Question

A body is rotated in the vertical plane by means of a thread of length $l$ with minimum possible speed. At the highest point $B$ the thread breaks and the body moves under gravity and strikes at $C$. Then,

• A. the horizontal range $AC$ is $2l$
• B. the horizontal range $AC$ is $\sqrt{2}l$
• C. tangential acceleration of the body at point $C$ is $\frac{2}{\sqrt{5}}g$
• D. tangential acceleration of the body at point $C$ will be $\frac{1}{\sqrt{5}}g$

Answer 1

Answer: (A), (C)

The correct options are
A the horizontal range $AC$ is $2l$
C tangential acceleration of the body at point $C$ is $\frac{2}{\sqrt{5}}g$

Distance $AC$ = (horizontal velocity at $B$) $\times$ (time taken by the body to cover horizontal displacement $AC$)

For minimum possible speed at $B$,
$mg+T=\frac{m{v}_B^2}{l}$
$T=0\Rightarrow v_B=\sqrt{gl}$

Time taken for body to reach $C$ $=\sqrt{\frac{2\times 2l}{g}}=\sqrt{\frac{4l}{g}}$

$AC=\left(\sqrt{gl}\right)\left(\sqrt{\frac{4l}{g}}\right)=2l$
$\therefore$ (a)


At point $C$, $v_x=\sqrt{gl},v_y=\sqrt{2g\times 2l}=\sqrt{4gl}$
$\tan \theta =\frac{v_x}{v_y}=\frac{1}{2}$
Tangential acceleration $=g\cos \theta =\frac{2}{\sqrt{5}}g$
$\therefore$ (c) is also correct.