Question

Derive and expression for maximum possible speed for vehicle to move horizontal unbanked road?​

Answer 1

Consider a vehicle of mass m moving with a speed v along the horizontal unbanked road of radius r

While taking a turn the vehicle performs circular motion

centripetal force is given by,

$F cp = mv^2 / r$

frictional force between the tyres and the road is, $F s = μN$

μ is the coefficient of friction between the road and the tyre

$N = mg$

therefore,$F s = μmg$

$F cp = F s$

$mv ^2 / r = μmg$

$v^2 = μrg$

$v = √μrg$

this is the maximum speed of the vehicle to move horizontal unbanked road.

 

Answer 2

Answer:

Consider a vehicle of mass m moving with a speed v along the horizontal unbanked road of radius r

While taking a turn the vehicle performs circular motion

centripetal force is given by,

F cp = mv^2 / rFcp=mv

2

/r

frictional force between the tyres and the road is, F s = μNFs=μN

μ is the coefficient of friction between the road and the tyre

N = mgN=mg

therefore,F s = μmgFs=μmg

F cp = F sFcp=Fs

mv ^2 / r = μmgmv

2

/r=μmg

v^2 = μrgv

2

=μrg

v = √μrgv=√μrg

this is the maximum speed of the vehicle to move horizontal unbanked road.

Explanation:

Consider a vehicle of mass m moving with a speed v along the horizontal unbanked road of radius r

While taking a turn the vehicle performs circular motion

centripetal force is given by,

F cp = mv^2 / rFcp=mv

2

/r

frictional force between the tyres and the road is, F s = μNFs=μN

μ is the coefficient of friction between the road and the tyre

N = mgN=mg

therefore,F s = μmgFs=μmg

F cp = F sFcp=Fs

mv ^2 / r = μmgmv

2

/r=μmg

v^2 = μrgv

2

=μrg

v = √μrgv=√μrg

this is the maximum speed of the vehicle to move horizontal unbanked road.