David Bauer Physics & Astronomy · UCLA

Physics 5A Chapter 20

Energy diagrams

Potential energy and force

Potential energy and force are equivalent ways of describing interactions. If we know the potential energy of a system in every configuration, we can determine the interaction forces between the objects in the system.

Recall that the work done by a conservative force is minus the change in the potential energy for that force:

\[W = -\Delta U.\]

For one-dimensional motion, the work done by a force $\vec{F}$ is

\[W = \int_{x_i}^{x_f} F_x(x)\,dx.\]

Combining these two equations, we find that the change in potential energy associated with a force is

\[\Delta U = -\int_{x_i}^{x_f} F_x(x)\,dx.\]

The fundamental theorem of calculus then tells us that the $x$-component of the force is

\[\boxed{F_x = -\frac{dU}{dx}}\]

In terms of a graph of the potential energy, this means

\[F_x = -\left(\text{slope of } U \text{ vs. } x \text{ graph}\right).\]

So if we know the potential energy of a system as a function of position, or equivalently if we have a graph of the potential energy, then the force is fully determined. In particular, the relationship $F_x = -dU/dx$ tells us that

\[\boxed{\text{conservative forces act in the direction of decreasing potential energy.}}\]

Intuitively, forces tend to point “downhill” in a potential energy graph.

Energy diagrams

An energy diagram is a graph of potential energy, typically plotted along with a horizontal line representing the total mechanical energy $E_\mathrm{mech}$ of the system. For example, here is an energy diagram for a mass attached to a spring, where the potential energy is the parabola $U(x) = \frac{1}{2}kx^2$:

A graph with horizontal axis labeled x and vertical axis labeled U of x. A blue upward parabola is centered at the origin with its minimum at the origin, representing the spring potential energy. A horizontal orange line, labeled E sub mech, sits partway up and crosses the parabola at two symmetric points. A legend at the right marks the blue curve as U and the orange line as E sub mech.

The total mechanical energy is the sum of the (macroscopic) kinetic and potential energies:

\[E_\mathrm{mech} = K + U.\]

For an isolated system with no dissipative forces, the total mechanical energy is constant: $\Delta E_\mathrm{mech} = 0$.

Given an energy diagram, we can find the kinetic energy at any point by subtracting the potential energy from the total mechanical energy:

\[K = E_\mathrm{mech} - U.\]

Here is the same energy diagram as before, but with a curve added showing the kinetic energy $K = E_\mathrm{mech} - U$ as a function of position:

The same spring potential energy graph. The blue parabola U has its minimum at the origin. A horizontal orange line E sub mech sits near the top and crosses the parabola near its outer edges. A green inverted parabola, the kinetic energy K, equals the vertical gap between the orange line and the blue curve: it is zero where the blue curve meets the orange line and rises to a maximum at the center, where the orange and green curves nearly touch above the minimum of U. A legend at the right labels the three curves U, E sub mech, and K.

For the mass on a spring, the kinetic energy is largest at the center of the parabola (where $U$ is smallest) and shrinks to zero where the potential energy curve rises up to meet the total energy line.

Turning points

We can use an energy diagram to find turning points of the motion. Recall that a turning point is where the speed is zero: the particle stops and changes direction.

Since $v = 0$ at a turning point, the kinetic energy is zero. So turning points are the points at which

\[\boxed{U(x) = E_\mathrm{mech}}\]

On an energy diagram, the turning points occur where the horizontal total energy line intersects the potential energy graph:

The spring potential energy graph: a blue parabola with its minimum at the origin and a horizontal orange total-energy line crossing it at two symmetric points. Black dots mark the two intersection points, labeled turning points where v equals zero, since there U equals E sub mech and the kinetic energy is zero. A third black dot marks the minimum of the parabola, labeled maximum speed, where U is smallest and the kinetic energy is greatest.

Forbidden regions

Since the kinetic energy can never be negative, the potential energy at a point sets the lower limit for the total mechanical energy. Any region where $U(x) > E_\mathrm{mech}$ is forbidden: the particle does not have enough energy to reach it. The motion stays in the allowed region between the turning points, as shown below.

The spring potential energy graph: a blue upward parabola with its minimum at the origin and a horizontal orange total-energy line E sub mech crossing it at two symmetric turning points, marked with black dots. The two vertical strips beyond the turning points, where the blue curve rises above the orange line so that U of x is greater than E sub mech, are shaded grey with diagonal hatching and labeled forbidden. The particle does not have enough energy to enter these regions, so its motion is confined to the allowed region between the turning points.

