Question

A car is traveling in a uniform circular motion on a section of road whose radius is r. The road is slippery, and the car is just on the verge of sliding.a.) If the car's speed were doubled, what would have to be the smalles radius in order that the car does not slid? Express your anwer in terms of r.b.) What would be your answer to part (a) if the car were replaced by one that weighed twice as much, the cars speed still being doubled?

Answer 1

Answer:

a)r' = 4 r

b)$V=\sqrt{\mu rg}$  

Explanation:

Given that car is moving with uniform speed.

At the verge of sliding

Friction force = Force due to circular motion

$\mu mg=\dfrac{mV^2}{r}$

So we can say that

$V=\sqrt{\mu rg}$          -----1

Where is the coefficient of friction

r is the radius of circular path

V is the velocity of car

a)

if the car speed become double

It means that new speed of car =2V

lets tale new radius of circular path is r'

$2V=\sqrt{\mu r'g}$                ---------2

From equation 12 and 2 we cay that

r' = 4 r

It means that we have to increase radius 4 times to avoid sliding

b)

$V=\sqrt{\mu rg}$  

In the above expression there is no any terms of mass ,it means that sliding speed does not depends on the weight of car.

So the sliding speed will remain same.

$V=\sqrt{\mu rg}$