Deriving formula for kinetic energy

[billard]

TL;DR Summary

Issue deriving 1/2 *mv^2 from some pre-assumed equations

Hello! I am new to the differential version of classical physics, and I am trying to work how to derive kinetic energy from some pre-assumed equations:

Assume that we know: ##\ddot{z} = 0## and ##m\ddot{\textbf{r}} \cdot \dot{\textbf{r}} = 0##This results in $$\frac{1}{2}m\dot{r}^2 = W = const.$$

How is the kinetic energy given here with our pre-assumptions? I am sure this is very simple, forgive me, I am a beginner.

 

[anuttarasammyak]

Work done per time is innerproduct of Force F and velocity v. F =ma. When no work done, you can deduce that KE is conserved.

 

[topsquark]

TL;DR Summary: Issue deriving 1/2 *mv^2 from some pre-assumed equations

Hello! I am new to the differential version of classical physics, and I am trying to work how to derive kinetic energy from some pre-assumed equations:

Assume that we know: ##\ddot{z} = 0## and ##m\ddot{\textbf{r}} \cdot \dot{\textbf{r}} = 0##This results in $$\frac{1}{2}m\dot{r}^2 = W = const.$$

How is the kinetic energy given here with our pre-assumptions? I am sure this is very simple, forgive me, I am a beginner.

If the applied force is constant, then ##\ddot{\textbf{r}}## is constant, and in particular, so is the direction of ##\dot{\textbf{r}}##, ala Newton's 2nd.

So
##\ddot{\textbf{r}} \cdot \dot{\textbf{r}} = \ddot{r} \dot{r} \, cos( \theta )##
where ##\theta## is the angle between the acceleration and the velocity is constant.

So now you have
##m\ddot{r} \dot{r} cos( \theta ) = ( m \, cos( \theta ) ) \ddot{r} \dot{r} = 0##

Now note that
##\dfrac{d}{dt} ( \dot{r} )^2 = 2 \ddot{r} \dot{r}##

Can you finish?

-Dan

 

[kuruman]

Start with ##K=\frac{1}{2}m(\mathbf{\dot r}\cdot \mathbf{\dot r}).##
Can you show that ##\dfrac{dK}{dt}=0~## if ##~\mathbf{\dot r}\cdot \mathbf{\ddot r}=0~?##

Here assume that ##\mathbf{r}=x~\mathbf{\hat x}+y~\mathbf{\hat y}+z~\mathbf{\hat z}.##