Question

A turn of radius 25 m is banked for the vehicle going at speed of 60km/hr. if the coefficient of static friction between the road and the tyre is 0.8. what are the possible speeds of a vehicle so that it neither slips down nor skids up?

Note:-Complete explaination required​

Answer 1

Answer:

Explanation:

When vihecle going in speed at a circular path, a centrifugal force applied at car to cause it slip but this cause is opposed by a static frictional force but static frictional force has a limit beyond that the car will slide or slip. To avoid sliding we make the roads banked.

On the banked road we will have two conditions

1.If the speed is below a certain speed ,vahicle will slide downside.

2.If the speed is still to gigh for a banked roard, it will slide  or slip towards out direction.

So what is the limit of speed for a banked road!

Speed v=60km/hr=60000/3600=16.67m/s

radius=25 m

static frictionμ=0.8

Bend angle of road θ=$tan^{-1}(\frac{v^{2} }{r*g}) =tan^{-1}(1.13)$

condition for max and min speed

$V(max)=\sqrt[]{\frac{rg(sin\theta+\mu cos\theta)}{(cos\theta-\mu sin\theta)} } =\sqrt[]{\frac{rg(tan\theta+\mu )}{(1-\mu tan\theta)} } \\and\\V(min)=\sqrt[]{\frac{rg(sin\theta-\mu cos\theta)}{(cos\theta+\mu sin\theta)} }=\sqrt[]{\frac{rg(tan\theta-\mu )}{(1+\mu tan\theta)} }$

hence

v(max)=70.18 m/s or 70.18x18/5=252km/hr

v(min)=6.5m/s=23.45km/hr