Question

What will be the maximum speed of a car on road-turn of radius $30m$ if the coefficient of friction between the tyres the rod is $0.75$?

• A. $10m{s}^{-1}$
• B. $15m{s}^{-1}$
• C. $8m{s}^{-1}$
• D. $12m{s}^{-1}$

Answer 1

Answer: (B)

$15m{s}^{-1}$

$Given$ :- Radius of curve $\left(R\right)$ = 30m

Coefficient of friction ( $\mu$ ) = 0.75

Acceleration due to gravity $\left(g\right)$ = 10 ${m/s}^2$

$ToFind$ :- Maximum speed of car on road turn

$Solution$ :- We know , $F_{Centripetal}$ = $\frac{m{v}^2}{R}$

Now , According to diagram , $f$ and $F$ are in same direction

$\therefore$ For the care not to slip in opp. direction of friction

$v_{max}\to f_{max}$

$⟹$ $f_{max}$ = $F_{Centripetal}$ = $\frac{m{v}^2}{R}$

We know , $f_{max}$ = $\mu N$

$\frac{m{v}^2}{R}$ = $\mu mg$ $(\because N=mg)$

$v_{max}$ = $\sqrt{\mu gR}$

$v_{max}$ = $\sqrt{75\times 3}$ $⟹$ $\underset{–}{15m/s}$

Hence , $\underset{–}{OptionB\left(15m/s\right)iscorrect.}$