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. $\sqrt{\frac{rg}{2}}$ and $\sqrt{rg}$
• B. $\frac{\sqrt{rg}}{2}$ and $\sqrt{rg}$
• C. $\frac{\sqrt{rg}}{2}$ and $2\sqrt{rg}$
• D. $0$ and infinite

Answer 1

Answer: (D)

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

$\Sigma f_{y^{\prime }}=N-mg\cos \theta -\frac{m{v}^2}{R}\sin \theta =0.$
$\Sigma f_{x^{\prime }}=\frac{m{v}^2}{R}\cos \theta +\mu N-mg\sin \theta =0$
$\Rightarrow \frac{m{v}^2}{R}\cos \theta +\mu \left(mg\cos \theta +\frac{m{v}^2}{R}\sin \theta \right)-mg\sin \theta =0$
$\Rightarrow v^2=\frac{(\mu Rg\cos \theta -Rg\sin \theta )}{(\cos \theta +\mu \sin \theta )}$
for $\theta =45^{\circ }$ and $m=1;v_{\min }=\frac{Rg-Rg}{1+1}=0$
$F.B.D$. for maximum speed (w.r.t. automobile)

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