Question

A bob of mass m is suspended by a light string of length L. It is imparted a horizontal velocity vo at the lowest point. IT COMPLETES A SEMI CIRCULAR TRAJECTORY IN VERTICAL PLANE..... REFER TO ATTACHMENT.

Answer 1

$\huge{\underline{\mathfrak{\red{Answer}}}}$

There are two external forces on the bob: Gravity and teñsion.

The teñsion does no work as displacement is always perpendicular to stŕing.

Total mechànical energy of the system is consèrved

Energy = E

Gravity = mg

Tēnsion = T

Taking potențial energy of the system to be at zēro at loweśt point A, then

At A,

${\boxed{\blue{E= \frac{1}{2} m {Vo}^{2}----i}}}$

From Newton's second law,

TA(tenšion at A) - mg

${\boxed{\red{ = \frac{ {mVo}^{2}}}{L}}} -----ii$

At highest point C, the string slaćks as tension at C becomes zero.

C,

${\boxed{\green{ E= \frac{1}{2} {mVc}^{2} + mg(2L)----iii}}}$

Let Vc be velocity at C then

From Newtons second law,

${\boxed{\pink{mg = \frac{ {mVc}^{2} }{L} ----iv}}}$

from iv we get

$Vc = \sqrt{gL}$

From iii

${\boxed{E = \frac{1}{2} m(gL) + 2mgL \\ = \frac{5}{2} mgL}}$

Using I,

${\bold{\red{ \frac{1}{2} {mVo}^{2} = \frac{5}{2} mgL}}}$

or

${\boxed{\bold{\red{Vo = \sqrt{5gL}}}}}$

AT B,

$E = \frac{1}{2} {mVв}^{2} + mg$

or

$\frac{1}{2} {mV}^{2} = E - mg(L) \\ = \frac{5}{2} mgL - mgL \: \: - - - - - using \: v$

Thus

${\boxed{\pink{\frac{1}{2} {mVв}^{2} = \frac{3}{2} mgL》 Vb = \sqrt{3gl}}}}$

The ratio of kinetic energy at B and C is

${\boxed{\blue{\frac{Kb}{Kc} = \frac{ \frac{1}{2} {mVb}^{2} }{ \frac{1}{2} {mVc}^{2} } = \frac{3gL}{gL}= \frac{3}{1}}}}$

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Now checking the options we get

$\mathfrak{\boxed{\red{Answer\:is\:Option\: C}}}$