The coefficient of friction between the tyres and the road is 0.25. The maximum speed with which car can be driven round a curve of radius 40 m without skidding is (assume g = 10 ms- 2)

Question

The coefficient of friction between the tyres and the road is 0.25. The maximum speed with which car can be driven round a curve of radius 40 m without skidding is (assume g = 10 ms- 2) (A) 40 ms-1 (B) 20 ms-1 (C) 15 ms-1 (D) 10 ms-1

Tardigrade Admin · Accepted Answer

For turning around a curve, centripetal force is provided by friction, so

\frac{m ν ^{2}}{r} < f

\frac{m ν ^{2}}{r} < μ m g

\Rightarrow

ν < μg r

=

ν_{ma x} = μg r

Substituting ,

μ

= 0.25,r = 40 m g = 10

m s^{- 2}

, we get

v_{ma x} = 10 m s^{- 1}

Q. The coefficient of friction between the tyres and the road is 0.25. The maximum speed with which car can be driven round a curve of radius 40 m without skidding is (assume g = 10 $m s^{- 2}$ )

40 $m s^{- 1}$

20 $m s^{- 1}$

15 $m s^{- 1}$

10 $m s^{- 1}$

For turning around a curve, centripetal force is provided by friction, so
$\frac{m ν ^{2}}{r} < f$
$\frac{m ν ^{2}}{r} < μ m g$
$\Rightarrow$ $ν < μg r$
= $ν_{ma x} = μg r$
Substituting , $μ$ = 0.25,r = 40 m g = 10 $m s^{- 2}$ , we get
$v_{ma x} = 10 m s^{- 1}$

Q. The coefficient of friction between the tyres and the road is 0.25. The maximum speed with which car can be driven round a curve of radius 40 m without skidding is (assume g = 10 ms−2)

40 ms−1

20 ms−1

15 ms−1

10 ms−1