Chapter 14, Problem 14.116RP

A chain of length I and mass m falls through a small hole in a plate. Initially, wheny is very small, the chain is at rest. In each case shown, determine (a) the acceleration of the first link A as a function of y, (b) the velocity of the chain as the last link passes through the hole. In case 1, assume that the individual links are at rest until they fall through the hole. In case 2, assume that at any instant all links have the same speed. Ignore the effect of friction.

To determine

The acceleration of the first link A as a function of y.

Answer to Problem 14.116RP

The value of acceleration of the first link A as a function of Y is$a=0.333g$

Explanation of Solution

Given information:

Mass per cent length = $\rho$

The value of length of the chain $=l$

Velocity is $=v$

Consider the case 1

And here we have to apply the principle of impulse and momentum

So, we must take:

$\Delta t\to 0$

$\begin{array}{l}\rho gy=\rho yl{t}_{\Delta t\to 0}\frac{\Delta v}{\Delta t}+\rho vl{t}_{\Delta t\to o}\frac{\Delta y}{\Delta t}+\rho l{t}_{\Delta t\to o}\frac{\Delta y\Delta v}{\Delta t}\\ \rho gy=\rho y\frac{dv}{dt}+\rho v\frac{dy}{dt}+0\end{array}$

$\begin{array}{l}\Delta y\Delta v=0\\ \rho gy=\rho \left[\frac{d}{dt}\left(yv\right)\right]\end{array}$

So, we have to multiply both sides with yv, we get:

$\begin{array}{l}\rho gyyv=\rho yv\left[\frac{d}{dt}\left(yv\right)\right]\\ \rho g{y}^{2}v=\rho yv\frac{d}{dt}\left(yv\right)\\ v=\frac{dy}{dt}\\ \rho g{y}^2\frac{dy}{dt}=\rho yv\frac{d}{dt}\left(yv\right)\\ \rho g{y}^{2}dy=\rho yvd\left(yv\right)\end{array}$

Applying integration on both sides, we get:

$\begin{array}{l}\rho g\int y^{2}dy=\rho \int yv\left(dyv\right)\\ \rho g\frac{\left(y^3\right)}{5}=\rho \frac{{\left(yv\right)}^2}{2}\\ y^2{v}^2=\frac{28y^3}{3}\\ v^2=\frac{2}{3}gy---(1)\end{array}$

And then differentiating both sides, we get:

$\begin{array}{l}2v\frac{dv}{dt}=\frac{2}{3}g\frac{dy}{dt}\\ 2vd=\frac{2}{3}gv\\ a=\frac{g}{3}\\ av=\frac{gv}{3}\end{array}$

Here $a=\frac{dy}{dt}$

So, the acceleration is:

$a=0.333g$

To determine

The velocity of the chain as the last link passes through the hole.

Answer to Problem 14.116RP

The value of velocity of the chain as the last link passes through the hole is$v=\sqrt{gl}$ .

Explanation of Solution

We must put y=l in first equation.

$\begin{array}{l}{v}_2=\frac{2}{3}gl\\ v^2=0.666gl\\ v=0.816\sqrt{gl}\end{array}$

Again, consider the second case.

Here initial kinetic energy is $T_1=0$

Initial potential energy is $V_1=0$

Final kinetic energy is $T_2=\frac{1}{2}m{v}^2$

Final potential energy is $V_2=-\rho gy\frac{y}{2}$

$\begin{array}{l}{T}_1+V_1=T_2+V_2\\ 0+0=\frac{1}{2}m{v}^2-\frac{\rho g{y}^2}{2}\\ \frac{1}{2}m{v}^2=\frac{\rho g{y}^2}{2}\\ m{v}^2=\rho g{y}^2\\ v^2=\frac{\rho g{y}^2}{m}\\ l=\frac{m}{\rho }\\ v^2=\frac{g{y}^2}{l}---(2)\end{array}$

We must differentiate the equation 2 with respect to Y.

$\begin{array}{l}2v\frac{dv}{dt}=\frac{g}{l}2y\\ v\frac{dv}{dt}=\frac{gy}{l}\end{array}$

We have a relation $a=v\frac{dv}{dt}$.

And the acceleration is $a=\frac{gy}{l}$.

Put y=l in equation 2, we get:

$\begin{array}{l}{v}^2=\frac{g{l}^2}{l}\\ v^2=gl\\ v=\sqrt{gl}\end{array}$