Question

A small ring is attached to the vertical ring as shown in figure. The small ring is given velocity $v_B=\sqrt{4gl}$ by a sharp hit, where $l$ is radius of the vertical ring. Find the normal force acting on the small ring at the topmost point, if the mass of the small ring is $m$ and mass of the vertical ring is $M$.

• A. $2mg$
• B. $2Mg$
• C. $mg$
• D. $Mg$

Answer 1

The correct option is C $mg$


Using energy conservation between $A$ and $B$:
$K_A+U_A=K_B+U_B$
(taking $B$ as a reference)

$\Rightarrow \frac{1}{2}m{v}_A^2+2mgl=\frac{1}{2}m{v}_B^2+0$
$\Rightarrow \frac{1}{2}m{v}_A^2+2mgl=\frac{1}{2}m\times 4gl=2mgl$
(given $v_B=\sqrt{4gl}$)
$\Rightarrow \frac{1}{2}m{v}_A^2=0$
$\Rightarrow v_A=0$

FBD of the small ring:


So at point $A$,
$N+\frac{m{v}_A^2}{R}=mg$
$N+0=mg\Rightarrow N=mg$