Question

Derive an expression for the linear velocity of particle at the lowermost position in V.C.M.​

Answer 1

To find:

The linear velocity of particle at the lowermost position in V.C.M. ?

Calculation:

First , let's calculate the minimum velocity at highest point :

$\therefore \: mg + T = \dfrac{m {u}^{2} }{r}$

$\implies\: mg + 0= \dfrac{m {u}^{2} }{r}$

$\implies\: mg = \dfrac{m {u}^{2} }{r}$

$\implies \: u = \sqrt{gr}$

Now, applying CONSERVATION OF MECHANICAL ENERGY:

• Let velocity at lowest point be v:

$\therefore \: KE1 + PE1 = KE2 + PE2$

$\implies \: \dfrac{1}{2} m {v}^{2} + 0 = \dfrac{1}{2} m {u}^{2} + mg(2r)$

$\implies \: \dfrac{1}{2} m {v}^{2} = \dfrac{1}{2} m {u}^{2} + mg(2r)$

$\implies \: \dfrac{1}{2} m {v}^{2} = \dfrac{1}{2} m {( \sqrt{gr}) }^{2} + mg(2r)$

$\implies \: \dfrac{1}{2} m {v}^{2} = \dfrac{1}{2} m(gr) + mg(2r)$

$\implies \: \dfrac{1}{2} {v}^{2} = \dfrac{1}{2} (gr) + g(2r)$

$\implies \: {v}^{2} = (gr) + 4gr$

$\implies \: {v}^{2} = 5gr$

$\implies \: v = \sqrt{5gr}$

So, minimum velocity at lowest point is √(5gr).