Wednesday

April 16, 2014

April 16, 2014

Posted by **Noah** on Wednesday, October 24, 2012 at 11:47am.

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 1:42pma = 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 7:43pmAc = 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=(|a|max)/(g)=4.069/9.8=.4152

- AP Physics -
**aziz**, Sunday, October 27, 2013 at 7:44pmAc = 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=(|a|max)/(g)=4.069/9.8=.4152

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