Question

Write the formula of maximum velocity to avoid skidding a vehicle moving on a level circular path.

Answer 1

Step 1: Gravitational attraction force($mg$)

Frictional force ($f$) – acting parallel to the slope of the road between the car’s tire and the road. The normal force (N) is a response force caused by the gravitational force imposed by the road on the vehicle.

Centripetal force ($\frac{m{v}^2}{r}$) – acting in the direction of the car’s circular path’s center.

Step 2: Derivation for safe speed

Let’s take the inclination angle $\theta$ of the road with respect to the x-axis

Now we can balance all the x-axis and y-axis factors of forces.

Balancing all the forces acting along the x-axis-

$f\cos \left(\theta \right)+N\sin \left(\theta \right)=\frac{m{v}^2}{r}$

Substituting the value $f=\mu N$ into the equation,

we get,

$\mu N\cos \left(\theta \right)+N\sin \left(\theta \right)=\frac{m{v}^2}{r}$

$N(\mu \cos \theta +\sin \theta )=\frac{m{v}^2}{r}\dots \dots \left(1\right)$

Balancing all the forces acting along the y-axis- $N\cos \theta =f\sin \theta +mg$

$N\cos \theta =\mu N\sin \theta +mg$

$N(\cos \theta -\mu \sin \theta )=mg\dots \dots ..\left(2\right)$

Frictional force- $f=\mu N$

Now we will use these three equations to find the value of speed ($v$).

Now we can divide equations ($1$) and ($2$),

and we get

$\frac{v^2}{rg}=\frac{(\mu +\tan \theta )}{(1-\mu \tan \theta )}$

$v=\sqrt{\frac{rg(\mu +\tan \theta )}{(1-\mu \tan \theta )}}$

This is an expression for the maximum speed of a car in a circular motion.