Potential Energy

Practice

practice problem 1

Write something.

solution

Answer it.

practice problem 2

Write something else.

solution

Answer it.

practice problem 3

Calculate the gravitational potential energy released by the collapse of the World Trade Center in New York City on 11 September 2001. Each 110 story tower had a mass of about 550,000,000 kg and a height of 415 m (not including the broadcast tower). Compare this to the energy released on 8 March 1993 when a truck carrying a fertilizer bomb exploded in the underground parking garage of this same complex. Assume an explosive yield equivalent to a half ton of TNT. (One ton of TNT has 4.184 × 109 J of chemical potential energy.)

solution

Since the mass of the towers is distributed throughout their volume and is not concentrated at a point, this problem is best solved using calculus. Assume a uniform linear mass density and integrate the potential energy formula over the height of the towers. Not surprisingly, the results show that the center of mass of a tower lies at its geometric center, halfway up.

h =  415 m
m =  2(550,000,000 kg) = 1.1 × 109 kg
h h
Ug =  ⌠
⌡		  ρgy dy =  m  g  ⌠
⌡		  y dy =  mg   h2  =  mgh
h h 2 2
0 0
Ug =  (1.1. × 109 kg)(9.8 m/s2)(415 m)
2
Ug =  2.2 × 1012 J 1 ton TNT  = 530 ton TNT
4.184 × 109 J

The gravitational energy released when the towers collapsed was thus about a thousand times greater than the chemical energy released when the truck bomb exploded.

practice problem 4

A region of space has the following two-dimensional potential energy function …

U(x, y) = x4 + y4 + 2x2y2 − 8x2 + 8y2 + 16

Find the …
  1. points of stable and unstable equilibrium
  2. range of bound states (not the right term for what I'm thinking of)

solution

  1. Stable equilibrium occurs at local minima. Unstable equilibrium occurs at local maxima. Minima and maxima are local extrema and occur where the value of the first derivative is zero. Let's find those points.
     U(x, y)  =    x4   +    y4   +   2 x2y2   −   8 x2   +   8 y2   +   16  
     
      ∂U  =   4 x3   +   0   +   4 xy2   −   16 x   +   0   +   0   = 0
      ∂x
      ∂U  =   0   +   4 y3   +   4 x2y   −   0   +   16 y   +   0   = 0
      ∂y

    Clean up and simplify the results.

    x3 + xy2 − 4x = 0
    y3 + x2y + 4y = 0

    Factor each equation.

    x(x2 + y2 − 4) = 0
    y(y2 + x2 + 4) = 0

    Therefore …

    x = 0 or x2 + y2 − 4 = 0

    and …

    y = 0 or y2 + x2 + 4 = 0

    The last of these conditions is impossible to satisfy using real numbers.

    y2 + x2 + 4 = 0

    This expression has no real solutions. Therefore y = 0 and nothing else. Which means that the statement …

    x = 0 or x2 + y2 − 4 = 0

    should really be stated as …

    x = 0 or x2 − 4 = 0

    and therefore

    x = 0, +2, −2.

    A point with zero derivative could be a minimum, but it could also be a maximum or a saddle point. Use the second derivative to determine the type of stationary point.

     U(x, y)  =    x4   +    y4   +   2 x2y2   −   8 x2   +   8 y2   +   16  
     
      ∂2U  =   12 x2   +   0   +   4 y2   −   16    +   0   +   0   
      ∂x2
      ∂2U  =   0   +   12 y2   +   4 x2   −   0   +   16    +   0   
      ∂y2

    Clean up and simplify the results.

    ∂2U  = 12x2 + 4y2 − 16   ∂2U  = 12y2 + 4x2 + 16
    ∂x2 ∂y2

    Now test each of the three stationary points starting first with (±2, 0).

    (−2, 0)   ∂2U  = 12(−2)2 + 4(0)2 − 16 = +32   ∂2U  = 12(0)2 + 4(−2)2 + 16 = +32
    ∂x2 ∂y2
    (+2, 0)   ∂2U  = 12(+2)2 + 4(0)2 − 16 = +32   ∂2U  = 12(0)2 + 4(+2)2 + 16 = +32
    ∂x2 ∂y2

    The second derivative in both coordinate directions is positive, which means the curvature is concave up in every direction and that these two points are local minima or points of stable equilibrium.

    At the origin, however, …

    (0, 0)   ∂2U  = 12(0)2 + 4(0)2 − 16 = −16   ∂2U  = 12(0)2 + 4(0)2 + 16 = +16
    ∂x2 ∂y2

    The second derivative is negative along the x-axis, which means the surface curves down, but it is positive along the y-axis, which means the surface curves up. What we have here is a saddle point. The location is unstable in the ±x directions and stable in the ±y directions.

    To summarize …

    coordinates stationary point equilibrium
    (−2, 0) local minimum stable in all directions
    (0, 0) saddle point unstable along the x-axis
    stable along the y-axis
    (+2, 0) local minimum stable in all directions
     
  2. We could continue trying to visualize this potential energy surface in our heads or we could use some appropriate technology and view a computer rendered image. I think we've laid out enough basics to give us a head start. Our potential energy surface has two wells lying on the x-axis on opposite sides of the origin and a saddle point in the middle.


    [magnify]

    I think it looks something like a pair of pants, so I've dubbed it the "trouser potential". This is my personal name and is not something you should memorize for a test.

    The whole thing is one giant well. The saddle point at the origin divides the potential surface into two domains for values below 16 of our arbitrary units.

     U(x, y)  =    x4   +    y4   +   2 x2y2   −   8 x2   +   8 y2   +   16  
     
     U(0, 0)  =   0   +   0   +   0   −   0   +   0   +   16  
     
     U(0, 0)  =   +16   
     

    Come back and finish this later.

The Physics Hypertextbook

  • No condition is permanent.