# Particle on a hoop

Gold Member
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 ?

Last edited:

Homework Helper
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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##.

• BvU
Gold Member
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.

Homework Helper
Gold Member
So how can I solve the problem.
Which parts of the solution at stackexchange are you having difficulty with?

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##?

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Gold Member
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

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.

Gold Member
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.

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.

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Mentor
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.

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