1
$\begingroup$

Does it make sense to model the kinetic energy as:

$$T=\frac{1}{2}mv\,|v| \ \ \ \text{instead} \ \text{of} \ \ \ T=\frac{1}{2}mv^2 \ \ \ ?$$

In the following youtube video, it is explained for an airfoil, why the classic Bernoulli Law description is not enough to explain how much lift is generated in reality. I am not interested in the topic of the video "per se", but in the construction of his explanation. The author uses the Bernoulli Equation to group the terms associated with the kinetic energy of the fluid. After some manipulation, using it to built the Drag Force proportional to a quadratic term of the velocity, giving intuition of why it is modeled as:

$$F_d=\frac{1}{2}\rho A\,C_d\,v^2$$

Now, I would like to review a simple and well known example of a physics model, where the drag force is considered, the classic nonlinear pendulum with friction.

If the classic nonlinear pendulum with the friction equation is reviewed, where the drag force is modeled proportional to the speed, as in Stokes' Law, in Wolfram-Alpha. It can be seen that it has decaying solutions as expected:

$$\ddot{x}+2\cdot0.021\,\dot{x}+0.2\sin(x)=0, \ \ \ \\ x(0)=\frac{\pi}{2}, \ \ \ \ \dot{x}(0)=0 \tag{Eq. 1}\label{Eq. 1}$$

Equation 1

If instead, the standard drag force $F_{\text{drag}}\propto (\dot{x})^2$ is used as is shown here for the equation:

$$\ddot{x}+0.021(\dot{x})^2+0.2\sin(x)=0, \ \ \ \ \\ x(0)=\frac{\pi}{2}, \ \ \ \ \ \dot{x}(0)=0 \tag{Eq. 2}\label{Eq. 2}$$

Equation 2

their solution isn't showing the expected decay one can see on the experimental pendulums.

This issue could be solved using an ansatz for the drag force $F_{\text{drag}}\propto \dot{x}|\dot{x}|$ (following this reference equation $1.127$), which as can be seen here for the equation:

$$\ddot{x}+0.021\dot{x}|\dot{x}|+0.2\sin(x)=0, \ \ \ \ \\ x(0)=\frac{\pi}{2}, \ \ \ \ \ \dot{x}(0)=0 \tag{Eq. 3}\label{Eq. 3}$$

Equation 3

their solution has recovered again the expected decaying behavior for a pendulum with friction.

Now, given the close relation between the kinetic energy and the drag force shown on the video, and due the failing of the classic form of the drag force to reproduce the decaying solutions of a pendulum with friction, I would like to know if make sense to consider the kinetic energy as:

$$T=\frac{1}{2}mv\,|v| \ \ \ \text{instead} \ \text{of} \ \ \ T=\frac{1}{2}mv^2$$

I hope you could explain why that is so, and if there are any examples, sharing them and explaining how the kinetic energy term $T\propto v\,|v|$ has arisen. For a negative answer, please elaborate, explaining why the standard equation of the drag force shown in Wikipedia fails at describing the pendulum with friction as shown in \eqref{Eq. 2} but it works for \eqref{Eq. 3} (in this paper at point $III$ is even solved piecewise).


Added Later

These are the calculation done in the mentioned youtube video:

$$\begin{array}{l} \text{velocities on upper/lower sides of the airfoil}\quad v_1=\frac{d_1}{t_1}; \quad v_2=\frac{d_2}{t_2}; \\ \text{assumption}\quad t_1=t_2\ \ \textit{(mistaken)}\\ \text{Bernoulli Equation} \qquad P_1 + \rho g h + \underbrace{\frac{1}{2}\rho v_1^2}_{\text{kinetic energy}} = P_2 + \rho g h + \underbrace{\frac{1}{2}\rho v_2^2}_{\text{kinetic energy}} \\ \Rightarrow \Delta P = \frac{1}{2}\rho\left(v_2^2-v_1^2\right) = \frac{1}{2}\rho\left(\left(\frac{v_2+v_1}{v_1}\right)\left(\frac{v_2-v_1}{v_1}\right)\right)v_1^2 = \frac{1}{2}\rho\left(\left(\frac{v_2}{v_1}+1\right)\left(\frac{v_2}{v_1}-1\right)\right)v_1^2 \\ \Rightarrow \Delta P = \frac{1}{2}\rho\left(\left(\frac{d_2}{d_1}+1\right)\left(\frac{d_2}{d_1}-1\right)\right)v_1^2 = \frac{1}{2}\rho\left(\left(\frac{d_2}{d_1}\right)^2-1\right)v_1^2 \\ \text{multiplying both sides by area} \Rightarrow \underbrace{A\Delta P}_{F_d} = \frac{1}{2}\rho A \left(\left(\frac{d_2}{d_1}\right)^2-1\right) v_1^2 \underbrace{\propto}_{\text{proportional}} \frac{1}{2}\rho A\,C_d\,v_1^2 \end{array}$$

since from the kinetic energy part of the Bernoulli Equation the video find the Drag Force for the example, I made the pairing with the version were the absolute value is used. Hope this better explain why the question arises.


