Posted by Noah on Wednesday, October 24, 2012 at 11:47am.
To test the performance of its tires, a car
travels along a perfectly ﬂat (no banking) circular track of radius 120 m. The car increases
its speed at uniform rate of
at ≡
d v
dt
= 2.99 m/s
2
until the tires start to skid.
If the tires start to skid when the car reaches
a speed of 18.2 m/s, what is the coeﬃcient of
static friction between the tires and the road?
The acceleration of gravity is 9.8 m/s
2

AP Physics  Damon, Wednesday, October 24, 2012 at 1:42pm
a = 2.99 m/s^2 is irrelevant I suspect
Ac = v^2/r = 18.2^2/120 = 2.76 m/s^2
m * Ac = friction force = m g mu
2.76 = 9.8 mu
mu = 2.76/9.8 = .282

AP Physics  Anonymous, Sunday, October 27, 2013 at 7:43pm
Ac = v^2/r = 18.2^2/120 = 2.76 m/s^2
at ≡d v/dt= 2.99 m/s2
sqrt(2.76^2+2.99^2)=4.069 m/s^2
μs=(amax)/(g)=4.069/9.8=.4152
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