Question

An automobile enters a turn of radius $R$. If the road is banked at an angle of $45^{\circ}$ and the coefficient of friction is $1$, the minimum and maximum speed with which the automobile can negotiate the turn without skidding is

• A. $\sqrt{\frac{r g}{2}}$ and $\sqrt{r g}$
• B. $\frac{\sqrt{r g}}{2}$ and $\sqrt{r g}$
• C. $\frac{\sqrt{r g}}{2}$ and $2 \sqrt{r g}$
• D. $0$ and infinite

Answer 1

Answer: (D)

$F.B.D$. for minimum speed (w.r.t. automobile):

$\Sigma f_{y'}=N-m g \cos \theta-\frac{m v^{2}}{R} \sin \theta=0 .$
$\Sigma f_{x'}=\frac{m v^{2}}{R} \cos \theta+\mu N-m g \sin \theta=0$
$\Rightarrow \frac{m v^{2}}{R} \cos \theta+\mu\left(m g \cos \theta+\frac{m v^{2}}{R} \sin \theta\right)-m g \sin \theta=0$
$\Rightarrow v^{2}=\frac{(\mu R g \cos \theta-R g \sin \theta)}{(\cos \theta+\mu \sin \theta)}$
for $\theta=45^{\circ}$ and $m=1 ; v_{\min }=\frac{R g-R g}{1+1}=0$
$F.B.D$. for maximum speed (w.r.t. automobile)

$\Sigma f_{x^{\prime}}=\frac{m v^{2}}{R} \cos \theta-m g \sin \theta$
$-\mu\left(m g \cos \theta+\frac{m v^{2}}{R} \sin \theta\right)=0$
for $\theta=45^{\circ}$ and $\mu=1$
$v_{\max }=\infty$ (infinite)