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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

10 years ago

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))

3 years ago

2 years ago

From enegy conservation we kniw that change in potential enegy is equal to that of change in kinetic energy i.e. energy can neither be created bor be destroyed

Δu=ΔK

mgl=1/2mu^{2} -1/2mv^{2}

gl= u^{2}/2 - v^{2}/2

v=√u^{2}-2gl

2 years ago

The portion of Velocity change –

change in velocity = v jcap – u icap

and the magnitude of change in velocity =>

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

=> __whole under-root (__Vsquare+Usquare)

So, we have to find the value of __(__Vsquare+Usquare) ,i.e.,

whole under-root [2*{U square – gl }] (ans.)

2 years ago

A stone tied to a string of lenght 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 its lowest position and has a speed miu. the magnitude of the change in its velocity as it reaches a position where the string is horizontal is:

11 months ago

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

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