Question

A heavy particle is suspended by a 1.5 m long string. It is given a horizontal velocity of √57 m/s. (a) Find the angle made by the string with the upward vertical, when it becomes slack. (b) Find the speed of the particle at this instant. (c) Find the maximum height reached by the particle over the point of suspension. Take g =10 m/s².

Answer 1

Given in the question :-

u = √57 m/s
l = 1.5 m.

Now for the condition
(a) Point where string slacked.

mv²/l = mg.cosθ
v² =lg.cosθ

Now change in Kinetic energy = work done , so Total energy at final point. 
=Kinetic energy + Potential Energy 
1/2 mv² + mgl(1+cosθ) 
1/2 mgl cosθ + mgl(1+cosθ) = 57m/2    {K.E= 57m/2 J}

Therefore the initial energy is equal to final energy 
lg.cosθ + 2gl(1+cosθ) = 57
3gl.cosθ= 57 - 2gl 
30 × 1.5.cosθ  = 57 - 2 × 10 × 1.5  = 27
45.cosθ= 27 
cosθ = 27/45 
cosθ= 3/5 
cosθ= 0.60 
θ = 53°

Now for the second condition :-
(a)
v² = lg cosθ
v = √lg cosθ
v = √ ( 1.5 × 10 × 0.60)
v = √9
v = 3 .

Now for the third condition :-

(c) at the time string act like slack,the particle start motion in a projectile.

Assume that the max . height on the point of suspension.= x
For the projectile , max height of projection is d = (vsinθ)²/ 2g

Now for suspension,
x = l cos θ +d
x = 1.5 × 0.60 + (vsinθ )²/2g
x = 0.9 + ( 0.8 × 3)² /20
x = 0.9 + 0.28
x = 1.18
x = 1.2

Hence the suspension is 1.2 .



Hope it Helps. :-)