Answer to Question #191386 in Mechanics | Relativity for Madiha

Question

A car is traveling in uniform circular motion on a section of road whose radius is r. If the road is slippery and the car is just on the verge of sliding. If the car's speed is doubled, what would be the smallest radius to stop the car from sliding?

Answer 1

Concept : turning of car on horizontal rough surface

Given : car is moving on circular horizontal rough track of radius r with speed v.

We need to find out the minimum radius of horizontal rough circular track if car's speed become double.

So

for motion of any body in circle, there will be a necessary force which acts towards centre and called centripetal force$=\frac{Mv^2}{r}$ .

here it is provided by friction force on the car.

when car is on the just verge of slipping, It becomes maximum or limiting friction.

$$f_m = \mu_s N$$

Where N is normal force between car and surface which is equal to weight of car. $(Mg)$ and $\mu_s$ depends nature of contact surfaces. (here it is constant)

$\mu_s Mg = \frac{Mv^2}{r}$

$\implies v^2 = \mu_sgr$

or $v^2$ is proportional to $r$.

So if car's speed is doubled$(2v)$ then minimum radius of circular path gets $4$ times to prevent the car from sliding.

Answer : $4r$