my phy proff forgot to post the class lectures online (we dont do lectures in class... long story) so i dont kno how to do any of the problems given on his online quiz... please help
The coefficient of friction for the tire-road interface of a car is 0.2936. What is the maximum speed in km/hr the car can have in a curve with a 27.61 m radius of curvature on a level road (not banked) before the car slides off the road
answer is 32.103193273093574, but how do u get it?
2007-02-20 01:22:57 · 2 answers · asked by Paul S 2 in Science & Mathematics ➔ Physics
Ok let´s begin.
First of all you have to know that in this case the force that keeps the car on the road is the friction force.
In a curve we also have the centripetal force.
So to know what´s the speed of the car we have to equal the centripetal force to the friction force.
Fcp = centripetal force
Fric = Friction force
Fcp = Fric
mass x velocity² / radius = cooeficient of friction x normal force
normal force in this case is = mass x gravity, know that we can say (gravity here is the gravity acceleration)
mass x velocity² / radius = cooeficient of friction x mass x gravity
organizing the equation we have
mass disappears because he have it on both sites of the equality
velocity² = cooeficient of friction x gravity x radiius
velocity = sqrt cooeficient of friction x gravity x radius
velocity = sqrt 0,2936 x 9,8 x 27,61
Velocity =~ 8,91 m / s =~ 32 km / h
Since I don´t know what´s the gravity acceleration value you have there I put 9,8 and found this result.
http://hyperphysics.phy-astr.gsu.edu/hbase/cf.html
check this site, it may help you.
I hope I helped you!
2007-02-20 02:20:26 · answer #1 · answered by Anonymous · 0⤊ 0⤋
F=mv^2/r,
where F=friction force
mgx0.2936=mxv^2/27.61
cancel off the m:
9.81x0.2936=v^2/27.61
So,v^2=79.51
v=? m/s
Dont have calculator with me now..Sorry
2007-02-20 01:41:25 · answer #2 · answered by yan 2 · 0⤊ 0⤋