Question

A body crossed the topmost point of a vertical circle with a critical speed. What will be the acceleration when the string is horizontal?

• A. 3g
• B. 6g
• C. g
• D. 2g

Answer 1

The correct option is A 3g

$v_A=v_C=\sqrt{gl}$
applying the law of conservation of energy
energy at point A $=$ energy at point B
$\frac{1}{2}m{v_A}^2+mgl=\frac{1}{2}m{v_B}^2\Rightarrow \frac{1}{2}mgl+mgl=\frac{1}{2}m{v_B}^2\Rightarrow v_B=\sqrt{3gl}$
and acceleration $a=\frac{v^2}{l}=\frac{(\sqrt{3gl}{)}^2}{l}=3g$