# A car 1500 kg is turning a flat curve while moving at constant speed. The radius of the curve is 35.0 m. If the coefficient of static friction between tires and dry pavement is 0.523. Determine the maximum speed of car to negotiate the turn successfully?

Mar 27, 2018

The centripetal force required for circular motion of the car is provided by the force of friction between the tires and pavement. Due to Newton's Third law of motion, the car experiences centrifugal force which tends to topple it off the road in the outward direction.

Force of friction along the radius $R$ is given as

${f}_{r} = {\mu}_{s} | \vec{N} |$
where $\vec{N}$ is normal reaction and $= m \vec{g}$, where $m$ is mass of car and $g$ is gravity $= 9.81 \setminus m {s}^{-} 2$

When the car having velocity ${v}_{\max}$ is on the verge of skidding force of friction must be equal to centripetal force which is given as

${F}_{c} = \frac{m {v}_{\max}^{2}}{R}$

Equating the two and inserting given values we get

$\frac{m {v}_{\max}^{2}}{R} = {\mu}_{s} m g$
$\implies {v}_{\max}^{2} = {\mu}_{s} g R$
$\implies {v}_{\max} = \sqrt{\left(0.523\right) \left(9.81\right) \left(35.0\right)}$
$\implies {v}_{\max} = 13.4 \setminus m {s}^{-} 1$