2nd added later

I found this paper:

where the autors on equations $(59)$ and $(60)$ introduce a Drag Force $F_d = b\ \dot{y} + c\ \dot{y}|\dot{y}|$ for a more accurate description of the air effects on the pendulum. I don't know if it add info about the Kinetic Energy question, but it make sense of the signs required for the proper description of the drag force.

$\endgroup$
8
  • 4
    $\begingroup$ As written this suggests that definitions of physical quantities are somehow not grounded in anything other than the whims of physicists. $\endgroup$ Jul 5 at 0:39
  • $\begingroup$ Sign is arbitrary if applied consistently to everything, but being able to take direction, even in one dimension (plus or minus) changes the meaning. $\endgroup$
    – g s
    Jul 5 at 0:51
  • 1
    $\begingroup$ I rapidly (but not closely) read the (very interesting) chapter you link to, and I don't see anything suggesting that $T$ should be modelled as you suggest. In fact, since there are damped oscillations, it seems that you need the usual definition of $T$ is you are to recover oscillations in the low drag limit. As explained in the text there are several models for drag so I don't think it would be wise use drag as a starting point to upend the definition of $T$. $\endgroup$ Jul 5 at 9:42
  • 1
    $\begingroup$ Caution as $T \geq 0$ is a requirement. $\endgroup$
    – JAlex
    Jul 5 at 12:11
  • 1
    $\begingroup$ Also, note that if you're going to make an expression like this work (say by taking $v = {\rm sign}({\vec v}\cdot {\hat j})$), you do make the value of the kinetic energy path-dependent. Which may be a thing you want, but generally is not how kinetic energy behaves. $\endgroup$ Jul 5 at 15:01

7 Answers 7

14
$\begingroup$

No, because kinetic energy is not a vector, so we cannot have it be proportional to a vector. Kinetic energy is a scalar quantity, and so $T=\frac12m\mathbf v\cdot|v|$ cannot be used. The drag force is a vector, and it's related to the direction of $\mathbf v$, so this is why we have the expression you give in your post.

In general, this shows issues with questions like these asking "why can't we use this expression for this physical quantity instead?" The problem is that you aren't linking this question to any other definition of kinetic energy, and so the question misses the purpose of definitions. "Kinetic energy" is the quantity whose change is equal to the net work done. From this, we find $T=\frac12mv^2$ is the expression that describes this definition.

If you have another understanding of kinetic energy, you should ask why $T=\frac12m\mathbf v\cdot|v|$ can't be used to describe the definition you have in mind. Otherwise there isn't much else to be said.

$\endgroup$
4
  • $\begingroup$ Isn't kinetic energy in vector form $$T = \int \boldsymbol{v} \cdot {\rm d} \boldsymbol{p} = \tfrac{1}{2} m \boldsymbol{v} \cdot \boldsymbol{v}$$ where $\boldsymbol{p} = m \boldsymbol{v}$ is the momentum? $\endgroup$
    – JAlex
    Jul 5 at 12:15
  • 1
    $\begingroup$ @JAlex Yes, but that's still not a vector quantity. I think I'm missing your point. Did you want me to explicitly state that $v^2=\mathbf v\cdot\mathbf v$? $\endgroup$ Jul 5 at 13:03
  • $\begingroup$ Yes I was eluding that scalar $T$ is calculated from the vectors of velocity and momentum. A result is that $T \geq 0 $ always, since $m>0$ always, something the proposed expression does not guarantee. $\endgroup$
    – JAlex
    Jul 5 at 13:52
  • $\begingroup$ Thanks for answering. As nonsense it is for the vectorial scenario as you (an others) have mentioned, for the scalar case @Qmechanic have linked in his answer a really interesting example where the form $1/2m\ v|v|$ is used for the kinetic energy on an answer. Hope you can see it and comment. $\endgroup$
    – Joako
    Jul 7 at 5:03
3
$\begingroup$