Equilibrium and stability

An energy diagram also tells us about points where a particle is in equilibrium. Recall that for a particle, equilibrium means the net force is zero. Assuming the net force $\vec{F}$ on the system is conservative, we can write the force in terms of the potential energy $U$ as $F_x = -dU/dx$. So at equilibrium, we have

\[\boxed{\frac{dU}{dx} = 0}\]

This means that points at which a particle is in equilibrium are critical points of the potential energy. If we place a particle at rest at one of these positions, it will stay there indefinitely as long as no other forces act on it.

A graph of a double-well potential energy U versus x. The blue curve has two valleys, a deeper minimum near x equals minus 3.3 and a shallower minimum near x equals 3, separated by a local maximum near x equals 0 that sits just above the axis. Black dots mark all three critical points, labeled equilibrium points, because the slope dU/dx is zero there.

An equilibrium may be stable or unstable:

  • At a stable equilibrium, the system tends to move back to equilibrium if we displace it slightly. Think about a pendulum, or a mass on a spring.
  • At an unstable equilibrium, the system accelerates away from equilibrium if we displace it slightly. Think about a ball balanced on top of a hill.

We can identify stable and unstable equilibrium points by looking for minima and maxima of the potential energy:

  • A minimum in potential energy is a point of stable equilibrium. If we move away from the equilibrium point, the force is directed back toward equilibrium (a restoring force). Remember: the force points “downhill” in a potential energy graph.

A graph of potential energy U versus x shaped as an upward parabola with its minimum at the origin, a stable equilibrium marked by a black dot. At two representative points, one on each side of the minimum, short red lines are drawn tangent to the curve to show the slope of U. A dashed guide drops from each tangent point down to the x-axis, where a red force vector labeled F is drawn along the axis. Since the force equals minus the slope of U, both vectors point back toward the equilibrium, the left one points right and the right one points left, a restoring force.

  • A maximum in potential energy is a point of unstable equilibrium. If we move the system away from the equilibrium point, the force is directed away from equilibrium.

A graph of potential energy U versus x shaped as a downward parabola with its maximum at the top on the vertical axis, an unstable equilibrium marked by a black dot. At two representative points, one on each side, short red lines are drawn tangent to the curve to show the slope of U. A dashed guide drops from each tangent point down to the x-axis, where a red force vector labeled F is drawn along the axis. Since the force equals minus the slope of U, both vectors point away from the equilibrium, the left one points left and the right one points right, so a displaced particle accelerates away.

Application: pendulum motion

Consider a pendulum formed by attaching a mass $m$ to a rigid bar of length $L$. The pendulum is free to rotate in a vertical circle about a fixed pivot.

A rigid-bar pendulum. A fixed pivot, drawn as a small hinge circle, sits at the top. A straight rigid bar of length L runs from the pivot down and to the right to a red point mass labeled m. A vertical dashed reference line drops straight down from the pivot, showing the vertical direction; the bar is displaced to one side of this vertical.

The pendulum mass moves in non-uniform circular motion, and we can measure its position using the angle $\theta$ of the bar from the vertical.

The same rigid-bar pendulum: a fixed pivot at top, a rigid bar of length L running down to the right to a red mass m, and a vertical dashed reference line straight down from the pivot. A small arc near the pivot, between the downward vertical and the bar, is labeled theta, the angle of the bar measured from the vertical.

The reason we’re considering a rigid bar is to allow the pendulum to swing above the horizontal without worrying about the string going slack.

The rigid-bar pendulum raised above horizontal. A horizontal dashed reference line passes through the pivot; the rigid bar of length L points up and to the right, above this horizontal line, ending at the red mass m. A vertical dashed reference line drops straight down from the pivot. A large arc, swinging from the downward vertical up to the bar and spanning more than ninety degrees, is labeled theta, showing the angle is still measured from the vertical even when the bar is above horizontal.

If we set $U = 0$ when the pendulum is at its lowest point, then the gravitational potential energy of the pendulum mass is

\[\begin{aligned} U &= mgh\\ &= mgL(1 - \cos\theta). \end{aligned}\]

A graph of the pendulum potential energy U of theta versus theta, from minus pi to pi. The blue curve is a raised cosine: it has a single minimum at theta equals zero, touching the horizontal axis at the bottom of the swing, and rises symmetrically to equal maxima at theta equals minus pi and plus pi, the inverted position. The horizontal axis is labeled theta with tick labels minus pi at the left end and pi at the right end.

Check your understanding: Where are the points of stable and unstable equilibrium? How would you describe the state of the pendulum at each of these points?

The raised-cosine pendulum potential U of theta, with a single minimum at theta equals zero and equal maxima at theta equals minus pi and plus pi. A black dot marks the minimum at theta equals zero; the two maxima are left unmarked.

