Ring-Race

Sorry for not marking $\alpha$ = $\angle$ DEC = $\angle$ CAB

Since they are rods( inelastic) , this would be the condition (thread becomes horizontal)

A to ceiling =' l ' , AB = a , BC = b , AC = c , CD = d , DE = e , CE = f

$a ^ {2}$ + $b ^ {2}$ = $c ^ {2}$

Diffrentiating with respect to time .

$\frac { dc }{ dt }$ = vcos $\alpha$ ( as b = const . )

Similarly , $d ^ {2}$ + $e ^ {2}$ = $f ^ {2}$

$\frac { de }{ dt }$ = sec $\alpha$ $\frac { df }{ dt }$

Since length of rope is constant -

l + c+ f = l + a + b + d

c+f = a + b + d

Diff . with respect to time -

$\frac { df }{ dt }$ = v( 1 - cos $\alpha$ )

$\frac { de }{ dt }$ = sec $\alpha$ $\frac { df }{ dt }$ = sec $\alpha$ v( 1 - cos $\alpha$ )

$\frac { de }{ dt }$ = Ans = 2 v sin^2 $\alpha /2$ / cos $\alpha$

A simple question of relative motion. Solving using wrt v1

I am bad at latex. So let cos(alpha)=cos(d)

v21= v2 w.r.t to v1 along the rope =v2-v1 ----(1)

Also using constraint we can see that

v1=v21 cos(d)

v21=v1/cos(d)------------(2)

Put (2) in (1)

v1/cos(d) = v2-v1

v2=v1(1+cos(d))/cos(d) =v1*2sin^2(d/2)/cos(d)

Thus a=b=c=2

a+b+c=6