Kinetic energy, the integral of vdp

[ponjavic]

I read that kinetic energy is:
$$\int(vdp)$$
where v is velocity and p is momentum
I'd like to see a calculation of this, I can't really get it right
$$\int(vdp)=vp+\int(pdv)$$
Is this right? If so then how do I continue?

 

[Justin Lazear]

$$\int v dp = \frac{1}{m} \int mv dp = \frac{1}{m} \int p dp = \frac{p^2}{2m}$$

Which is one of the familiar forms of KE.

--J

 

[ponjavic]

bah that's a mean definition, is that 1/m understandable or is it just how it is
I mean if you hadn't seen it before would you be able to solve it?

$$\int(udv)=uv+\int(vdu)$$
u=p du=dp dv=dp v=p
$$\int(pdp)=p^2+\int(pdp)$$
What am I doing wrong?
how is $$\int(pdp)=\int(p)=\frac{p^2}{2}$$ ??
Still can't solve it...

 

[BobG]

The idea is to get consistent variables. You either have to integrate:

$$\int (p) dp$$ or $$\int (v) dv$$

Since p=mv, it's easy to convert v into p. Except you don't want to change the value of your equation, so you can only multiply by 1.

$$\frac{m}{m}\int (v) dp = \frac{1}{m} \int (mv) dp$$

 

[ponjavic]

Oh my god, that's a good way to forget the definition of the basic integration theorem
I thought of $$\int(p)dp$$ as $$\int(pdp)dp$$ totally forgot the meaning of dx in $$\int(f(x))dx$$ well at least I feel a bit better now and it is good that I got this sorted out before my cambridge interview, thank you!

Could someone explain why I get
$$\int(pdp)=p^2+\int(pdp)$$
Why the left side does not equal the right.
I believe I have done a correct integration by parts...
This integral is a hypothetical $$\int(pdp)dp$$