Question

Derive the expression for normal component of acceleration as an=(v2/p) While the particle is moving around curved path of a radius r and v is the velocity in m/s.

Answer 1

To derive:

Expression for normal component of acceleration when a particle is moving around a curved path of radius r and velocity v.

Derivation:

From the diagram , we can say ;

$\rm{\Delta ABC \sim \Delta PQR}$

Using properties of similar triangle, we can say;

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

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

Dividing both sides by ∆t ;

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

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

$\rm{ = > \: \dfrac{ \Delta v }{\Delta t} = \dfrac{v}{r} \times v }$

$\rm{ = > \: \dfrac{ \Delta v }{\Delta t} = \dfrac{ {v}^{2} }{r} }$

$\boxed{\rm{ = > \: a_{c} = \dfrac{ {v}^{2} }{r} }}$

[Hence Derived]