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theorems:work-energy_theorem [2021/04/24 22:54]
cleonis Capital letters in the title
theorems:work-energy_theorem [2021/07/18 10:57] (current)
cleonis [Concrete] Reverted to the derivation with an intermediary stage, it is transparent whereas the version-with-less-steps wasn't
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 ====== Work-Energy theorem ====== ====== Work-Energy theorem ======
  
-<tabbox Intuitive> ​ 
- 
-<note tip> 
-Explanations in this section should contain no formulas, but instead colloquial things like you would hear them during a coffee break or at a cocktail party. 
-</​note>​ 
   ​   ​
 <tabbox Concrete> ​ <tabbox Concrete> ​
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 This exemplifies that it is very useful to have an equation that expresses force acting over distance. ​ This exemplifies that it is very useful to have an equation that expresses force acting over distance. ​
  
-In the case of the paddle wheel setup used by Joule the friction limits the rate at which the weight descends. Other experiments corroborate that the amount of potential energy release on descent does not depend on the velocity of descent. If an object falls without friction the velocity increases linear with time. That corroborates that for free fall also the potential energy release is released ​linear with height.+In the case of the paddle wheel setup used by Joule the friction limits the rate at which the weight descends. Other experiments corroborate that the amount of potential energy release on descent does not depend on the velocity of descent. If an object falls without friction the velocity increases linear with time. That corroborates that for free fall also the potential energy release is linear with height.
  
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 (1.1) is the derivative of (1.2), in that sense the two are not independent,​ but it's still possible to take advantage of the fact that there are two equations there. (1.1) is the derivative of (1.2), in that sense the two are not independent,​ but it's still possible to take advantage of the fact that there are two equations there.
  
-If an object is subject to uniform acceleration a over a distancs $(s - s_0)$ then how much change of velocity will that cause?+If an object is subject to uniform acceleration ​$aover a distancs $(s - s_0)$ then how much change of velocity will that cause?
  
 So, we want to start with acceleration and distance traveled as a given, and obtain velocity. That means we need to work towards an expression that contains position, acceleration and velocity, but not time.  So, we want to start with acceleration and distance traveled as a given, and obtain velocity. That means we need to work towards an expression that contains position, acceleration and velocity, but not time. 
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 $$ s = s_0 + v_0\frac{(v-v_0)}{a} + \tfrac{1}{2} a \frac{{(v-v_0)}^2}{a^2} \qquad (1.3) $$ $$ s = s_0 + v_0\frac{(v-v_0)}{a} + \tfrac{1}{2} a \frac{{(v-v_0)}^2}{a^2} \qquad (1.3) $$
  
-We mulitply ​everything out, and move terms for position and acceleration to the left. +We multiply ​everything out, and move terms for position and acceleration to the left.
  
 $$ a(s-s_0) = vv_0 - {v_0}^2 + \tfrac{1}{2}v^2 - vv_0 + \tfrac{1}{2} {v_0}^2 ​ \qquad (1.4) $$ $$ a(s-s_0) = vv_0 - {v_0}^2 + \tfrac{1}{2}v^2 - vv_0 + \tfrac{1}{2} {v_0}^2 ​ \qquad (1.4) $$
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 Repeating the three equations that are sufficient to define the Work-Energy theorem: Repeating the three equations that are sufficient to define the Work-Energy theorem:
- 
  
 $$ v = v_0 + at \qquad (1.1) $$ $$ v = v_0 + at \qquad (1.1) $$
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 $$ F=ma $$ $$ F=ma $$
  
- +$F=ma$ and the work-energy theorem are very closely connected indeed. By taking the derivative with respect to position $F=ma$ is recovered immediately
-$F=ma$ and the work-energy theorem are very closely connected indeed.+
  
 \\  \\ 
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 The following derivation is more abstract. This derivation is from the outset designed to accomodate an arbitrary acceleration profile. The following derivation is more abstract. This derivation is from the outset designed to accomodate an arbitrary acceleration profile.
  