Drag force
The reason for failure of the drag force equation, is that the drag force should be directed against the velocity, $\dot{x}$. The Stoke's term, linear in velocity (as well as any odd power of velocity) satisfies this condition $$ F_{2n+1}(\dot{x})=-\gamma\dot{x}^{2n+1}=-F_{2n+1}(-\dot{x}). $$ On the other hand, an even power of $\dot{x}^2$ requires a bit more involved math, as, e.g., $$ F_{2n}(\dot{x})=-\gamma\text{sign}(\dot{x})\dot{x}^{2n}= =-\gamma\dot{x}|\dot{x}|^{2n-1}=-F_{2n}(-\dot{x}). $$

Energy
The reasons that the energy is a scalar are quite different - it is the first integral of the equations of motion. E.g., if we take Newton's equation for a particle in a conservative field $$ m\ddot{\mathbf{r}}=-\nabla U(\mathbf{r}), $$ multiply it by $\dot{\mathbf{r}}$ and perform some algebraic transformations, we obtain: $$ 0=\left[m\ddot{\mathbf{r}} +\nabla U(\mathbf{r})\right]\dot{\mathbf{r}}= m\ddot{\mathbf{r}}\dot{\mathbf{r}} +\nabla U(\mathbf{r})\dot{\mathbf{r}}= \frac{d}{dt}\left[\frac{m\dot{\mathbf{r}}^2}{2}+U(\mathbf{r})\right]=\frac{d}{dt}E(t) $$ That is, the quantity $E(t)=\frac{m\dot{\mathbf{r}}^2}{2}+U(\mathbf{r})$ does not change with time (i.e., conserved) along the trajectories described by the equation of motion. This is just a mathematical fact - there is no much flexibility in adjusting it.

$\endgroup$
3
  • $\begingroup$ Thanks for the answer. I think I have a confusion with the direction of the "speed": in some Bernoulli examples is the fluid which is moving over an stationary object, different in direction of the pendulum example where is an object moving through a stationary fluid... since the force must be against the speed of the "moving object" in the frame of reference of the fluid... Does it means that Wikipedia example should be instead $|F_d| = \frac{1}{2}\rho\ A\ C_d\ v^2$??? $\endgroup$
    – Joako
    Jul 7 at 1:30
  • $\begingroup$ I have added in the question a paper where an accurate drag force is introduced, which behaves as you mention (I believe it could be improved, which I am asking here, if you are interested to look at it). $\endgroup$
    – Joako
    Jul 7 at 15:39
  • $\begingroup$ Also, in another answer by @Qmechanic, is linked an answer where a kinetic energy of the form $\propto v|v|$ is used, don't knowing if is right but is interesting. Hope you can see it and comment. $\endgroup$
    – Joako
    Jul 7 at 15:44
3
$\begingroup$
  1. The drag force $${\bf F}_{\rm drag}~=~-f(v^2)~ {\bf v}, \qquad f(v^2)~>~0, \tag{1} $$ is always directed opposite of the velocity ${\bf v}$:

    1. Linear friction/drag corresponds to a constant $f$-function.

    2. The $f$-function is a square root for quadratic drag. In 1D this becomes an absolute value, cf. OP's post.

  2. It is unclear what a quadratic drag force ${\bf F}_{\rm drag}\propto-\dot{x}|\dot{x}|$ has to do with a signed kinetic energy $\frac{1}{2}m\dot{x}|\dot{x}|$ in 1D, cf. OP's post.

    Nevertheless, a signed kinetic energy is e.g. useful in the 1D double-integrator, cf. this Phys.SE post.

$\endgroup$
1
  • $\begingroup$ Thanks for the answer. I have added now the equations of the video were the pairing between the kinetic energy and the drag force is introduced. Is really interesting the answer to the post you mentioned, where it is introduced a kinetic energy of the form $1/2\ m\ v\ |v|$, but I don't understand the physical interpretation: It is like the energy of a dissipative force in an open system? (so it could be negative since it extract energy of the process). Hope you can elaborate into this. $\endgroup$
    – Joako
    Jul 7 at 4:57
2
$\begingroup$

Kinetic energy is not a vector but a scalar. Energy as a scalar makes sense as it is only with that definition a conservative quantity:

Consider a normal pendulum where the total energy remains constant: $$ E_{tot} = E_{kin} + E_{pot} = \frac{1}{2}mv^2 + mgh = \mathrm{const}$$

Now, if one would use your definition of kinetic energy, such equation would not work anymore: for one, the potential energy is not a vector, and if it were, what direction would it be? Downward? Either way, the total energy of the system would constantly change and would be opposite when the pendulum swings left to right as compared to swinging back from right to left. While this is true for velocity or momentum, it is not useful in terms of energetic discussions.

