Forces of constraint

  • I
  • Thread starter Silviu
  • Start date
  • #1
624
11
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!
 

Answers and Replies

  • #2
tnich
Homework Helper
1,048
336
$$z=f(r)$$
$$f=\phi-\omega t$$
I think your problem is that f represents a distance in the first of these equations and an angle in the second.
 
  • #3
624
11
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.
 
  • #4
tnich
Homework Helper
1,048
336
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.
 
  • #5
624
11
##\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...
 
  • #6
vanhees71
Science Advisor
Insights Author
Gold Member
17,061
8,168
What you calculate with your formula is not the force but the "generalized force". Since your generalized coordinate is an angle this generalized force is a torque.

Of course, you book is correct too. To get the force, you consider the piece with the Lagrange parameter as a potential for the constraint forces (in this case time dependent, but that doesn't matter). Then the force is given as the gradient of this "potential". Now you have to define the gradient in terms of your generalized coordinates. This is fortunately done in any textbook, because you have cylinder coordinates, and the gradient is
$$\vec{\nabla} \Phi(r,\varphi,z)=\vec{e}_r \partial_r \Phi + \frac{\vec{e}_{\varphi}}{r} \partial_{\varphi} \Phi+\vec{e}_z \partial_z \Phi.$$
Note that you don't need to take into account the derivatives of ##\lambda##, because the constraint equation makes this contribution obviously vanish.
 
  • Like
Likes CenMo and Orodruin

Related Threads on Forces of constraint

  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
4
Views
31K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
0
Views
1K
Replies
1
Views
2K
  • Last Post
Replies
2
Views
89
  • Last Post
Replies
4
Views
970
  • Last Post
Replies
12
Views
2K
Replies
2
Views
763
Replies
1
Views
2K
Top