Find Height of Particle on a Hoop

[Arman777]

Homework Statement

Particle on a hoop - Finding the height that particle falls

Relevant Equations

Lagrangian

There is a particle with mass $m$ sliding from the hoop with radius R. Its asked to find the height of the hoop which the particles falls. Now I did the hard part (I guess) and find the Lagrangian of the system. Which is given as

$$-mR\dot{\theta}^2 + mg\sin\theta = \lambda$$
and
$$\ddot{\theta} = -\frac{g}{R}\cos\theta$$

but I cannot find the height which it falls. I have thought that when it falls from the hoop the constrain force will be $0$ (i.e, $\lambda = 0$). But nothing seems to be working. Any ideas ?

 

[TSny]

I have seen a solution in this page
https://math.stackexchange.com/questions/1347302/particle-on-a-hemisphere-lagrange

But the solution does not make sense to me. Especially this part

\begin{equation}
\lambda=2mR\frac{g}{R}\cos\theta + mg\cos\theta = mg(3\cos\theta-2)
\end{equation}

This is clearly equal to $3mg\cos\theta$ so I don't know where that $-2$ is coming from.

Important Note: The definition of theta is different in my question and the answer given above.

The solution at stackexchange has an error. The integration constant, $c$, in the solution is not zero. Instead, $c = 2g/R$. This is noted in a footnote at the end of the solution. This correction accounts for the -2 in the answer for $\lambda$.

 

[Arman777]

The solution at stackexchange has an error. The integration constant, $c$, in the solution is not zero. Instead, $c = 2g/R$. This is noted in a footnote at the end of the solution. This correction accounts for the -2 in the answer for $\lambda$.

So how can I solve the problem.

 

[TSny]

So how can I solve the problem.

Which parts of the solution at stackexchange are you having difficulty with?

 

[etotheipi]

Does the particle have a radius, or is it a point? I guess if it were point-like you wouldn't really care to use the Lagrangian formulation, so I'll just assume you're dealing with a rolling disc of radius, say, $r$.

Use generalised coordinates $\mathbf{q} = (\rho, \varphi, \xi)$, where $\rho$ & $\varphi$ are the plane polar coordinates of the centre of the disc and $\xi$ the angle by which the disc has turned about its centre. Introduce the holonomic constraint $f(\mathbf{q}) = \rho - (R + r) = 0$ and the non-holonomic constraint $g(\dot{\mathbf{q}}) = (R+r)\dot{\varphi} - r\dot{\xi} = 0$. Now, write the Lagrangian$$\mathcal{L}(\mathbf{q}, \dot{\mathbf{q}}) = \frac{1}{2}m (\dot{\rho}^2 + \rho^2 \dot{\varphi}^2) + \frac{1}{2} I \dot{\xi}^2 - mg\rho \cos{\varphi}$$Lagrange's equations read$$\begin{align*}

\frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{\rho}} \right) - \frac{\partial \mathcal{L}}{\partial \rho} - a \frac{\partial f}{\partial \rho} - b \frac{\partial f}{\partial \dot{\rho}} &= 0 \implies m\ddot{\rho} - m \rho \dot{\varphi}^2 + mg\cos{\varphi} - a = 0 \\ \\

\frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{\varphi}} \right) - \frac{\partial \mathcal{L}}{\partial \varphi} - a \frac{\partial f}{\partial \varphi} - b \frac{\partial f}{\partial \dot{\varphi}} &= 0 \implies m\rho^2 \ddot{\varphi} + 2m \rho \dot{\rho} \dot{\varphi} - mg\rho \sin{\varphi}-b(R+r) = 0 \\ \\

\frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{\xi}} \right) - \frac{\partial \mathcal{L}}{\partial \xi} - a \frac{\partial f}{\partial \xi} - b \frac{\partial f}{\partial \dot{\xi}} &= 0 \implies I \ddot{\xi} + br = 0
\end{align*}$$Can you see how to eliminate $b$ and solve for the parameter $a$?

 

[Arman777]

Which parts of the solution at stackexchange are you having difficulty with?

I am just looking for my own solution.

Homework Statement:: Particle on a hoop - Finding the height that particle falls
Relevant Equations:: Lagrangian

but I cannot find the height which it falls.

Does the particle have a radius, or is it a point?

Its a point particle

 

[etotheipi]

Its a point particle

In that case I guess you can just take $r = I = b= 0$, ignore the third Lagrange equation and also ignore the second constraint.

 

[Arman777]

In that case I guess you can just take $r = I = 0$ ignore the third Lagrange equation, and also ignore the second constraint.

I have already written the lagrangian. I just need to find the height it falls. How can I find it from the given equations that I have already written.

 

[etotheipi]

Okay, you have written $-mR\dot{\theta}^2 + mg\sin{\theta} =\lambda$ and $\ddot{\theta} = -g/R \cos{\theta}$. You should differentiate the first equation with respect to time,$$-2mR \dot{\theta} \ddot{\theta} + mg \dot{\theta}\cos{\theta} = \frac{d\lambda}{d\theta} \dot{\theta}$$Now cancel the $\dot{\theta}$'s and replace the $\ddot{\theta}$,$$2mg\cos{\theta} + mg\cos{\theta} = 3mg\cos{\theta} = \frac{d\lambda}{d\theta}$$Solve with initial condition $\lambda(\pi/2) = mg$, corresponding to the normal force at top of the hill.

 

[berkeman]

I have already written the lagrangian. I just need to find the height it falls. How can I find it from the given equations that I have already written.

@Arman777 -- You have been getting excellent help in this thread so far, but you continue to show no work or effort.

This thread is locked for Moderation. Please send me a PM with your work so far so we can reopen the thread. Thank you.

 

[berkeman]

OP says he has solved the problem and thanks you all for the hints and tips. thread will remain closed.