Question

A car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of 2.05 m/s2. The car makes it one quarter of the way around the circle before it skids off the track. From these data, determine the coefficient of static friction between the car and track.

Answer 1

The coefficient of static friction is
`\mu_s = 0.6897`

Step-by-step explanation:

From the question we are told that

The tangential acceleration is
`a = 2.05 \ m/s^2`

Generally the circumference of the circle is mathematically represented as

`C = 2 \pi r \\`

and given that the car travels one quarter of the circular path before it skids off the track , then the distance traveled along the track is mathematically represented as

`d = (1)/(4 ) * C = (2\pi r)/(4) = (\pi r)/(2)`

Generally from kinematic equations

`v ^2 = u^2 + 2 * a * d`

Here u is the initial velocity of the car which is 0 m/s

while v is the velocity of the car that keeps it from skiing off the track

So

`v ^2 = 0 + 2 * a * (\pi r)/(2)`

=>
`v = √(a \pi r)`

Generally the centripetal force acting on the car is mathematically represented as

`F_c = (m v^2)/(r)`

=>
`F_c = (m (√(a \pi r ) )^2)/(r)`

=>
`F_c = m \pi a`

Generally the frictional force between the track and the tires of the car is mathematically represented as

`F_f = \mu_s * m * g`

Generally the tangential force acting on the car is mathematically represented as

`F_f = m * a`

Generally the resultant force acting on the car is mathematically \

`F_r = √(F_t + F_c)`

=>
`F_r = √( (ma)^2 + (\pi ma)^2 )`

=>
`F_r = √( m^2a^2+ \pi^2 m^2a^2 )`

=>
`F_r = √((ma)^2 (1 + \pi^2) )`

=>
`F_r = ma √( 1 + \pi^2 )`

Generally the point before the car skids off the track, frictional force is equal to the resultant force and this is mathematically represented as

`ma √( 1 + \pi^2 ) = \mu_s * m * g`

=>
`\mu_s = (a)/(g) * √(1 + \pi^2)`

=>
`\mu_s = (2.05)/(9.8) * √(1 + 3.142 ^2)`

=>
`\mu_s = 0.6897`