Question

A heavy particle hanging from fixed point by a light inextensible string of length l is projected horizontally with speed √gl . Find the speed of to the particle and the inclination of the string to the vertical at the instant of the motion when the tension in the string is equal to the weight of the particle•••

[ANSwER WiTH ExplaNatiON•••]

WrONG AnSWeRS WiLL BE RePoRTED☠️❌❌❌​

Answer 1

Let T = mg is at angle θ.

$\implies \sf F_R \ = \ T \ - \ mg \ cos \theta$ ----- (i)

If v is the speed of particle at B,

So, $\sf F_R \ = \ \dfrac {mv^2}{l}$ ------ (ii)

From the above (i) and (ii) equation,

We get,

$\implies \sf T \ - \ mg \ cos \theta \ = \ \dfrac {mv^2}{l}$ ------- (iii)

So, At the particle B, T = mg.

$\implies \sf mg \ (1 \ - \ cos \theta) \ = \ \dfrac {mv^3}{l}$

$\implies \sf V^2 \ = \ gl \ (1 \ - \ cos \theta)$ ------- (iv)

Now, By conserving the point A and point B energy,

$\implies \sf \dfrac {1}{2} mv_0^2 \ = \ mgl \ ( 1 \ - \ \cos \theta \ + \ \dfrac {1}{2} mv^2)$

$\implies \sf gl \ = \ 2 gl ( 1 \ - \ cos \theta) \ + \ gl (1 \ - \ cos \theta)$

$\implies \sf cos \theta \ = \ \dfrac {2}{3}$ -------- (v)

Now,

Put the value of cos in the equation (iv),

We get,

$\implies \sf V \ = \ {\sqrt \dfrac {gl}{3}}$

$\qquad \large \orange {\underline {\boxed {\sf V \ = \ {\sqrt \dfrac {gl}{3}}}}}$

Answer 2

I have attached the full solution

hope it helped you