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Q1) Write down the Lagrangian of the system in terms of y(t) Q2) Obtain the Eqn of motion Q3)Using Lagrange Multiplier method find the forces of constraints 1) We have a constraint such that $$f=y-r\theta=0 $$

And the lagrangian is

$$L=1/2m[\dot{y}^2+\frac{R^2\dot{\theta}^2}{2}]+mgy$$

from here I have to get rid of the $\theta$ by using constraint. Then I get

$$L=1/2m[\dot{y}^2+\frac{\dot{y}^2}{2}]+mgy$$

$$\ddot{y}=2g/3$$ where g is gravity of earth.

Now the question asks to find the constraints so

I dont use the constraint in the lagrangian and I just wrote the Multipler equation which it is

$$\frac{\mathrm d}{\mathrm dt} \frac{\partial L}{\partial \dot{q_j}} - \frac{\partial L}{\partial q_j} = \sum {\lambda_j}\frac{\partial {f_j}}{\partial q_i}$$

From here I get two equations

$$m\ddot{y}-mg=\lambda$$ $$mR^2\ddot{\theta}=-\lambda R$$

But I dont know how to proceed to find the force of constraint, thanks

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closed as off-topic by Chris, ZeroTheHero, John Rennie, Kyle Kanos, GiorgioP May 1 at 17:00

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  • $\begingroup$ I'm guessing your Lagrangian is incorrect - which could easily be a typesetting problem. If I use $L=\frac{1}{2}m(\dot{y}^{2}+r^{2}\dot{\theta}^{2})-mgy$ as the Lagrangian instead, I get a constraint force of $\ddot y=-\frac{g}{2}$. But you had right idea - in the end you need to eliminate $\lambda$ to find the force of constraint. $\endgroup$ – Cinaed Simson May 1 at 18:55
  • $\begingroup$ Its not a type. This is a given lagrange for the system $\endgroup$ – Reign May 2 at 15:38
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Your last equations should be \begin{align} &m\ddot{y} = \lambda + mg\\ &\frac{mR^2}{2}\ddot{\theta} = - \lambda R \end{align} The equations can be slightly simplified to \begin{align} &m\ddot{y} = \lambda + mg\\ &mR\ddot{\theta} = - 2\lambda \end{align}

The constraint $y - R \theta = 0$ can be multipled by the mass $m$, which yields $$my - R^2\theta = 0$$ Now, differentiate it twice with respect to time $t$ $$m\ddot{y} - mR\ddot{\theta} = 0 $$ Next, plug in the twice differentiated constraint the expressions for the second derivatives from the system of differential equations $$0 = m\ddot{y} - mR\ddot{\theta} = \lambda + mg - (-2\lambda) = 3\lambda + mg$$ Now, solve for $\lambda$ $$\lambda = -\frac{1}{3}mg$$ Finally, your constrained equations become \begin{align} &m\ddot{y} = \lambda + mg = -\frac{1}{3}mg + mg = \frac{2}{3}mg\\ &mR\ddot{\theta} = - \lambda = \frac{1}{3}mg \end{align} which after cancelling out the mass turn into \begin{align} &\ddot{y} = \frac{2g}{3}\\ &\ddot{\theta} = \frac{g}{3R} \end{align}

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