
#1
Nov2906, 05:29 AM

P: 286

Problem statement:
Consider a pendulum consisting of two parts: a uniform rod of mass m, length l, negligible thickness and with one end fixed; and a uniform disk of mass [tex]\mu[/tex] and radius [tex]\rho[/tex]. The rod is moving in a plane, and the disk is attached at a point P on its boundary to the nonfixed end of the rod, in such a way that it can freely rotate about P in the plane in which the rod is moving. Obtain the Lagrangian and the equations of motion. I suppose I should divide this into two parts; one for the rod and one for the disk. For the rod I get [tex]T_1 = \frac{1}{2} I_1 \dot{\theta}_1^2 + \frac{1}{2} m(\dot{x}_1^2 + \dot{y}_1^2)[/tex] ,where the index 1 is the rod, and [tex]V_1 = mgh = mg \frac{l}{2} (1  \cos \theta_1)[/tex]. The moment of inertia for the rod is [tex]I_1 = \frac{1}{3} m l^2[/tex]. So far so good, I think.. But how should I do with the disc? Should I treat this the same way and just use a superposition of the two lagrangians? And how do I get the moment of inertia? 



#2
Nov2906, 06:23 AM

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P: 11,866

Can you post a picture ? I'm unable to imagine the setup.
Daniel. P.S. There's no such thing as a superposition of lagrangians. 



#3
Nov2906, 06:29 AM

P: 286

This is what I think it should look like. I guess it's like a double pendulum with two rigid bodies.




#4
Nov2906, 08:49 AM

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P: 3,033

Rigid body motion 



#5
Nov2906, 09:40 AM

P: 286

So the kinetic energies are just
[tex]T_{rod} = \frac{1}{2} I_O \dot{\theta}_1^2[/tex] and [tex]T_{disc} = \frac{1}{2} I_P \dot{\theta}_2^2[/tex] or should I include translation movement aswell? 



#6
Nov2906, 11:28 AM

P: 286

I get the equations of motion to be
[tex]0 = \mu \left[ l \ddot{\theta}_1 + R \dot{\theta}_2 \sin{(\theta_1  \theta_2)} (\dot{\theta}_1  \dot{\theta}_2 + \dot{\theta}_1 \dot{\theta}_2) + R \ddot{\theta}_2 \cos{(\theta_1  \theta_2)} + g \sin{\theta_1} \right] + \frac{1}{2} mg \dot{\theta}_1 \sin{\theta_1}[/tex] and [tex]0 = R \ddot{\theta}_2 + l \ddot{\theta}_1 \cos{(\theta_1  \theta_2)}  l \dot{\theta}_1^2 \sin{(\theta_1  \theta_2)} + g \sin{\theta_2} [/tex] where I have used R instead of [tex]\rho[/tex] Could this be correct? 



#7
Nov2906, 01:32 PM

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#8
Nov2906, 02:04 PM

P: 286

Yes I used
[tex]T_{rod} = \frac{1}{2} I_O \dot{\theta}_1^2 [/tex] and [tex]T_{disc} = \frac{1}{2} I_P \dot{\theta}_2^2 + \frac{1}{2} m (\dot{x}_2^2 + \dot{y}_2^2) [/tex] where [tex]I_O[/tex] is the moment of inertia about the upper end of the rod, [tex]I_P[/tex] is the moment of inertia about the pivot point P and [tex]x_2[/tex] and [tex]y_2[/tex] are the coordinates of the c.m. of the disc. 



#9
Nov2906, 07:31 PM

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#10
Nov3006, 02:19 AM

P: 286

But if I'm using I about P, what should I include then?




#11
Nov3006, 09:28 AM

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When you used x and y translational velocities for part of the kinetic energy, you were already including some of the energy of rotation about P. If you use I about point P, you will be couble counting some of the energy contribution. 



#12
Nov3006, 09:32 AM

P: 286

So
[tex]T_{rod} = \frac{1}{2} I_O \dot{\theta}_1^2 [/tex] and [tex]T_{disc} = \frac{1}{2} I_{CM} \dot{\theta}_2^2 + \frac{1}{2} m (\dot{x}_2^2 + \dot{y}_2^2) [/tex] is correct? 



#13
Nov3006, 12:32 PM

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For what it's worth, I found it easier to do the algebra using the angle between the radius from P to the disk center and the vertical. That would be the sum of your angles I believe. When the algebra is done it can be expressed in terms of your angles. 



#14
Nov3006, 01:28 PM

P: 286

That's how I've done it. =) Thanks for your help!




#15
Dec1106, 08:32 PM

P: 1

which one is right?
Could you explain a little? 


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