Question

state and expression for minimum speed required for particle to complete vertical circular motion at a) Lowest position b) Top most position ​

Answer 1

Answer:

When String is Horizontal (h = r): This is the minimum velocity when the string is horizontal at the point M of the vertical circle required for a body looping a loop. If the velocity 'v' of the body is such that, v ≤ √2gr, then the body oscillates about point L, the lowest point of the vertical circle.

Answer 2

To derive:

Expression for minimum speed required for particle to complete vertical circular motion at a) Lowest position b) Top most position .

Solution:

Let us assume that the velocity of the ball at the lowest position be u and the radius of the vertical circle be r :

Now , at highest point , at critical conditions , the velocity should be such that the string becomes loose.

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

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

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

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

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

So, the minimum velocity of the particle at the highest point of the vertical circle should be √(gr).

Now, we will apply the principle of CONSERVATION OF MECHANICAL ENERGY to find out the minimum velocity at the lowest point:

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

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

$\implies \: m {v}^{2} + 4mgr = m {u}^{2}$

$\implies \: m {( \sqrt{gr}) }^{2} + 4mgr = m {u}^{2}$

$\implies \: mgr + 4mgr = m {u}^{2}$

$\implies \: 5mgr = m {u}^{2}$

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

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

So, minimum velocity of the particle at the lowest point of the vertical circle should be √(5gr).