Question

Discuss motion of a body in vertical circle & find minimum velocity required at lowest point for looping of loop​

Answer 1

Answer:

Given:

A particle is performing vertical circle motion.

To find:

Minimum velocity at lowest point to complete the circle.

Concept:

First of all the minimum velocity at highest point has to √(gl) to complete the circular loop.

Calculation:

We will now apply Conservation of Mechanical energy :

Let velocity at lowest point be v and length of rope be l

$\therefore \: ke1 + pe1 = ke2 + pe2$

$= > \frac{1}{2} m ( { \sqrt{gl})}^{2} + mg(2l) = \frac{1}{2} m {v}^{2} + 0$

$= > \frac{mgl}{2} + 2mgl = \frac{1}{2} m {v}^{2}$

$= > {v}^{2} = 5gl$

$= > v = \sqrt{5gl}$

So final answer :

$\boxed{ \red{v = \sqrt{5gl}}}$

Answer 2

$\huge{\underline{\underline{\red{\mathfrak{AnSwEr :}}}}}$

$\small{\underline{\blue{\sf{Given :}}}}$

• A body is in verticle circle motion.

$\rule{200}{1}$

$\small{\underline{\green{\sf{Solution :}}}}$

As we know that minimum velocity is √gl

$\footnotesize \large \bigstar {\boxed{\sf{Energy \: at \: Highest \: Point \: = \: Energy \: at \: lowest \: point}}} \\ \\ \implies {\sf{K.E_1 \: + \: P.E_1 \: = \: K.E_2 \: + \: P.E_2}} \\ \\ \implies {\sf{\dfrac{1}{2}mv_1 ^2 \: + \: mgh \: = \: \dfrac{1}{2} mv_2 ^2 \: + \: mgh}}$

At highest point

• v1 = √lg
• h = 2l

At Lowest point

• v2 = v
• h = 0

$\rule{200}{2}$

_________________[Put Values]

$\implies {\sf{\dfrac{1}{2} m(\sqrt{gl})^2 \: + \: mg(2l) \: = \: \dfrac{1}{2} mv^2 \: + \: mg(0)}} \\ \\ \implies {\sf{\dfrac{1}{2}mgl \: + \: 2mgl \: = \: \dfrac{1}{2} mv^2 \: + \: 0}} \\ \\ \implies {\sf{\dfrac{mgl \: + \: 4mgl}{2} \: = \: \dfrac{1}{2} m v^2}} \\ \\ \implies {\sf{\dfrac{5 \cancel{m}gl}{\cancel{2}} \: = \: \dfrac{1}{\cancel{2}} \cancel{m}v^2}} \\ \\ \implies {\sf{v^2 \: = \: 5gl}} \\ \\ \implies {\sf{v \: = \: \sqrt{5gl}}} \\ \\ \large {\boxed{\sf{Velocity \: at \: Lowest \: Point \: = \: \sqrt{5gl}}}}$