this force in the equation Fs(max) = μn, where n is the normal force, calculated as 19,600 N. The resulting coefficient of static friction is approximately 0.26. The conversation also highlights that if the car could maintain a higher speed without slipping, the calculated coefficient would increase.
Naoki
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The Known:
Mass=2000kg
Velocity=50m/s
Radius=1000m

With the given information the lab worksheet I'm working on is asking me to deduce the coefficient of friction so that I can manipulate the data and answer some questions. The problem I have is that I am not entirely sure that this could be done with just the given information. BTW, this is supposed to be a car on a track.

My attempt at the solution is:

Using Fs= mv^2/r I can get 2000*50^2/1000=5000N

so the question I have is can I use this as the following Fs(max) in this equation to figure out the coefficient of static friction?

Fs(max)=μn where n is the normal force or m*g witch in this case would be 2000*9.8=19,600kgs.

If were to attempt that is would be Fx(max)/n = μ 5000/19600=.26

Is my logic sound? Please help.
 
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Hi Naoki. Would you like to have another go at posting your question. None of the details have appeared in your message.
 
could you look again, I just edited it. thanks
 
Naoki said:
If were to attempt that is would be Fx(max)/n = μ 5000/19600=.26

Is my logic sound? Please help.
That looks right, assuming it can go no faster without slipping. :smile:

http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif
 
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I don't know if that is the max speed without slipping. I know that it is the Force of friction with the known. Would that change the answer?
 
Were the car able to go even faster, still without slipping, then your calculations would come up with a greater value for μ for that surface.
 
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