Assuming a Lagrangian proportional to the following terms:
##L \sim ( \partial_\mu \sigma) (\partial^\mu \sigma) - g^{m\bar{n}} g^{r\bar{p}} (\partial_\mu g_{mr} ) ( \partial^\mu g_{\bar{n}\bar{p}} ) ~~~~~ \to (1) ##
Where ##\mu =0,1,2,3,4## and m, n,r, p and ##\bar{n}, \bar{p}, \bar{m}## and...
The final answer should have a negative b^2⋅r(dot)^2⋅r term but I have no idea how that term would become negative. Also I know for a fact that my Lagrangian is correct.
I start out by substituting rcos(Θ) and rsin(Θ) for x and y respectively. This gives me z=(b/2)r^2. The Lagrangian of this system is (1/2)m(rdot^2+r^2⋅Θdot^2+zdot^2)-mgz. (rdot and such is the time derivative of said variable). I then find the time derivative of z, giving me zdot=br⋅rdot and...
Why ##\frac{\partial (0.5*m*\dot{x}^2-m*g*x)}{\partial x}=-mg##? why ##\frac{\partial \dot{x}}{\partial x}=0##?
Why ##\frac{\partial (0.5*m*\dot{x}^2-m*g*x)}{\partial \dot{x}}=m*\dot{x}## ? why ##\frac{\partial x}{\partial \dot{x}}=0##?
Does it assume that speed is same at every location?
I...
First of all, disclaimer: This isn't an official assignment or anything, so I'm not even sure if there is a resonably simple solution.
Consider the following sketch.
(Forgive me if it isn't completely clear, I didn't want to fiddle around for too long with tikz...)
Let us assume that we can...