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...acceleration of 1.8m/s/^2.
The car makes it 0.6 of the way around the circle before skidding off the track.
acceleration of gravity=9.8m/s^2
What is the coeffcient of static friction between the car and track??

2007-10-01 13:21:18 · 1 answers · asked by michael 1 in Science & Mathematics Physics

1 answers

The centripetal force is
m*v^2/r

or in radians/sec
m*w^2

the acceleration in radians/s^2 is
1.8/r
the angular displacement is
.6*2*pi=.5*1.8*t^2/r
the speed at the point is
w=1.8*t/r
which is
g*u
u=1.8*t/(g*r)
t=sqrt(2*.6*2*pi*r/1.8)

solve for u
=0.133

I used pi=3.14

j

2007-10-02 06:34:43 · answer #1 · answered by odu83 7 · 0 0

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