Answer

At $\theta = 0$, the potential energy has a minimum. This is a stable equilibrium: the mass hangs straight down below the pivot, and a small displacement produces a restoring torque back toward the bottom.

\[\boxed{\theta = 0 \text{ is stable equilibrium}.}\]

At $\theta = \pm 180^\circ$, the potential energy has a maximum. This is unstable equilibrium: the mass is balanced directly above the pivot, and a small displacement produces a torque that makes the pendulum fall away from the top.

\[\boxed{\theta = \pm 180^\circ \text{ is unstable equilibrium}.}\]

Check your understanding: What motion does the pendulum have in each of the following three energy diagrams?

(a)

The raised-cosine pendulum potential U of theta with a horizontal orange total-energy line at an intermediate height that crosses the curve at two symmetric points. Black dots mark these intersections as turning points, where U equals E sub mech and the speed is zero. The pendulum oscillates back and forth between the two turning points.

(b)

The raised-cosine pendulum potential U of theta with a horizontal orange total-energy line drawn above both maxima of the curve. Because E sub mech exceeds the maximum potential energy everywhere, the line never meets the curve, so there are no turning points: the kinetic energy stays positive and the pendulum rotates all the way around.

(c)

The raised-cosine pendulum potential U of theta with a horizontal orange total-energy line drawn exactly at the height of the two maxima, so the line is tangent to the curve at theta equals minus pi and plus pi. Black dots mark these tangent points. This is the critical energy: the pendulum just barely reaches the inverted position with vanishing speed, approaching the top asymptotically.

Answer

In an energy diagram, the kinetic energy is

\[K = E_\mathrm{mech} - U.\]

The motion is allowed only where $E_\mathrm{mech} \ge U$, and turning points occur where $U = E_\mathrm{mech}$.

(a) The total energy line intersects the potential energy curve at two angles. These are turning points, so the pendulum oscillates back and forth between them. The speed is zero at the turning points and largest at the bottom, where $U$ is smallest.

\[\boxed{\text{The pendulum oscillates between two turning angles}.}\]

(b) The total energy is above the maximum value of $U$. There are no turning points, so the pendulum has enough energy to rotate all the way over the top.

\[\boxed{\text{The pendulum makes full rotations}.}\]

(c) The total energy is exactly equal to the maximum value of $U$. This is the critical case: the pendulum just reaches the upright position with zero speed. With exactly this energy, it approaches the top with vanishing speed; any small change in energy or disturbance changes whether it falls back or rotates over the top.

\[\boxed{\text{The pendulum just barely reaches the upright position}.}\]

(One interesting feature of this case is that it theoretically takes an infinite amount of time to reach the top, because the speed approaches zero as it gets there.)

Example: Particle in a cubic potential

A single conservative force $\vec{F}$ acts on a particle with mass $2.0\ \mathrm{g}$ moving along the $x$-axis. The potential energy function $U(x)$ for this force is plotted below as a function of $x$.

A graph of potential energy U of x in joules versus position x in meters. The blue curve is a cubic: it rises steeply from the lower left, reaches a local maximum of about 1.0 joule at x equals 3 meters, falls to cross the axis near x equals 5 meters, dips to a local minimum near x equals 6.3 meters at about minus 0.85 joule, then rises steeply again past x equals 8 meters. Light grey horizontal gridlines mark U equals 1.0, 0.5, minus 0.5, and minus 1.0 joule, and the horizontal axis is ticked from 2 to 8 meters. Four points on the curve are marked with black dots: A at x equals 2 meters and U equals 0.5 joule on the rising left branch, B near x equals 5 meters on the axis where U equals zero, C at the local minimum near x equals 6.3 meters, and D at x equals 8 meters and U equals 0.5 joule on the rising right branch.

(a) At each of the four labeled points, determine the direction (left or right) of the force on the particle.

(b) If the particle is released at point D at rest, estimate the turning points of the particle’s motion. Where does it reach its maximum speed?

(c) Estimate the $x$-component of the force on the particle at point B.

(d) What minimum speed does the particle need at point A in order to reach point D?

Solution

(a) For a one-dimensional conservative force,

\[F_x = -\frac{dU}{dx}.\]

The force points “downhill” on the potential energy graph. Therefore:

  • At A, the graph slopes upward as $x$ increases, so $dU/dx > 0$ and the force points left.
  • At B, the graph slopes downward as $x$ increases, so $dU/dx < 0$ and the force points right.
  • At C, the graph has a minimum, so $dU/dx = 0$ and the force is zero.
  • At D, the graph slopes upward as $x$ increases, so $dU/dx > 0$ and the force points left.

