Question

A particle is suspended by thread of length $l$ from a fixed point. If it is launched from the bottom with a horizontal speed of $v=\sqrt{5gl}$, then tension in the thread when the particle is at the topmost point

• A. $T=mg$
• B. $T=0$
• C. $T=5mg$
• D. $T=6mg$

Answer 1

The correct option is B $T=0$

At top most point
$\frac{1}{2}m{v}_o^2=\frac{1}{2}m{v}^2+mgh$
$v^2=v_o^2-2gl=5gl-2g\left(2l\right)=gl$
$v=\sqrt{gl}$
At top

$\uparrow \frac{m{v}^2}{r}$
$\downarrow mg\downarrow T$ $\therefore T+mg=\frac{m{v}^2}{r}$
$T+mg=\frac{m}{r}gl$
$\therefore T=0$