Work Energy and Power
10.0 Work done by a distributed mass
10.0 Work done by a distributed mass
The problem of distributed mass can be easily solved by following some simple steps.
- Choose the frame of reference
- Find the position of center of gravity $\left( {COG} \right)$ for both initial and final arrangement
- Find the distance of $COG$ from the frame of reference for both initial and final arrangements
- Calculate the total potential energy about the frame of reference for both the arrangements
- Calculate the change in potential energy and the work done
- Apply work energy theorem
Question 22. A chain of mass $m$ and length $\pi R$ is placed inside a frictionless semi circular tube of radius $R$. Assuming that the chain starts slipping after a slight disturbance. Find out the velocity with which the last ring of the chain emerges out of the semicircular tube.
Solution: The center of gravity $\left( {COG} \right)$ of a semicircular chain is $\frac{2R}{\pi}$ units above the center
Distance between ${\left( {COG} \right)_i}$ and the frame of reference $=+\frac{2R}{\pi}$
Distance between ${\left( {COG} \right)_f}$ and the frame of reference $=-\frac{\pi R}{2}$
So, the initial and final potential energy is, $$\begin{equation} \begin{aligned} {U_i} = mg\left( {\frac{{2R}}{\pi }} \right) \\ {U_f} = mg\left( { - \frac{{\pi R}}{2}} \right) \\\end{aligned} \end{equation} $$
Change in potential energy $\left( {\Delta U} \right)$ , $$\begin{equation} \begin{aligned} \Delta U = {U_f} - {U_i} \\ \Delta U = mg\left( { - \frac{{\pi R}}{2} - \frac{{2R}}{\pi }} \right) \\ \Delta U = - mgR\left( {\frac{\pi }{2} + \frac{2}{\pi }} \right) \\\end{aligned} \end{equation} $$
As gravitational force is conservative in nature, therefore, $$\begin{equation} \begin{aligned} W = - \Delta U \\ W = mgR\left( {\frac{\pi }{2} + \frac{2}{\pi }} \right) \\\end{aligned} \end{equation} $$
Applying work energy theorem we get, $$\begin{equation} \begin{aligned} \sum W = \Delta KE \\ mgR\left( {\frac{\pi }{2} + \frac{2}{\pi }} \right) = K{E_f} - K{E_i} \\ mgR\left( {\frac{\pi }{2} + \frac{2}{\pi }} \right) = \frac{1}{2}m{v^2} - 0 \\ v = \sqrt {2gR\left( {\frac{\pi }{2} + \frac{2}{\pi }} \right)} \\\end{aligned} \end{equation} $$
Question 23. A chain of length $L$ and mass $m$ is kept on a horizontal table. Mass per unit length of chain varies as $\lambda (x) = Ax$. Find the velocity when the last ring of the chain is about to fall off.
Solution: Let us find the center of gravity $\left( {COG} \right)$ of the chain whose mass per unit length varies $\lambda (x) = Ax$.
Mass per unit length at a distance $x$ is $Ax$.
Small mass at a distance $x$ is given by, $$\begin{equation} \begin{aligned} dm = \lambda dx \\ dm = Axdx...(i) \\\end{aligned} \end{equation} $$
Integrating the above equation with proper limits we get, $$\begin{equation} \begin{aligned} \int\limits_0^m {dm} = \int\limits_0^L {Axdx} \\ m = \frac{{A{L^2}}}{2}...(ii) \\\end{aligned} \end{equation} $$
Position of centre of mass is, $${x_{com}} = \int\limits_0^L {xdm} $$
From equations $(i)$ and $(ii)$ we get, $${x_{com}} = \frac{{\int\limits_0^L {xAxdx} }}{m}$$ $${x_{com}} = \frac{{\frac{{A{L^3}}}{3}}}{{\frac{{A{L^2}}}{2}}}$$ $${x_{com}} = \frac{{2L}}{3}...(iii)$$
So, the initial and final potential energy is, $$\begin{equation} \begin{aligned} {U_i} = 0 \\ {U_f} = - mg\left( {\frac{{2L}}{3}} \right) \\\end{aligned} \end{equation} $$
Change in potential energy $\left( {\Delta U} \right)$ is, $$\begin{equation} \begin{aligned} \Delta U = {U_f} - {U_i} \\ \Delta U = - mg\left( {\frac{{2L}}{3}} \right) - 0 \\ \Delta U = - mg\left( {\frac{{2L}}{3}} \right) \\\end{aligned} \end{equation} $$
As gravitational force is conservative in nature, therefore, $$\begin{equation} \begin{aligned} W = - \Delta U \\ W = mg\left( {\frac{{2L}}{3}} \right) \\\end{aligned} \end{equation} $$
Applying work energy theorem we get, $$\begin{equation} \begin{aligned} \sum W = \Delta KE \\ mg\left( {\frac{{2L}}{3}} \right) = K{E_f} - K{E_i} \\ mg\left( {\frac{{2L}}{3}} \right) = \frac{1}{2}m{v^2} - 0 \\ v = \sqrt {\frac{{4gL}}{3}} \\\end{aligned} \end{equation} $$
