Question

A particle of mass m is suspended from the ceiling through a string of length L. The particle revolves in a horizontal circle of radius L/2. The speed of the particle is

Options
1) √gL
2) √gL/12
3) √gL/√12
4) √gL/(12)^1/4​

Answer 1

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From the fig., we get,

$\sf{\longrightarrow \sin\theta=\dfrac{R}{L}}$

$\sf{\longrightarrow \cos\theta=\dfrac{\sqrt{L^2-R^2}}{L}}$

$\sf{\longrightarrow \tan\theta=\dfrac{R}{\sqrt{L^2-R^2}}}$

Since the net vertical force acting on the particle is zero, we have,

$\sf{\longrightarrow T\cos\theta=mg}$

$\sf{\longrightarrow \dfrac{T\sqrt{L^2-R^2}}{L}=mg}$

$\sf{\longrightarrow T=\dfrac{mgL}{\sqrt{L^2-R^2}}}$

The particle revolves in a horizontal circle, so it has centripetal force which is provided by the horizontal component of tension in the string.

$\sf{\longrightarrow T\sin\theta=\dfrac{mv^2}{R}}$

$\sf{\longrightarrow\dfrac{mgL}{\sqrt{L^2-R^2}}\cdot\dfrac{R}{L}=\dfrac{mv^2}{R}}$

$\sf{\longrightarrow v=\sqrt{\dfrac{gR^2}{\sqrt{L^2-R^2}}}}$

$\sf{\longrightarrow v=\sqrt{gR\cdot\dfrac{R}{\sqrt{L^2-R^2}}}\quad\quad\dots(1)}$

$\sf{\longrightarrow\underline{\underline{v=\sqrt{gR\tan\theta}}}}$

In the question, $\sf{R=\dfrac{L}{2}.}$

So, from (1),

$\sf{\longrightarrow v=\sqrt{g\cdot\dfrac{L}{2}\cdot\dfrac{L}{2\sqrt{L^2-\left(\dfrac{L}{2}\right)^2}}}}$

$\sf{\longrightarrow v=\sqrt{g\cdot\dfrac{L}{2}\cdot\dfrac{L}{2\sqrt{L^2-\dfrac{L^2}{4}}}}}$

$\sf{\longrightarrow v=\sqrt{g\cdot\dfrac{L}{2}\cdot\dfrac{L}{2\sqrt{\dfrac{3L^2}{4}}}}}$

$\sf{\longrightarrow v=\sqrt{g\cdot\dfrac{L}{2}\cdot\dfrac{2L}{2L\sqrt3}}}$

$\sf{\longrightarrow v=\sqrt{\dfrac{gL}{2\sqrt3}}}$

$\sf{\longrightarrow v=\sqrt{\dfrac{gL}{\sqrt{12}}}}$

$\sf{\longrightarrow\underline{\underline{v=\dfrac{\sqrt{gL}}{12^{\frac{1}{4}}}}}}$

Hence (4) is the answer.