Question

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

• 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 $\frac{L}{3}(1-\cos \theta )$

The correct option is C

Given,

A particle having $mass=m$

$length=L$

Mass of the bead is $2m$

Since no force acts on the horizontal direction

So, the center of mass the beard and the particle should remain at the same x-coordinate.

We have a mass of bead is 2m and make an angle $\theta$ with the wire

Initial x-coordinate of the center of mass from the original position of the bead is:

$=\frac{2m\times 0+m\times 1}{3}$

$=\frac{L}{3m}...1$

Now the final position of the center of mass if bead has moved by a distance:

$=\frac{[2mx+m(x+Lcosa\left)\right]}{3m}$

$=\frac{3x+Lcosa}{3m}...2$

By solving equation 1 and 2 then we get,

$3x+Lcosa=L$

$x=\frac{L(1-cosa)}{3}$

This is the required solution.