Question

An unbanked circular highway curve on level ground makes aturn

Answer 1

Answer:

Explanation:

Given

Velocity of traffic $v=60\ mi/hr\approx 26.82\ m/s$

Maximum value of Centripetal force is one-tenth of weight

such that $(F_c)_{max}=\frac{mg}{10}$

to make a safe turn centripetal force with radius of curvature r is given by

$F_c=\frac{mv^2}{r}$

$(F_c)_{max}=\frac{mg}{10}=F_c=\frac{mv^2}{r}$

$\frac{mg}{10}=\frac{mv^2}{r}$

$r=\frac{10v^2}{g}$

$r=\frac{10\times 26.82^2}{9.8}$

$r=733.99\approx 734\ m$