Question

$\huge{\underline{\underline{\red{\mathfrak{Question :}}}}}$
Why does a cyclist inwards while negotiating a curve ? Explain with Diagram .

Obtain an expression for angle which a cyclist will have to make with verticle ( for safe negotiation) , while taking circular turn.
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Answer 1

Answer:

A cyclist bends inwards on a banked road so as to take a component of Reaction force as centripetal force which would help in it's turning at fast velocities.

Proof:

First look at the diagram to understand better.

Let Reaction force be R , θ is the angle of banked road, r is the radius of circle and mass is m.

If we take equilibrium of the various forces acting on the cycle.

R sin(θ) = mv²/r ...........eq(1)

R cos(θ) = mg .........eq(2)

Now eq(1) ÷ eq(2) :

tan(θ) = v²/rg

Answer 2

$\huge\bold\green{ANSWER}$

When riding on a curved road, there is a force that keeps the cyclist on the path which is Centripetal force.This force is provided by the friction on the tires.

However since the cyclist is cycling at a high speed, an extra force is needed to keep the cyclist on the path. This force is provided by the horizontal component of the normal force on the cyclist.

This normal force is the force countering gravity. By bending, the direction of the normal force tilts resulting in a vertical as well as horizontal component of the force.

$\huge\sf{Expression}$

$\sf{r \: \sin (\theta) = \frac{m {v}^{2} }{r} }$

_____________________(1)

$\sf{r \: \cos( \theta) = mg }$

_____________________(2)

Dividing eqn. (1) by eqn. (2) , we get

$\bold{\frac{r \: \sin( \theta) }{r \: \cos( \theta) } = \frac{ \frac{m {v}^{2} }{r} }{mg} }$

$\bold{ \tan( \theta) = \frac{ {v}^{2} }{rg} }$

$\huge\bold \green{ \theta = {tan}^{ - 1}( \frac{ {v}^{2} }{rg}) }$