0

Consider $q_m$ to be the generalised coordinates where m = 1,2,3....

in Cartesian coordinates $x(q_m, t)$, we can express the velocity as $$\dot{x_{p,i}} = \dfrac{\partial x_{p,i}}{\partial q_m} \dot{q_m} + \dfrac{\partial x_{p,i}}{\partial t} dt$$

This seems fine...

If $x$ is considered to be a function of $q_m$ and $\dot{q_m}$, then how does the statement below hold water,

$$\dfrac{\partial{\dot x_{p,i}}}{\partial \dot{q_m}} = \dfrac{\partial x_{p,i}}{\partial q_m}$$

This is from the textbook of https://books.google.co.in/books?id=1gxk4oq9trYC&printsec=frontcover&source=gbs_ge_summary_r&cad=0#v=onepage&q&f=false, page 78.

Section of the Book

Raptor
  • 109
  • 1
  • 1
  • 9
  • more details..? $q_{m}$ for the position coordinate? What is $x$ represent? Thanks – Redsbefall Oct 31 '17 at 09:42
  • $q_m$, would be the position general coordinates, ie, can be Cartesian , cylindrical, etc... And $x$, would be the position in rectangular coordinates.. – Raptor Oct 31 '17 at 09:51
  • You are asking why $\frac{\partial \dot{x} }{\partial \dot{q}} = \frac{ \partial x}{ \partial q}$? or $ \frac{ \dot{x} }{ \dot{q}} = \frac{ x}{ q}$? – Redsbefall Oct 31 '17 at 10:16
  • I think the first one is because $ \dot{x} $ depends on $\dot{q}$ linearly, with gradient $ \frac{ \partial x}{ \partial q} $. You can see it is linear from eqn (31.1). – Redsbefall Oct 31 '17 at 10:18
  • The second one, ie why \dfrac{\partial \dot{x}}{\partial \dot{q}} = \dfrac{\partial x}{\partial q} , want to know how this would work.. – Raptor Oct 31 '17 at 10:24
  • In your question, you may want to add $\partial$ on the 2nd one. The books says $q_{m}$ and $\dot{q_{m}}$ are independent variables. – Redsbefall Oct 31 '17 at 10:30

0 Answers0