Question

A bead can slide on a smooth straight wire and a particle of mass $m$ is attached to the bead by a light string of length $L$. The particle is held in contact with the wire with the string taut and is then let fall. The bead has mass $2m$. When the string makes an angle $\theta$ with the wire find the distance moved by the bead

• A. $L(1-\cos \theta )$
• B. $\frac{L}{2}(1-\cos \theta )$
• C. $\frac{L}{3}(1-\cos \theta )$
• D. $\frac{L}{6}(1-\cos \theta )$

Answer 1

The correct option is C $\frac{L}{3}(1-\cos \theta )$

no force acts in horizontal direction.

center of mass of the bead and particle should remain at the same X-coordinate

initial (X- coordinate) of C.O.M

$(2m\times 0+m\times L)/3=\frac{L}{3}$

final position = C.O.M

moved by a distance $x$,

$[2mX+m(X+L\left(\cos a\right)\left)\right]/3m$

$=(3X+L\cos a)/3$

equating we get

$(3X+L(\cos a)=L$

$X=\frac{L(1-\cos a)}{3}$ Required answer