A turn of radius 20 m is banked for the vehicles going at a speed of 36km/h. If the coefficient of static friction between the road and the tyre is 0.4, what are the possible speeds of a vehicle so that it neither slips down nor skids up?

Options

A. between `10 m//s` and `15 m//s`
B. between `10 m//s` and `15 sqrt (1/6) m//s`
C. between `10 sqrt (1/6) m//s` and `15 m//s`
D. None of these

Answer

C

Explanation

Given `v=36 km/hr`
`=10 m/s`
`r=20m, mu=0.4`
`The road is banked with angle
`theta=tan^-1 v^2/(rg)`
`=tan^-1 (100/(20xx10-)`
`=tan^-1 (1/2)`
or `tantheta=0.5`

when the car travels at a maximum speed, so that to slips upward `muR_1` acts downward As shown it ure.
`R_1-mgcostheta=(mv_1^2)/rsintheta=0`........i
and `muR_1+vmgsintheta-(mv_1^2)/rcostheta=0`.....ii
Solving equations we get,
`v_1=sqrt(rg(mu+tantheta)/(1-mutantheta))`
=sqrt(20xx10xx0.9/0.8`
`=15 m/s=54 km/hr

Similarly it fcan be proved that
Similarly it can be proved that
`v_2=sqrt(rg(tantheta-mu)/(sqrt(1-mutantheta)))`
`=sqrt(20xx10xx0.1/1.2`
`=4.08m/s=14.7km/hr`
So the possible speeds asre between 14.7 km/hr and 54 km/hr.