Question

A heavy particle is suspended by a string of length l. The particle is given a horizontal velocity $v_0$. The string becomes slack at some angle and the particle proceeds on a parabola. Find the value of $v_0$ if the particle passes through the point of suspension.

• A. $\left[gl\right(2+\sqrt{2}){]}^{\frac{1}{2}}$.
• B. $\sqrt{3gl}$
• C. $\left[gl\right(2+\sqrt{3}){]}^{\frac{1}{2}}$.
• D. None of these

Answer 1

The correct option is C

$\left[gl\right(2+\sqrt{3}){]}^{\frac{1}{2}}$.

Suppose the string becomes slack when the particle reaches the point P (figure) suppose the string OP makes an angle $\theta$ with the upward vertical. The only force acting on the particle at P is its weight mg. The radial component of the force is mg cos$\theta$. As the particle moves on the circle upto P,

mg cos$\theta =m\left(\frac{v^2}{l}\right)$

Or, $v^2=glcos\theta$ ---------------(i)

Where v is its speed at P.Using conservation of energy, $\frac{1}{2}m{v}_0^2=\frac{1}{2}m{v}^2+mgl(1+cos\theta )$

Or, $v^2=v_0^2-2gl(1+cos\theta )$ --------------------(ii)

From (i) and (ii), $v_0^2-2gl(1+cos\theta )=glcos\theta$

Or, $v_0^2=gl(2+3cos\theta )$ -----------------(iii)

Now onwards the particle goes in a parabola under the action of gravity.As it passes through the point of suspension O, the equation for horizontal and vertical motion give,

$lsin\theta =\left(vcos\theta \right)t$

and - $lcos\theta =\left(vsin\theta \right)t-\frac{1}{2}g{t}^2$

or, $-lcos\theta =\left(vsin\theta \right)t-\left(\frac{lsin\theta }{vcos\theta }\right)-\frac{1}{2}g{\left(\frac{lsin\theta }{vcos\theta }\right)}^2$

or, $-co{s}^2\theta =si{n}^2\theta -\frac{1}{2}g\frac{lsi{n}^2\theta }{v^{2}cos\theta }$

or, $-cos\theta =1-co{s}^2\theta -\frac{1}{2}\frac{glsi{n}^2\theta }{glco{s}^2\theta }$ [From (i)]

or, $1=\frac{1}{2}ta{n}^2\theta$

or, $tan\theta =\sqrt{2}$

From (iii), $v_0=\left[gl\right(2+\sqrt{3}){]}^{\frac{1}{2}}$.