# A srone 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 in the lowest position , and has a speed u. The magnitude of the change in its velocity as it reaches a position where the sting is horizontal is : a) root (u^2 - 2gL) b) root (2gL) c) root (u^2 - gL) d) root (2(u^2 - gL)) please help by answering this Q and the method to do this Question

A srone 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 in the lowest position , and has a speed u. The magnitude of the change in its velocity as it reaches a position where the sting is horizontal is :

a) root (u^2 - 2gL)

b) root (2gL)

c) root (u^2 - gL)

d) root (2(u^2 - gL))

please help by answering this Q and the method to do this Question

## 7 Answers

applying energy conservation

1/2mu^2=mgl+1/2mv^2

we get v=root(u^2-2gl)

in vector form we can express as v vector=root(u^2-2gl)j and u vector =ui

difference is v-u =root(u^2+v^2)=root(2(u^2-gl))

^{2}-1/2mv

^{2}

^{2}/2 - v

^{2}/2

^{2}-2gl

__whole uder-root__( Vsquare + U square + 2 * V * U * Cos90 )

__whole under-root__(Vsquare+Usquare +0 ) (as cos90 = 0)

__whole under-root (__Vsquare+Usquare)

__(__Vsquare+Usquare) ,i.e.,

Please find below the solution to your problem.

applying energy conservation

1/2mu^2=mgl+1/2mv^2

we get v=root(u^2-2gl)

in vector form we can express as v vector

=root(u^2-2gl)j and

u vector =ui

difference is v-u

= root(u^2+v^2)

= root(2(u^2-gl))

Thanks and Regards