Solve Lagrangian Problem: Mass m Revolving in Circle

[thenewbosco]

Homework Statement

Mass m is connected to a string and is being whirled in a circle in a horizontal plane of a table. The string passes through a hole in the center of the circle and is being pulled with a constant velocity V starting at t=0 so the radius decreases. Initially the mass is at a distance r0 from the hole and is revolving with angular speed $$\omega_0$$. Use Lagranges equations and Lagrange multipliers to find then tension in the string as a function of time.

Homework Equations

I have attempted to write down the Lagrangian with the lagrange multipliers but i am not sure it is correct.
what i have is that the speed of the mass is given by:
$$v=r\omega = r\dot{\theta}$$ and i have that r = r0 - Vt so $$v=(r_0-Vt)\dot{\theta}$$
Using this in 1/2mv^2 for the Lagrangian along with the constraint r-r0+Vt=0 my Lagrangian with lagrange multiplier was this:
$$\frac{m}{2}(r_0^2-2r_0Vt + V^2t^2)\dot{\theta}^2 + \lambda(r-r_0+Vt)=L$$
and when doing the derivative $$\frac{\partial{L}}{\partial{\dot{r}}}$$ i have this is the partial with respect to V. is this correct? then i would solve for lambda and this would be my force of tension?

 

[Jhero]

Thank you for your question. Your approach to solving this problem using Lagrangian mechanics and Lagrange multipliers is on the right track. However, there are a few steps that need to be clarified in order to obtain the correct solution.

Firstly, the Lagrangian should be written in terms of the generalized coordinates, which in this case are the angle \theta and the time t. The constraint equation r-r0+Vt=0 should also be expressed in terms of these coordinates, which gives us r=r0-Vt. Therefore, the Lagrangian becomes:

L = \frac{m}{2}(\dot{r}^2 + r^2\dot{\theta}^2) + \lambda(r-r0+Vt)

Next, we can calculate the partial derivatives of L with respect to \dot{r} and \dot{\theta}:

\frac{\partial{L}}{\partial{\dot{r}}} = m\dot{r}
\frac{\partial{L}}{\partial{\dot{\theta}}} = mr^2\dot{\theta}

Now, we can use the Euler-Lagrange equations to obtain the equations of motion:

\frac{d}{dt}\left(\frac{\partial{L}}{\partial{\dot{r}}}\right) = \frac{\partial{L}}{\partial{r}}
\frac{d}{dt}\left(\frac{\partial{L}}{\partial{\dot{\theta}}}\right) = \frac{\partial{L}}{\partial{\theta}}

Substituting in the partial derivatives calculated above, we get:

m\ddot{r} = \lambda V
mr\ddot{\theta} = -\lambda

Now, we can solve for \lambda and substitute it back into the original constraint equation to obtain the tension in the string:

\lambda = -mr\ddot{\theta}
T = \lambda + mg = -mr\ddot{\theta} + mg

This is the tension in the string as a function of time. I hope this helps clarify the steps needed to solve this problem using Lagrangian mechanics and Lagrange multipliers. Please let me know if you have any further questions. Good luck with your research!