Question

A block is tied to one end of a light string of length l whose other end is fixed to a rigid support. The block is given a speed of $\sqrt{\frac{7gl}{2}}$ from the lowermost position. Find the height at which the block leaves the circle.

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Let the velocity at the bottom =

$v_b\Rightarrow \sqrt{\frac{7gl}{2}}$

AS $v_b<\sqrt{5gl}$

The block will leave the circle at some point P, where the radius OP makes an angle $\theta$ with the upward vertical.

From A to P:Loss in KE - Gain in GPE

$\frac{1}{2}m{v}_b^2-\frac{1}{2}m{v}^2=mg(l+lcos\theta )$

From the force diagram:T + mg cos = $\frac{m{v}^2}{l}$

As the block leave the circle at P,T = 0

mg cos $\theta =\frac{m{v}^2}{l}$

Putting value of $v^2$ from (ii) in (i)

$\frac{1}{2}m{v}_b^2-\frac{1}{2}m\left(glcos\theta \right)=mgl(1+cos\theta )\Rightarrow \frac{1}{2}m\frac{7}{2}gl=\frac{mgl}{2}(2+3cos\theta )$

$\Rightarrow cos\theta =\frac{1}{2}\Rightarrow \theta ={60}^{\circ }$

Hence the block leaves the circle and string slack at a height h = l + l cos${60}^{\circ }$ = 1.5 l from the bottom.The velocity at that moment v is v = $\sqrt{lgcos\theta }=\sqrt{\frac{lg}{2}}$

After the string becomes slack, the block moves as a projectile in parabolic path.Now, further height attained = $\frac{v^{2}si{n}^2\theta }{2g}=\frac{1g}{4g}\left(\frac{3}{4}\right)=\left(\frac{3}{16}\right)l$

$\Rightarrow$ Total height attained from the bottom is : $\frac{3}{2}l+\frac{3}{16}l=\frac{27}{16}l$