Question

Derive an expression for maximum speed of a car negotiating safe circular turn on a level circular road​

Answer 1

Answer:

Hope u got it ♥ _________

Answer 2

To derive:

An expression for maximum speed of a car negotiating safe circular turn on a level circular road.

Solution:

First of all , we need to following forces that act on a body while an object is making a turn on a circular road :

• mg (weight) acts downwards.
• N (normal reaction) acts upwards.
• mv²/r (centripetal force) acts radially inwards.

Now, let's assume that the coefficient of friction between the tyres and the surface of the road is $\mu$.

Now, at critical conditions , the centripetal force will be equal and opposite to the frictional force provided by the road. This will prevent slipping of the tyres.

$\sf \therefore \: \dfrac{m {v}^{2} }{r} = f$

$\sf \implies \: \dfrac{m {v}^{2} }{r} = \mu(N)$

$\sf \implies \: \dfrac{m {v}^{2} }{r} = \mu(mg)$

$\sf \implies \: \dfrac{ {v}^{2} }{r} = \mu(g)$

$\sf \implies \: {v}^{2} = \mu rg$

$\sf \implies \: v = \sqrt{ \mu rg}$

So, the maximum allowable speed during the turn will be $\sf v = \sqrt{ \mu rg}$ such that the car can turn safely.

$\star$ Hope It Helps.