We can summarize these results in the table below:

Point Direction of force ($F_x$)
A Left
B Right
C Zero force
D Left

(b) If the particle is released at D from rest, then its total mechanical energy is

\[E_\mathrm{mech} = U(D) \approx 0.5\ \mathrm{J}.\]

The turning points occur where $U(x) = E_\mathrm{mech}$. One turning point is D itself:

\[x_D \approx 7.8\ \mathrm{m}.\]

Drawing the horizontal line $E_\mathrm{mech} = 0.5\ \mathrm{J}$ across the graph, we see that it intersects the curve again to the right of the local maximum at

\[x \approx 4.1\ \mathrm{m}.\]

The particle cannot reach point A in this motion (or any point left of $x \approx 4.1\ \mathrm{m}$), because the potential energy barrier near $x = 3.0\ \mathrm{m}$ is above its total energy.

The same cubic potential energy graph, now with a horizontal orange total mechanical energy line drawn at U equals 0.5 joule, the value of U at point D. The line meets the curve at three places: at point A on the far left branch, at about x equals 4.1 meters on the rising side of the well, and at point D at about x equals 7.8 meters. The middle and right intersections are marked with red dots and labelled as the turning points, with dotted guides dropping to the horizontal axis at x approximately 4.1 meters and x approximately 7.8 meters. A particle released from rest at D is confined between these two turning points, because the potential energy barrier of about 1.0 joule near x equals 3 meters rises above the energy line and blocks the region containing A.

The speed is largest where the kinetic energy

\[K = E_\mathrm{mech} - U\]

is largest. That occurs where $U$ is smallest in the allowed region, at point C:

\[\boxed{\text{maximum speed at C, near } x = 6.3\ \mathrm{m}.}\]

(c) The force is minus the slope of the graph at B. To estimate that slope, we draw the tangent line to the curve at B and take its rise over run: picking any two points $(x_1, U_1)$ and $(x_2, U_2)$ on the tangent line,

\[\frac{dU}{dx}\bigg|_B \approx \frac{\Delta U}{\Delta x} = \frac{U_2 - U_1}{x_2 - x_1}.\]

The same cubic potential energy graph with a straight red tangent line drawn through point B, where the curve crosses the horizontal axis near x equals 4.8 meters. The tangent slopes downward to the right, matching the slope of the curve at B. Its upper left end stops exactly on the U equals plus 0.5 joule gridline at about x equals 4.2 meters, and its lower right end stops exactly on the U equals minus 0.5 joule gridline at about x equals 5.4 meters. The tangent therefore drops 1.0 joule over a horizontal run of about 1.2 meters, giving a slope of roughly minus 0.8 joules per meter.

Extending the tangent line to the $U = 0.5\ \mathrm{J}$ and $U = -0.5\ \mathrm{J}$ gridlines makes its endpoints easy to read off from the graph:

\[(x_1, U_1) \approx (4.2\ \mathrm{m},\ 0.5\ \mathrm{J}), \qquad (x_2, U_2) \approx (5.4\ \mathrm{m},\ -0.5\ \mathrm{J}).\]

The slope is then

\[\begin{aligned} \frac{dU}{dx}\bigg|_B &\approx \frac{U_2 - U_1}{x_2 - x_1}\\ &= \frac{(-0.5\ \mathrm{J}) - (0.5\ \mathrm{J})}{5.4\ \mathrm{m} - 4.2\ \mathrm{m}}\\ &= \frac{-1.0\ \mathrm{J}}{1.2\ \mathrm{m}}\\ &\approx -0.8\ \mathrm{J/m}. \end{aligned}\]

Since $1\ \mathrm{J/m} = 1\ \mathrm{N}$,

\[F_{x,B} = -\frac{dU}{dx}\bigg|_B \approx +0.8\ \mathrm{N}.\]

The positive sign means the force points to the right:

\[\boxed{F_{x,B} \approx +0.8\ \mathrm{N}.}\]

(d) To get from A to D, the particle must pass over the potential energy maximum near $x = 3.0\ \mathrm{m}$. From the graph,

\[U(A) \approx 0.5\ \mathrm{J}, \qquad U_\mathrm{max} \approx 1.0\ \mathrm{J}.\]

The minimum speed is the speed that makes the total mechanical energy equal to the top of the barrier:

\[\frac{1}{2}mv_\mathrm{min}^2 + U(A) = U_\mathrm{max}.\]

Therefore

\[\begin{aligned} v_\mathrm{min} &= \sqrt{\frac{2\left(U_\mathrm{max} - U(A)\right)}{m}}\\ &= \sqrt{\frac{2\left(1.0\ \mathrm{J} - 0.5\ \mathrm{J}\right)}{0.0020\ \mathrm{kg}}}\\ &= 22\ \mathrm{m/s}. \end{aligned}\]

Thus

\[\boxed{v_\mathrm{min} \approx 22\ \mathrm{m/s}.}\]