Question

OR
Derive expression for the maximum secure speed of a vehicle in a circular path.
17
find the cynression of centrifugal acceleration for a uniform circular motion​

Answer 1

To derive:

Maximum secure speed of a vehicle in circular path .

Derivation:

Let's assume that :

• Mass of vehicle is m.
• Max secure velocity is v.
• Radius of circular track is r.
• Coefficient of friction between the road and the tyres is $\mu$.
• Gravitational acceleration be g.

So, the basic concept is this:

• The speed of the vehicle will be such that the frictional force acting radially inward will be able to cancel out the centrifugal force acting radially outwards.

$\rm \therefore \: F_{c} = f$

$\rm \implies\: \dfrac{m {v}^{2} }{r} = \mu(normal \: reaction)$

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

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

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

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

So, max safe velocity will be :

$\boxed{ \bold{\: v = \sqrt{\mu rg}}}$

Expression for Centrifugal Acceleration:

Since the triangles are similar , we can say that:

$\rm \therefore \: \dfrac{\Delta v}{v} = \dfrac{\Delta s}{r}$

Now , centrifugal acceleration:

$\rm\therefore \: a = \dfrac{\Delta v}{\Delta t}$

$\rm\implies \: a = \dfrac{( \frac{v}{r} \times \Delta s)}{\Delta t}$

$\rm\implies \: a = \dfrac{v}{r} \times \dfrac{ \Delta s}{\Delta t}$

$\rm\implies \: a = \dfrac{v}{r} \times v$

$\rm\implies \: a = \dfrac{ {v}^{2} }{r}$

So, the final expression:

$\boxed{ \bold{ \: a = \dfrac{ {v}^{2} }{r}}}$