Question

A thin light inextensible frictionless cord of length I wearing a small bead is tied between two nails that are in the same level a distance 0.5 I apart. initially the bead is held close to nail and released. find speed of the bead immediately after the cord becomes taut.

Answer 1

$\sqrt{\frac{3}{4}gl}$

Answer 2

The bead is released from rest close to nail A. We assume it is released from the position of nail A $(0,0)$ and falls vertically under gravity. The cord, attached to A and B, is initially slack. The cord becomes taut when the position P of the bead satisfies AP + PB = $l$. Since A is at $(0,0)$ and B is at $(0.5l, 0)$, and P is at $(0, y)$, the condition is $\sqrt{0^2+y^2} + \sqrt{(0.5l)^2+y^2} = l$. Assuming $y<0$ (falling downwards), this is $-y + \sqrt{0.25l^2+y^2} = l$. Solving for $y$, we get $y = -0.375l = -\frac{3}{8}l$.

The bead falls from initial height $h_i = 0$ to final height $h_f = -0.375l$. By conservation of energy, the loss in potential energy equals the gain in kinetic energy.

$mgh_i - mgh_f = \frac{1}{2}mv^2$.

$mg(0) - mg(-0.375l) = \frac{1}{2}mv^2$.

$0.375mgl = \frac{1}{2}mv^2$.

$\frac{3}{8}gl = \frac{1}{2}v^2$.

$v^2 = \frac{3}{4}gl$.

$v = \sqrt{\frac{3}{4}gl}$.

The speed of the bead immediately after the cord becomes taut is $\sqrt{\frac{3}{4}gl}$.