Question

What minimum velocity should be given to a particle of mass m, tied to a massless string, in the lowermost point so that it passee through the center after loosing vertical circular motion between points B and C ?​

Answer 1

Let the length of the string be $\sf{L.}$

The kinetic energy of the particle at the lowermost point,

• $\sf{K_1=\dfrac{1}{2}\,mu^2}$

And potential energy is zero there, as we take it as base position.

• $\sf{U_1=0}$

Let $\sf{K_2}$ be the kinetic energy of the particle at the point where the particle starts to fall towards the center.

And the potential energy of the particle at that point,

• $\sf{U_2=mgL(1+\sin\theta)}$

We know the total mechanical energy of the particle undergoing vertical circular motion is conserved.

Therefore,

$\sf{\longrightarrow K_1+U_1=K_2+U_2}$

$\sf{\longrightarrow K_2=K_1+U_1-U_2}$

$\sf{\longrightarrow K_2=\dfrac{1}{2}\,mu^2-mgL(1+\sin\theta)}$

In general the kinetic energy $\sf{K_2}$ should be non - negative.

$\sf{\longrightarrow K_2\geq0}$

$\sf{\longrightarrow \dfrac{1}{2}\,mu^2-mgL(1+\sin\theta)\geq0}$

$\sf{\longrightarrow \dfrac{1}{2}\,mu^2\geq mgL(1+\sin\theta)}$

$\sf{\longrightarrow u\geq\sqrt{2gL(1+\sin\theta)}}$

Hence the minimum velocity at the lowermost point will be,

$\sf{\longrightarrow\underline{\underline{u_{min}=\sqrt{2gL(1+\sin\theta)}}}}$

With this velocity at the lowermost point, the particle gets stopped after traversing an angular displacement $\sf{\dfrac{\pi}{2}+\theta}$ and then it moves towards the center as the tension in the string, $\sf{T<0}$ too.