1
$\begingroup$

I was reading the definition of work done in terms of a kinetic energy.

It read that when a force is applied to a particle moving with constant velocity v, its kinetic energy changes as follows:- $$\frac{dK}{dt} = \frac{1}{2}\frac{d(m\vec v.\vec v)}{dt} = \frac{1}{2}\frac{d(mv^2)}{dt} = m\vec v\frac{d\vec v}{dt}$$ The speed of a particle does not change when the force applied on it is perpendicular to the velocity and as a result, the kinetic energy of the particle also doesn't change.

Hence the force acting on the particle must have a tangential component.
Therefore:- $$\frac{dK}{dt} = \vec F.\vec v$$

My query:

I know that the symbol $\vec v$ represents velocity(which is a vector) and $v^2$ or $\vec v.\vec v$ is a scalar. But then why is $\frac{d\vec v}{dt}$ equal to zero when the force applied on a particle is perpendicular since the velocity clearly is changing,at least its direction is.

$\endgroup$
3
  • $\begingroup$ What about $\frac{dK}{dt} =m\vec v \cdot \frac{d\vec v}{dt}$ in the first equation? $\endgroup$
    – Farcher
    Mar 28, 2017 at 15:33
  • $\begingroup$ sorry it should be $m\vec v\frac{d\vec v}{dt}$ $\endgroup$ Mar 28, 2017 at 17:05
  • 1
    $\begingroup$ I just saw that this has a bounty -- what about the current answers doesn't satisfy you? $\endgroup$
    – Javier
    Apr 5, 2017 at 0:00

4 Answers 4

5
+50
$\begingroup$

The last expression should be a dot product:

$$\frac{dK}{dt} = m\vec v \cdot \frac{d\vec v}{dt}$$

So when a force is applied perpendicularly to the direction of motion, $\frac{d\vec v}{dt}$ is a vector that is perpendicular to $\vec v$, so their dot product is zero, giving you zero change in the kinetic energy as expected.

$\endgroup$
1
  • 1
    $\begingroup$ This is the correct answer. $\dot{\vec{v}}$ isn't zero by $\vec{v} \cdot \dot{\vec{v}}$ is $\endgroup$ Mar 28, 2017 at 17:31
0
$\begingroup$

the magnitude of the vector $\vec v $ is constant, so is it's square $v^2= \vec v \cdot \vec v$ , therefore the derivative over time is zero (the derivative of a constant is zero) therefore from the expression: $$\frac{d(\vec v\cdot\vec v)}{dt} = \vec v \cdot \frac{d\vec v}{dt} +\vec v\cdot \frac{d\vec v}{dt} = 0$$ it implies that $\vec v $ is orthogonal to $\frac{d\vec v}{dt}$(from the definitions of orthogonal vector) if $\vec v$ is different to zero.

$\endgroup$
0
$\begingroup$

\begin{equation} \frac{d\tilde{v}}{dt} \neq 0. \end{equation}

It is not necessary that the vector change in velocity be zero to prove constant Kinetic energy. If the Force is perpendicular to the velocity direction, then $\tilde{F} \cdot \tilde{v} = 0$, but $\tilde{F} \neq 0,$ in general. The other way to see this is

\begin{equation} \frac{dK}{dt} = \tilde{F} \cdot \tilde{v} = \frac{1}{2} \frac{d}{dt} (v^2) = 0, \end{equation}

if the speed is constant. Thus it's a matter of force being perpendicular to the motion, not zero.

$\endgroup$
-1
$\begingroup$

that first equation seems fishy; you're getting a 'vector kinetic energy' by differentiating the velocity? it is the 'speed' or simply the magnitude of velocity 'v' (without the vector sign) that needs to be differentiated.

the speed is constant so that turns out to be zero!

$\endgroup$
4
  • $\begingroup$ He's not getting a vector kinetic energy - $\mathbf{v}\frac{d\mathbf{v}}{dt}$ is meaningless because of the missing dot product. It is also worth mentioning the speed of the particle is constant because the force acts to move the particle in a circle $\endgroup$
    – binaryfunt
    Apr 10, 2017 at 0:14
  • $\begingroup$ i don't see anything wrong with my answer though. The person said that he 'read' that expression of KE in some source and I merely pointed out that the expression was fishy.. $\endgroup$ Apr 10, 2017 at 0:18
  • $\begingroup$ For the record, it wasn't I who downvoted $\endgroup$
    – binaryfunt
    Apr 10, 2017 at 0:43
  • $\begingroup$ ik that since i didn't notice a change in my points :P. But that wasn't the point; there IS no vector kinetic energy. I should've put the KE in apostrophe to avoid confusion, sorry about that if that's what you pointed out. $\endgroup$ Apr 10, 2017 at 0:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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