Question

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

Answer 1

The speed of the particle is v = √gl/3

Explanation:

Let T = mg when angle is θ

h = l (1−cosθ)

Using conservation of energy at A and B

1 / 2 m(u^2 − v^2) = mgh

u = gl, v = speed of particle at B

v^2 = u^2 − 2gh

T − mg cosθ = mv^2 / l

mg − mgcosθ = mv^2 / l

v^2 = gl( 1 − cosθ)    -----(1)

gl ( 1 − cosθ) = gl − 2gl(1 − cosθ)  

cosθ = 2 / 3 , Substituting in (1)

Therefore v = √gl/3

So the speed of the particle is v = √gl/3

Answer 2

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