Work Energy and Power
9.0 Types of equilibrium
9.0 Types of equilibrium
When a body is in translatory equilibrium then, $${\overrightarrow F _{net}} = 0$$
For conservative forces we know, $${\overrightarrow F _{net}} = - \frac{{dU}}{{dr}}$$
So, at equilibrium, $$\frac{dU}{dr}=0$$
Let us look at the graph between $U$ and $r$. At all three point $A, B$ and $C$, gradient $\left( {\frac{{dU}}{{dr}}} \right)$ is zero.
Therefore, points $A, B$ and $C$ are at equilibrium.
As all the above three points are in equilibrium but have different potential energy $\left( {dU} \right)$. So, the equilibrium is divided into three types on the basis of potential energy.
There are three types of equilibrium
- Stable equilibrium: This type of equilibrium occurs when the body returns to its original position after being disturbed. This happens because the center of gravity of the initial position is lower than that of the displaced position. Point $B$ is at stable equilibrium as the potential energy $\left( {U} \right)$ is minimum.
- Unstable equilibrium: This type of equilibrium occurs when the body never returns to its original position after being disturbed. This happens because the center of gravity of the initial position is higher than that of the displaced position. Point $C$ is an unstable equilibrium as the potential energy $\left( {U} \right)$ is maximum.
- Neutral equilibrium: This type of equilibrium occurs when the center of gravity of the initial and final position is at the same level. Point $A$ is at neutral equilibrium as the potential energy $\left( {U} \right)$ is constant.
Let us discuss the types of equilibrium in details.
| Type | Stable equilibrium | Unstable equilibrium | Neutral equilibrium |
| Illustration |   |   |   |
| Gradient | $\frac{dU}{dr}=0$, or gradient of $U-r$ graph is zero | $\frac{dU}{dr}=0$, or gradient of $U-r$ graph is zero | $\frac{dU}{dr}=0$, or gradient of $U-r$ graph is zero |
| Center of gravity $\left( {COG} \right)$ | $\left( {COG} \right)$ of an equilibrium position is lower than that of the displaced position | $\left( {COG} \right)$ of an equilibrium position is higher than that of the displaced position | $\left( {COG} \right)$ of an equilibrium position and displaced position remains at the same level |
| Restoring force | When displaced from its equilibrium position a net restoring force starts acting on the body which has a tendency to bring the body back to its equilibrium position. | When displaced from its equilibrium position, a net force starts acting on the body which moves the body away from the equilibrium position. | When displaced from its equilibrium position the body has neither the tendency to come back nor to move away from the original position. |
| Potential energy | Potential energy in this equilibrium position is minimum.$$\frac{{{d^2}U}}{{d{r^2}}} > 0$$ | Potential energy in this equilibrium position is maximum. $$\frac{{{d^2}U}}{{d{r^2}}} < 0$$ | Potential energy constant. $$\frac{{{d^2}U}}{{d{r^2}}} = 0$$ |
| Examples | Pendulum oscillation | Ball rolling down an inclined wedge | Ball rolling on a horizontal surface |
| From Fig. 38 | Point $B$ is stable equilibrium position | Point $C$ is unstable equilibrium position | Point $A$ is neutral equilibrium position |
Example 21. The potential energy for a body varies $U(x)=3px^2+6x+3$. Find the value of $p$ such that the body is in,
- Stable equilibrium
- Unstable equilibrium
- Neutral equilibrium
Solution: $$U(x)=3px^2+6x+3$$
Double differentiating the above equation with respect to $x$ we get, $$\begin{equation} \begin{aligned} \frac{{dU(x)}}{{dx}} = 6px + 6 \\ \frac{{{d^2}U(x)}}{{d{x^2}}} = 6p \\\end{aligned} \end{equation} $$
For stable equilibrium, $$\begin{equation} \begin{aligned} \frac{{{d^2}U(x)}}{{d{x^2}}} = 6p > 0 \\ p > 0 \\\end{aligned} \end{equation} $$
For unstable equilibrium, $$\begin{equation} \begin{aligned} \frac{{{d^2}U(x)}}{{d{x^2}}} = 6p < 0 \\ p < 0 \\\end{aligned} \end{equation} $$
For neutral equilibrium, $$\begin{equation} \begin{aligned} \frac{{{d^2}U(x)}}{{d{x^2}}} = 6p = 0 \\ p = 0 \\\end{aligned} \end{equation} $$