$\endgroup$
1
  • $\begingroup$ Thanks for answering. As nonsense it is for the vectorial scenario as you (an others) have mentioned, for the scalar case @Qmechanic have linked in his answer a really interesting example where the form $1/2m\ v|v|$ is used for the kinetic energy on an answer. Hope you can see it and comment. $\endgroup$
    – Joako
    Jul 7 at 5:06
1
$\begingroup$

No, it can't. Unlike the quantity $T=\frac{1}{2}m|\mathbf{v}|^2$, which we call kinetic energy, the quantity $\frac{1}{2}m|\mathbf{v}|\mathbf{v}$ is useless. It does not appear in any physical application nor it can be used to simplify any physics problem. You must understand that all physical quantities are named simply because they are useful, nothing more than that. If a quantity is useless, it will not have a name.

$\endgroup$
1
  • $\begingroup$ Thanks for answering. As nonsense it is for the vectorial scenario as you (an others) have mentioned, for the scalar case @Qmechanic have linked in his answer a really interesting example where the form $1/2m\ v|v|$ is used for the kinetic energy on an answer. Hope you can see it and comment. $\endgroup$
    – Joako
    Jul 7 at 5:05
1
$\begingroup$

assume one dimensional case hence the velocity v is scalar $~v=\dot x~$

and the kinetic energy T is:

$$ T=\frac m2\,\dot x^2$$

you obtain the equation of motion

$$m\,\ddot x=-F_d$$ where $~F_d~$ is the drag force

this is also the Newton equation of motion

now your kinetic energy ansatz

$$T=\frac m2\,\dot x\,|\dot x|$$

the equation of motion is

$$ m\,\ddot x=-F_d \quad\text{if}\quad \dot x > 0$$ and $$ m\,\ddot x=+F_d \quad\text{if}\quad \dot x < 0$$

but the second equation is wrong , hence your ansatz for the kinetic energy is wrong.


if the velocity v is a vector then $~T=\frac m2 \vec v\cdot\vec v~$ scalar

but with your ansatz $~T=\frac m2 |v|\,\vec v~$ the kinetic energy is not a scalar

$\endgroup$
3
  • $\begingroup$ Thanks for answering. As nonsense it is for the vectorial scenario as you (an others) have mentioned, for the scalar case @Qmechanic have linked in his answer a really interesting example where the form $1/2m\ v|v|$ is used for the kinetic energy on an answer. Hope you can see it and comment. $\endgroup$
    – Joako
    Jul 7 at 5:05
  • $\begingroup$ @Joako in the answer from Qmechanic the force is multiply with $~sgn(\dot x)~$, so your kinetic energy ansatz is correct if $~F_d\mapsto F_d\,sgn(v)~$ thus if $~ v > 0~,T=\frac m2\,v^2~,U=-F_d\,x~$ and if $~ v < 0~,T=-\frac m2\,v^2~,U=F_d\,x~$ both cases the equation of motion are $~m\,\ddot x+F_d=0~$ $\endgroup$
    – Eli
    Jul 7 at 14:35
  • $\begingroup$ Sorry but I don't fully understand what you are explaining: I have incorporated in the question a paper where an accurate modeling of the drag force is used, where the signs are as on the Stokes' Law and the $v|v|$ scenario... Does this fulfill what you are saying about the force on the answer of @Qmechanic?... Does the kinetic energy used there make sense or it is just a control theory construction? $\endgroup$
    – Joako
    Jul 7 at 15:21
1
$\begingroup$

Two no's:

  1. Kinetic energy form is such $K=\frac {1}{2}mv^2$,- not because we like it to see like that, but because of relationship between elementary mechanical work and force applied $\to dW = F \cdot dr$ , the rest is just a conclusion of integration procedure, i.e. mathematical proof.

  2. If you would make kinetic energy a vector in the form $\vec v \cdot |v|$, then energy conservation law would not hold anymore. Consider an ideal elastic collision of billiard ball with a steady wall. When incoming to wall - it would have kinetic energy K, and after perpendicular collision $-K$, so energy would swap for no reason. (However speed, and momentum vectors can swap). Scalar kinetic energy is very useful tool with respect to energy conservation law. If you would break that,- you would have to re-formulate 99% thermodynamics laws as well.

$\endgroup$
1
  • $\begingroup$ Thanks for answering. As nonsense it is for the vectorial scenario as you (an others) have mentioned, for the scalar case @Qmechanic have linked in his answer a really interesting example where the form $1/2m\ v|v|$ is used for the kinetic energy on an answer. Hope you can see it and comment. $\endgroup$
    – Joako
    Jul 7 at 5:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.