Question

A bead can slide on a smooth circular wire frame of radius $r$ which is fixed in the vertical plane. The bead is displaced slightly from the highest point of the wire frame. The speed of the bead subsequently as a function of the angle $\theta$ made by the bead with the vertical line is

• A. $\sqrt{2gr\sin \theta }$
• B. $\sqrt{2gr(1-\sin \theta )}$
• C. $\sqrt{2gr(1-\cos \theta )}$
• D. $\sqrt{2gr\cos \theta }$

Answer 1

The correct option is C $\sqrt{2gr(1-\cos \theta )}$

FBD of the bead when angle is $\theta$:


We have,
$h=h_A-h_B=r-r\cos \theta =r(1-\cos \theta )$
Using energy conservation between $A$ and $B$:
$K_A+U_A=K_B+U_B$
Since, bead is slightly displaced, $K_A=0$
$0+mg{h}_A=\frac{1}{2}m{v}_B^2+mg{h}_B$
$\Rightarrow mg(h_A-h_B)=\frac{1}{2}m{v}_B^2$
$\Rightarrow mgr(1-\cos \theta )=\frac{1}{2}m{v}_B^2$
$\Rightarrow v_B=\sqrt{2gr(1-\cos \theta )}$