A car travels along a curved circular turn of radius $R$ banked at an angle $\alpha$ to the horizont

Question

A car travels along a curved circular turn of radius $R$ banked at an angle $\alpha$ to the horizontal. Let $v_0 = \sqrt{gR 	an\alpha}$ be the design speed of the banked turn (where no lateral friction is needed). Which of the following statements regarding the static friction force between the road and the tires are correct?

Professor PC Thomas Classes · Answer

At the ideal banking speed $v_0 = \sqrt{gR 	an\alpha}$, the horizontal component of the normal reaction ($N \sin\alpha$) provides the exact centripetal force required ($m v_0^2 / R$), so no lateral friction is required, confirming A. If $v > v_0$, greater centripetal force is needed, so static friction must act down the incline towards the center of curvature, confirming B. If $v < v_0$, friction acts up the incline to prevent slipping down, confirming C. For outward skidding condition ($v > v_0$), balancing vertical and horizontal forces gives $\frac{v^2}{gR} = \frac{	an\alpha + \mu}{1 - \mu 	an\alpha}$. For $\alpha = 45^\circ$, $	an\alpha = 1$ and $v_0 = \sqrt{gR}$. When $v = 2v_0$, we have $\frac{v^2}{gR} = 4$. Solving $\frac{1+\mu}{1-\mu} = 4$ yields $1+\mu = 4 - 4\mu \implies 5\mu = 3 \implies \mu = 0.6$, confirming D.