Question

A heavy particle hanging from a fixed point by a light inextensible string of length l is projected horizontally with speed √(gl). Find the speed of 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 1

Let

• Tension in the string becomes equal to the weight of the particle when particle reaches B and deflection of the string from vertical is ө
• Now resolving mg along the string and perpendicular to the string.
• We get the force of radial on the particle at B i.e.

$F_{R} = T - mg \: cos\theta.....(1)$

If v be the speed of the speed of the particle at B, then

$F_{R} = \frac{mv^{2}}{l}.....(2)$

From (1) and (2), we get

$T - mg \:cod\theta = \frac{mv^2}{l}.....(3)$

Since at B, $T = mg$

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

$\implies \:v^2 = gl(1-cos\theta).....(4)$

Conserving Point A and B energy, we have

$\frac{1}{2}mv^{2}_{0} = mgl(1-cos\theta)+\frac{1}{2}mv^2$

Where,

• $v_{0} = \sqrt{gl}$
• $v = \sqrt{gl(1-cos\theta)}$

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

$\implies \:cos\theta = \frac{2}{3}.....(5)$

Putting cosө value in eq(4), we get

${\boxed{v = \sqrt{\frac{gl}{3}}}}$

Answer 2

$\large\bold{\underline{\underline{Question:-}}}$

A heavy particle hanging from a fixed point by a light inextensible string of length l is projected horizontally with speed √(gl). Find the speed of 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

$\large\bold{\underline{\underline{Answer:-}}}$

Let T = mg when angle is θ

h = l (1−cosθ)

Using conservation of energy at A and B

1 / 2 m(u² − v²) = mgh

u = gl, v = speed of particle at B

v² = u² − 2gh

T − mg cosθ = mv² / l

mg − mgcosθ = mv²/ l

v² = gl( 1 − cosθ) -----(1)

gl ( 1 − cosθ) = gl − 2gl(1 − cosθ)

cosθ = 2 / 3 , Substituting in (1)

Therefore v $= \frac{ \sqrt{gl} }{3}$

So the speed of the particle is $\huge V= \frac{ \sqrt{ gl} }{3}$