Question

A car with speed v and an identical car with speed 2v both travel the same circular section of an unbanked road. If the frictional force required to keep the faster car on the road without skidding is F, then the frictional force required to keep the slower car on the road without skidding is __-

Answer 1

Answer:

$F'=\dfrac{F}{4}$

Explanation:

Let m is the mass of both cars. The first car is moving with speed v and the other car is moving with speed 2v. The only force acting on both cars is the centripetal force.

For faster car on the road,

$F=\dfrac{mv^2}{r}$

v = 2v

$F=\dfrac{m(2v)^2}{r}$

$F=4\dfrac{m(v)^2}{r}$..........(1)

For the slower car on the road,

$F'=\dfrac{mv^2}{r}$............(2)

Equation (1) becomes,

$F=4F'$

$F'=\dfrac{F}{4}$

So, the frictional force required to keep the slower car on the road without skidding is one fourth of the faster car.

Answer 2

The frictional force required to keep the slower car on the road without skidding is $\dfrac{F}{4}$ .

Given data:

The speed of car 1 is, v.

The speed of car 2 is, 2v.

Let m be the masses of both cars. Then, frictional force acting on each car is due to the centripetal force, to avoid skidding. Then for car 1,

$F_{1}=\dfrac{mv^{2}}{r} ............................................................(1)$

Here, r is the radius of curved path.

And, friction force for car 2 is,

$F_{2}=\dfrac{m(2v)^{2}}{r} \\F_{2}=\dfrac{4mv^{2}}{r}\\F=\dfrac{4mv^{2}}{r}............................................................(2)$

Taking ratio of equation (1) and (2) as,

$\dfrac{F_{1}}{F}=\dfrac{\dfrac{mv^{2}}{r} }{\dfrac{4mv^{2}}{r}} \\\dfrac{F_{1}}{F}=\dfrac{1}{4} \\{F_{1}=\dfrac{F}{4}$

Thus, the frictional force on the slower car is one-fourth times the frictional force on faster car.

Learn more about centripetal force here:

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