 +The two relations that constitute the mathematics of the Work-Energy theorem:
  
-In the course of the derivation the following two relations will be used:+$$ ds = v \ dt \qquad (2.1) $$
  
-$$ ds = v \ dt  \qquad (2.1)  $$ +$$ dv = \ dt \qquad (2.2) $$
- +
-$$ dv = \frac{dv}{dt}dt  \qquad (2.2)   ​$$+
  
  
 The integral for acceleration from a starting point $s_0$ to a final point $s$. The integral for acceleration from a starting point $s_0$ to a final point $s$.
- 
  
 $$ \int_{s_0}^s a \ ds \qquad (2.3)  $$ $$ \int_{s_0}^s a \ ds \qquad (2.3)  $$
  
-Use (11) to change the differential ​from *ds* to *dt*Since the differential is changed the limits ​change accordingly.+Intermediary step: change ​of the differential ​according ​to (2.1), with corresponding change of limits.
  
 $$ \int_{t_0}^t a \ v \ dt \qquad (2.4) $$ $$ \int_{t_0}^t a \ v \ dt \qquad (2.4) $$
  
-Rearrange ​the order, and write the acceleration $a$ as $\tfrac{dv}{dt}$+Change ​the order:
  
-$$ \int_{t_0}^t v \ \frac{dv}{dt} \ dt  ​\qquad (2.5) $$+$$ \int_{t_0}^t v \ \ dt \qquad (2.5) $$
  
-Use (2.2) for a second ​change of differential,​ again the limits ​change accordingly.+Change of differential according to (2.2), with corresponding ​change of limits.
  
-$$ \int_{v_0}^v v \ dv  \qquad (2.6)  $$+$$ \int_{v_0}^v v \ dv \qquad (2.6) $$
  
 Putting everything together: Putting everything together:
  
-$$ \int_{s_0}^s a \ ds = \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2 ​ \qquad (2.7) $$+$$ \int_{s_0}^s a \ ds = \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2 \qquad (2.7) $$
  
  
-Combining ​with $F=ma$ ​gives the work-energy theorem +Multiply both sides with $m$ and apply $F=ma$ ​to arrive at the work-energy theorem:
- +
-$$ \int_{s_0}^s F \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2 ​ \qquad (2.8) $$+
  
 +$$ \int_{s_0}^s F \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2 \qquad (8) $$
  
 +\\ 
  
 This form of the work-energy theorem emphasizes that in order to define a potential energy an integral must be defined. ​ This form of the work-energy theorem emphasizes that in order to define a potential energy an integral must be defined. ​
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 </​note>​ </​note>​
  
-<tabbox Why is it interesting?> ​  ​+<tabbox Why is it interesting?> ​ 
 + 
 +As stated in the lead up to the derivation: the Work-Energy theorem is both old and new.   
 + 
 +The Work-Energy theorem is the basis of the formulation of mechanics that was developed by Joseph Louis Lagrange, called '​Lagrangian mechanics'​. But at the time the Work-Energy theorem did not yet exist in its current form. The modern formulation of the Work-Energy theorem is not present in Lagrange'​s work. The form that Lagrange used is called d'​Alembert'​s virtual work. 
 + 
 +**Kinetic energy and Lagrangian mechanics** 
 + 
 +Motion of objects is represented with coordinates of a coordinate system. In a system of cartesian coordinates a velocity vector can be decomposed in components along the axes of the cartesian coordinate system. In order to represent momentum in a cartesian coordinate system with three dimensions the momentum is decomposed along the three axes of the coordinate system. Likewise the force is decomposed along the three axes. Newton'​s second law is valid for each component of motion. 
 + 
 +$ \vec{F_x} = m \vec{a_x} $ \\  
 +$ \vec{F_y} = m \vec{a_y} $ \\  
 +$ \vec{F_z} = m \vec{a_z} $ \\  
 + 
 +Kinetic energy is proportional to the square of velocity. This means that the sum of the component kinetic energies (along the coordinate axes) is equal to the kinetic energy of the undecomposed velocity (pythagoras'​ theorem). This enables the following: in Lagrangian mechanics the direction of the velocity vector is discarded. The kinetic energy is treated as a //scalar//.  
 + 
 +The directional information of the velocity vector can be discarded because that information is still available in the expression for the //potential energy//. The potential energe is the integral of the force over distance. When the calculation is for motion in three dimensions of space then the expression for the potential energy has three components, one for each spatial dimension. 
 + 
 +Kinetic energy can be treated as a scalar without loss of expressive power; that is what Joseph Louis Lagrange capitalized on when he developed Lagrangian mechanics. 
 + 
 +The second innovation from Lagrange was systematic use of generalized coordinates. The importance of being able to use generalized coordinates cannot be overstated. 
 +  ​
  
 /​*<​tabbox FAQ>​*/ ​ /​*<​tabbox FAQ>​*/ ​
theorems/work-energy_theorem.1619297655.txt.gz · Last modified: 2021/04/24 22:54 by cleonis