Question

A car is traveling in uniform circular motion on a section of flat roadway whose radius is . The road is slippery, and the car is just on the verge of sliding off the roadway. If the car’s speed were now doubled, what would be the smallest radius of the curve it could drive on without slipping?

Answer 1

The smallest radius will be four (4) times the initial radius

Step-by-step explanation:

The car maintains a constant angular speed. According to Newton's Second Law F = m a

1.
`F_(r)=m*A_(n)`

2.
`A_(n)=(v^2)/(R_(p))`

Replacing 2 in 1

3.
`F_(r)=m*(v^2)/(R_(p))`

Where:

Fr= Frictional force

Rp= Initial Radius

An= Centripetal Acceleration

M= Mass

V= Velocity

Also we have that:

4.
`F_(r)=\mu *W=\mu*m*g`

μ= Coefficient of friction between the car and the surface

M= Mass

W= Weight

G= Gravity

r is cleared from equation 3

5.
`R_(p)=m*(v^2)/(F_(r))`

Replacing 4 in 5

6.
`R_(p)=m*(v^2)/(\mu*m*g)`

Simplifying

7.
`R_(p)=(v^2)/(\mu*g)`

Now we have a new velocity equal to twice the initial velocity, We replace it by 2v in equation 7

8.
`R_(n)=((2v)^2)/(\mu*g)`

Computing

9.
`R_(n)=(4v^2)/(\mu*g)`

Replacing 5 in 9

`R_(n)=4*R_(p)`