1. Aim: How do we analyze energy diagrams and stability of equilibrium?

2. Reminder of Chain Rule
We will have to calculate the derivative of more complicated potential energy functions. df/dx=df/du*du/dx

3. Equilibrium Types

4. Example 1
Consider the potential energy function shown. A particle is released at point A with the total given energy, and it is not subject to any nonconservative forces.
Identify all points of stable, unstable, and neutral equilibrium.
Stable Equilibrium = D,H
Unstable Equilibrium=E
Neutral Equilibrium = C,F
b) Where is the object moving with the greatest speed? Zero speed?
The greatest speed occurs at point D. At A and J, the particle has zero speed.
c) Is the motion bounded?
Yes
d) In what region or point is the magnitude of the force maximized?
In between points C and D
e) Within which region(s) is the force negative?
At points I and I. Also, between D and E

6. Problem 1
An object of mass m = 4kg has a potential energy function, U(x) =(x-2)-(2x-3)3
Determine the positions of points A and B, the equilibrium points.
If the object is released from point B, can it reach point A or C? Explain.
The particle is released at point C. Determine its speed as it passes point A.

7. Example on previous slide

8. 1a) dU/dx= 1-3(2x-3)2(2)=1-6(2x-3)2=1-6(4x2-12x+9) dU/dx=1-24x2+72x-54=-24x2+72x-53 F=24x2-72x+53 Let F =0 and solve for x using the quadratic formula. We get x = 1.3 m and x =1.7 m 1b) The particle can reach position A but does not have enough energy to reach position C 1c) Solve for potential energy at C, U(x)=(x-2)-(2x-3)3 U(.5)=6.5 J so E = 6.5 J since the particle has no kinetic energy at C Now solve for potential energy at A by plugging in 1.3 into the potential energy function, U(1.3)=-.636 J. Since E=6.5 J, we use E=KE+U to find that KE =7.1 J We use KE=1/2 mv2 to solve for v since we know m=4 kg and know KE =7.1J, thus v =1.89 m/s

9. Problem 2

11. Problem 3
A 2 kg mass moves under the influence of the potential energy U(x)=1/3(x-4)3 –x2/4 +5
Find the force exerted on the mass at x = 2m.
The mass is placed at x=4 and released, giving it a total energy of 1 J.
b) Describe the qualitative motion
c)Determine the approximate speed of the mass
d) Write an equation that would allow you to determine where the mass will momentarily come to rest.