1 / 13

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

Aim: How do we analyze energy diagrams and stability of equilibrium?. Reminder of Chain Rule. We will have to calculate the derivative of more complicated potential energy functions. df/dx=df/du*du/dx. Equilibrium Types. Example 1.

bchaparro
Download Presentation

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

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

  5. 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.

  6. Example on previous slide

  7. 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

  8. Problem 2

  9. 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.

More Related