A stone tied to a string of length `L` is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time the stone is at lowest position and has a speed `u` . Find the magnitude of the change in its velocity as it reaches a position, where the string is horizontal.

Options

A. `sqrt(2(u^(2) - gl))`
B. `sqrt(u^(2) - gl)`
C. `u - sqrt(u^(2) - 2gl)`
D. `sqrt(2 gl)`

Answer

A

Explanation

Using conservation of mechanical energy at initial and final position.

`Delta k + Delta U = 0`
`((1)/(2) mu'^(2) - (1)/(2) mu^(2)) + (mgl) = 0`
or `u'^(2) = u^(2) - 2 gl`
or `u' = sqrt(u^(2) - 2gl)` (i)
So, the magnitude of change in velocity
`Delta vecu = vecu_(f) - vecu_(i) = vecu_(f) + (vecu_(i))`

`|Delta vecu| = sqrt(u'^(2) + u^(2)) = sqrt((u^(2) - 2 gh) + u^(2))`
`= sqrt(2(u^(2) - gl))`.