Name: Centripetal and Centrifugal Force
Uploaded: 2019-07-19
Duration: 2 min 42 s
Description: The meaning of centrifugal force is the apparent force that is felt by an object moving in a curved path that acts outwardly away from the centre of rotation

Question

A turn of radius

20 m

is banked for the vehicles going at a speed of

36 km h^{- 1}

. If the coefficient of static friction between the road and the tyre is

0.4

, what are the possible speeds of a vehicle so that it neither slips down nor skids up?

Accepted Answer

Given,

radius, $R = 20 m$ ,

ideal velocity of vehicle, $v = 36 km h^{- 1} = 36 \times \frac{5}{18} m s^{- 1} = 10 m s^{- 1}$ ,

coefficient of static friction, $μ = 0.4$ .

Angle of banking for ideal speed is given by,

$\tan θ = \frac{v^{2}}{r g} = \frac{(10)^{2}}{20 \times 10}$

$\Rightarrow \tan θ = 0.5$ .

Now, two cases are possible according to the question.

Case $1$ : Minimum speed

In this case, the vehicle will have a tendency to move down the incline. So, limiting friction will act up the incline. Limiting friction is given by,

$f = μ N$ .

Now, we have to resolve forces into radial and vertical components.

Along vertical:

$N \cos θ + μ N \sin θ = m g \dots (1)$

$N \sin θ - μ N \cos θ = \frac{m v_{\min}^{2}}{R} \dots (2)$

Dividing equation $(2)$ by $(1)$ , we get,

$\frac{\sin θ - μ \cos θ}{\cos θ + μ \sin θ} = \frac{v_{\min}^{2}}{R g}$ ,

$\Rightarrow v_{\min} = \sqrt{R g (\frac{\tan θ - μ}{1 + μ \tan θ})}$

$\Rightarrow v_{\min} = \sqrt{20 \times 10 \times (\frac{0.5 - 0.4}{1 + 0.4 \times 0.5})} = 4.08 m s^{- 1}$

$\Rightarrow v_{\min} = 14.68 km h^{- 1}$ .

Case $2$ : Maximum speed

In this case, the vehicle will have a tendency to move up the incline, so the limiting friction will act down the incline. Limiting friction is given by,

$f = μ N$ .

Now, we have to resolve forces into radial and vertical components.

Along vertical:

$N \cos θ - μ N \sin θ = m g \dots (3)$

$N \sin θ + μ N \cos θ = \frac{m v_{\min}^{2}}{R} \dots (4)$

Dividing equation $(2)$ by $(1)$ , we get,

$\frac{\sin θ + μ \cos θ}{\cos θ - μ \sin θ} = \frac{v_{\max}^{2}}{R g}$ ,

$\Rightarrow v_{\max} = \sqrt{R g (\frac{\tan θ + μ}{1 - μ \tan θ})}$

$\Rightarrow v_{\max} = \sqrt{20 \times 10 \times (\frac{0.5 + 0.4}{1 - 0.4 \times 0.5})} = 15 m s^{- 1}$

$\Rightarrow v_{\max} = 54 km h^{- 1}$ .

A turn of radius $20 m$ is banked for the vehicles going at a speed of $36 km h^{- 1}$ . If the coefficient of static friction between the road and the tyre is $0.4$ , what are the possible speeds of a vehicle so that it neither slips down nor skids up?

Important Questions on Circular Motion

Learn and Prepare for any exam you want !

A turn of radius 20 m is banked for the vehicles going at a speed of 36 km h-1. If the coefficient of static friction between the road and the tyre is 0.4, what are the possible speeds of a vehicle so that it neither slips down nor skids up?

Important Questions on Circular Motion

Learn and Prepare for any exam you want !

A turn of radius $20 m$ is banked for the vehicles going at a speed of $36 km h^{- 1}$ . If the coefficient of static friction between the road and the tyre is $0.4$ , what are the possible speeds of a vehicle so that it neither slips down nor skids up?