Question

Derive an expression for the linear velocity of particle at the lowermost position in vcm.
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Answer 1

Answer:

$v = \sqrt{2gh}$  (or)   $v = \sqrt{2gh} = \sqrt{2gR(1-cos(x)}$  

Explanation:

VCM, or Vertical Circular Motion is an important topic to understand to get your fundamentals clear.

In VCM, you have to use conservation of energy to find velocity, and centripetal force to calculate tension. Here's how:

If we assume that the particle starts from an height 'h':

$mgh = 1/2mv^2$

As initially total energy is P.E, and finally it all becomes K.E.

[NOTE: Energy (or) Work doesn't depend upon whether the path was circular, or a straight drop, or looped motion. It only depends upon the initial and final energy, so we can use this formula. ]

Simplifying it, we get:  $v = \sqrt{2gh}$

If we assume that the particle is the tip of a pendulum and we have to find velocity as a function of the radius of the pendulum, using simple trigonometry, we find the relation between 'h' and 'R':

h = R - Rcosθ = R(1 - cosθ), where θ is the initial angle made by the pendulum string with the y - axis, or the vertical line.

So,       $v = \sqrt{2gh} = \sqrt{2gR(1-cos(x)}$               [θ is denoted as 'x']

Hope it helps!