# Forces of constraint

• I
Hello! I have this Lagrangian: $$L=\frac{1}{2}m\dot{r}^2(1+f'(r)^2)+\frac{1}{2}m\dot{\phi}^2r^2-mgf(r)+\lambda(\phi-\omega t)$$ This represents the motion of a point-like object of mass m along a curved wire with shape $$z=f(r)$$ The wire rotates with constant angular velocity around the z axis $$\omega=\dot{\phi}$$ and I need to find the component of constraint force on the bead in the ##e_{\phi}## direction. Hence why I have that last term in the Lagrangian. Solving the Euler-Lagrange equation for ##\phi##, gives me (using the fact that ##\ddot{\phi}=0##) $$\lambda = 2mr\dot{r}\omega$$ Up to here my answer is like the one in the book. Now, from what I understand, to get the constraint force, you apply this formula: $$F=\lambda \frac{\partial f}{\partial q}$$ where in my case $$f=\phi-\omega t$$ and $$q=\phi$$ If I do this I get $$F_{\phi}=2mr\dot{r}\omega$$ which is wrong as it doesn't have units of force. In their solution they do $$F_{\phi}=\frac{1}{r}\frac{\partial L}{\partial \phi}=2m\dot{r}\omega$$ which seems correct. So what is wrong with what am I doing? Is the formula I am using wrong, or am I applying it the wrong way? Thank you!

tnich
Homework Helper
$$z=f(r)$$
$$f=\phi-\omega t$$
I think your problem is that f represents a distance in the first of these equations and an angle in the second.

I think your problem is that f represents a distance in the first of these equations and an angle in the second.
Shouldn't ##\lambda## take care of this automatically? Like shouldn't I get the right units, just by applying the formula? The Lagrangian I obtained is the same as the one they obtained so I assume it is correct. Edit: the ##f## that appears in ##z=f(r)## is not the same as the ##f=\phi - \omega t##, in case I wasn't clear there. In the second case I just gave it a general name, to reflect the way I applied the formula.

tnich
Homework Helper
Shouldn't ##\lambda## take care of this automatically? Like shouldn't I get the right units, just by applying the formula? The Lagrangian I obtained is the same as the one they obtained so I assume it is correct. Edit: the ##f## that appears in ##z=f(r)## is not the same as the ##f=\phi - \omega t##, in case I wasn't clear there. In the second case I just gave it a general name, to reflect the way I applied the formula.
##\lambda## has units consistent with your original equation, so the problem is probably somewhere else.

##\lambda## has units consistent with your original equation, so the problem is probably somewhere else.
Yeah, but I am not sure what I am doing wrong...

vanhees71
$$\vec{\nabla} \Phi(r,\varphi,z)=\vec{e}_r \partial_r \Phi + \frac{\vec{e}_{\varphi}}{r} \partial_{\varphi} \Phi+\vec{e}_z \partial_z \Phi.$$