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Q. A circular road of radius $1000\, m$ has banking angle $45^{\circ}$. The maximum safe speed (in $ms ^{-1}$ ) of a car having a mass $2000\, kg$ will be (if the coefficient of friction between tyre and road is $0.5)$

Laws of Motion

Solution:

From FBD N $=\frac{ mg }{\sqrt{2}}+\frac{ mv ^2}{ R \sqrt{2}}$ and $\frac{ mg }{\sqrt{2}}+ f =\frac{ mv ^2}{ R \sqrt{2}}$
$ f =\mu N \Rightarrow \frac{\mu mg }{\sqrt{2}}+\frac{\mu m v^2}{ R \sqrt{2}}+\frac{m g}{\sqrt{2}}=\frac{m v^2}{R \sqrt{2}} $
$ \Rightarrow \mu mgR +\mu mv ^2+ mgR = mv ^2 $
$ \Rightarrow \operatorname{mgR}(\mu+1)= mv ^2(1-\mu) $
$ \Rightarrow v ^2= gR \frac{(\mu+1)}{1-\mu}=10 \times 1000 \times \frac{(1.5)}{0.5}=(\sqrt{3})^2 \times(100)^2 $
$ \therefore v =172 $