Potential Energy
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 × 10 9 J of chemical potential energy.)
solution
State the givens and the unknown.
| h = | 415 m |
| m = | 2(550,000,000 kg) = 1.1 × 10 9 kg |
| Ug = | ? |
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 ( λ = m/h ) 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.
| Ug = | (1.1. × 10 9 kg)(9.8 m/s 2 )(415 m) |
| 2 |
Answer out (in joules).
Convert to the TNT equivalent.
| Ug = | 2.2 × 10 12 J |
| 4.184 × 10 9 J/ton TNT |
| Ug = 530 ton tnt |
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 (since 500 is 1,000 times 0.5).
practice problem 4
A region of space has the following two-dimensional potential energy function…
U(x, y) = x 4 + y 4 + 2x 2 y 2 − 8x 2 + 8y 2 + 16
- points of stable and unstable equilibrium
- range of bound states (not the right term for what I'm thinking of)
solution
- 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) = | x 4 + y 4 + 2x 2 y 2 − 8x 2 + 8y 2 + 16 |
| ∂U | = 4x 3 + 0 + 4xy 2 − 16x + 0 + 0 = 0 |
| ∂x |
| ∂U | = 0 + 4y 3 + 4x 2 y − 0 + 16y + 0 = 0 |
| ∂y |
Clean up and simplify the results.
| x 3 + xy 2 − 4x = 0 |
| y 3 + x 2 y + 4y = 0 |
Factor each equation.
| x(x 2 + y 2 − 4) = 0 |
| y(y 2 + x 2 + 4) = 0 |
| x = 0 |
| or |
| x 2 + y 2 − 4 = 0 |
| y = 0 |
| or |
| y 2 + x 2 + 4 = 0 |
The last of these conditions is impossible to satisfy using real numbers.
This expression has no real solutions. Therefore y = 0 and nothing else. Which means that the statement…
| x = 0 |
| or |
| x 2 + y 2 − 4 = 0 |
should really be stated as…
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) = | x 4 + y 4 + 2x 2 y 2 − 8x 2 + 8y 2 + 16 |
| ∂ 2 U | = 12x 2 + 0 + 4y 2 − 16 + 0 + 0 |
| ∂x 2 |
| ∂ 2 U | = 0 + 12y 2 + 4x 2 − 0 + 16 + 0 |
| ∂y 2 |
Clean up and simplify the results.
| ∂ 2 U | = 12x 2 + 4y 2 − 16 |
| ∂x 2 |
| ∂ 2 U | = 12y 2 + 4x 2 + 16 |
| ∂y 2 |
Let's test each of the three stationary points. Start with (−2, 0)…
| ∂ 2 U | = 12( −2 ) 2 + 4(0) 2 − 16 = +32 |
| ∂x 2 |
| ∂ 2 U | = 12(0) 2 + 4( −2 ) 2 + 16 = +32 |
| ∂y 2 |
| ∂ 2 U | = 12(+2) 2 + 4(0) 2 − 16 = +32 |
| ∂x 2 |
| ∂ 2 U | = 12(0) 2 + 4(+2) 2 + 16 = +32 |
| ∂y 2 |
The second derivative in both coordinate directions at both locations 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 (0, 0), however …
| ∂ 2 U | = 12(0) 2 + 4(0) 2 − 16 = −16 |
| ∂x 2 |
| ∂ 2 U | = 12(0) 2 + 4(0) 2 + 16 = +16 |
| ∂y 2 |
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 |
| U(x, y) = | x 4 + y 4 + 2x 2 y 2 − 8x 2 + 8y 2 + 16 |
| U(0, 0) = | 0 + 0 + 0 − 0 + 0 + 16 |
| U(0, 0) = | +16 |
