Forest UCM Energy Line1D - Revision history

Foretony at 12:18, 1 October 2014

2014-10-01T12:18:07Z

2014-09-26T15:44:17Z

spring example

2014-09-26T15:43:35Z

Free fall

2014-09-26T15:43:07Z

Free fall

2014-09-26T15:42:43Z

Free fall

2014-09-26T15:41:02Z

Free fall

2014-09-26T15:15:12Z

Free fall

2014-09-26T15:13:48Z

spring example

2014-09-26T12:46:50Z

spring example

2014-09-26T12:43:04Z

spring example

@@ Line 1: / Line 1: @@
 The equation of motion for a system restricted to 1-D is readily solved from conservation of energy when the force is conservative.
-: <math>T + U(x) =</math> cosntant <math>\equiv E</math>
+: <math>T + U(x) =</math> constant <math>\equiv E</math>
 : <math>\Rightarrow T = E - U(x)</math>

@@ Line 39: / Line 39: @@
 :<math>x = \frac{1}{2} gt^2</math>
-== spring example==
+== spring example (problem 2.8)==
 Consider the problem of a mass attached to a spring in 1-D.

@@ Line 33: / Line 33: @@
 :: <math> = \int \pm \sqrt{\frac{1}{2gx}} dx  = \int \pm (2gx)^{-\frac{1}{2}}dx  </math>
 :: <math> =  \pm (2g)^{-\frac{1}{2}} 2\sqrt x =  \sqrt{\frac{2x}{g} } </math>
 == spring example==

@@ Line 32: / Line 32: @@
 :: <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> =  \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==

@@ Line 32: / Line 32: @@
 :: <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>
 == spring example==

@@ Line 29: / Line 29: @@
 :<math>E_{tot} = T + U = 0 -0 = 0</math>
-: <math>t = \int \pm \sqrt{\frac{m}{2\left (E-U(x) \right )}} dx = \int dt  </math>
+: <math>t = \int \pm \sqrt{\frac{m}{2\left (E-U(x) \right )}} dx  </math>
-:: <math> = \int \pm \sqrt{\frac{m}{2\left (0-(-mgx) \right )}} dx = \int dt  </math>
+:: <math> = \int \pm \sqrt{\frac{m}{2\left (0-(-mgx) \right )}} dx  </math>
+:: <math> = \int \pm \sqrt{\frac{1}{2gx}} dx  </math>
 == spring example==

← Older revision		Revision as of 15:13, 26 September 2014
Line 14:		Line 14:

	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.		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
		+
		+	<math>U(x) = -mgx</math>
		+
		+	;Note: The potential is highest at x=0 and becomes negative as x increases
		+
		+	The initial total energy is
		+
		+	:<math>E_{tot} = T + U = 0 -0 = 0</math>
		+	: <math>t = \int \pm \sqrt{\frac{m}{2\left (E-U(x) \right )}} dx = \int dt </math>
		+	:: <math> = \int \pm \sqrt{\frac{m}{2\left (0-(-mgx) \right )}} dx = \int dt </math>
		+
		+
		+

@@ Line 48: / Line 48: @@
 :<math>\sqrt{\frac{2E}{m}} \sin \theta = \sqrt{\frac{2E}{m}} \sin {\left (\omega t + \theta_0 \right )}</math>
 :<math>x = \sqrt{\frac{2E}{m}} \sin {\left (\omega t + \theta_0 \right )}</math>
 [[Forest_UCM_Energy#Energy_for_Linear_1-D_systems]]

@@ Line 46: / Line 46: @@
 :<math> \theta = \omega t + \theta_0 </math>
 :<math> \sin \theta = \sin {\left (\omega t + \theta_0 \right )}</math>
-:<math>\sqrt{\frac{2E}{m}} \sin \theta = \sqrt{\frac{2E}{m}} \sin {\omega t + \theta_0}</math>
+:<math>\sqrt{\frac{2E}{m}} \sin \theta = \sqrt{\frac{2E}{m}} \sin {\left (\omega t + \theta_0 \right )}</math>
-:<math>x = \sqrt{\frac{2E}{m}} \sin {\omega t + \theta_0}</math>
+:<math>x = \sqrt{\frac{2E}{m}} \sin {\left (\omega t + \theta_0 \right )}</math>
 [[Forest_UCM_Energy#Energy_for_Linear_1-D_systems]]