Difference between revisions of "Forest UCM Energy Line1D"
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:: <math> = \int \pm \sqrt{\frac{m}{2\left (0-(-mgx) \right )}} dx </math> | :: <math> = \int \pm \sqrt{\frac{m}{2\left (0-(-mgx) \right )}} dx </math> | ||
:: <math> = \int \pm \sqrt{\frac{1}{2gx}} dx = \int \pm (2gx)^{-\frac{1}{2}}dx </math> | :: <math> = \int \pm \sqrt{\frac{1}{2gx}} dx = \int \pm (2gx)^{-\frac{1}{2}}dx </math> | ||
− | :: <math> = \pm (2g)^{-\frac{1}{2}} \sqrt x = \sqrt{\frac{2x}{g} } </math> | + | :: <math> = \pm (2g)^{-\frac{1}{2}} 2\sqrt x = \sqrt{\frac{2x}{g} } </math> |
== spring example== | == spring example== |
Revision as of 15:43, 26 September 2014
The equation of motion for a system restricted to 1-D is readily solved from conservation of energy when the force is conservative.
- cosntant
The ambiguity in the sign of the above relation, due to the square root operation, is easily resolved in one dimension by inspection and more difficult to resolve in 3-D.
The velocity can change direction (signs) during the motion. In such cases it is best to separte the inegral into a part for one direction of the velocity and a second integral for the case of a negative velocity.
Free fall
Consider a rock dropped at t=0 from a tower of height h.
The potential energy stored in the rock at any instant is given by
- Note
- The potential is highest at x=0 and becomes negative as x increases
The initial total energy is
spring example
Consider the problem of a mass attached to a spring in 1-D.
The potential is given by
let
- and
then
- = amplitude of oscillating motion