Question

A particle starts from rest on a horizontal semi circular path at point 'A' and goes to point with uniform rate of change of speed. The ratio of
centripetal and tangential acceleration at the half way point 'Bis
​

Answer 1

Given:

A particle starts from rest on a horizontal semi circular path at point 'A' and goes to point with uniform rate of change of speed.

To find:

The ratio of centripetal and tangential acceleration at the half way point B is ?

Calculation:

Let tangential acceleration be a_(t) and centripetal acceleration be a_(c). Let the value of tangential acceleration be "a".

Now , applying 3rd equation of kinematics:

$\rm\therefore \: {v}^{2} = {u}^{2} + 2(a_{t})s$

$\rm\implies \: {v}^{2} = {(0)}^{2} + 2(a)(\pi r)$

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

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

$\rm\implies \: a_{c} = 2a\pi$

Therefore required ratio:

$\rm \therefore \: a_{c} : a_{t} = 2a\pi : a$

$\rm \implies\: a_{c} : a_{t} = 2\pi : 1$

So, the final answer is:

$\boxed{ \bold{\: a_{c} : a_{t} = 2\pi : 1}}$