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Rigid body motion

by Logarythmic
Tags: body, motion, rigid
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Logarythmic
#1
Nov29-06, 05:29 AM
P: 285
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 non-fixed 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?
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dextercioby
#2
Nov29-06, 06:23 AM
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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.
Logarythmic
#3
Nov29-06, 06:29 AM
P: 285
This is what I think it should look like. I guess it's like a double pendulum with two rigid bodies.
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OlderDan
#4
Nov29-06, 08:49 AM
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Rigid body motion

Quote Quote by Logarythmic View Post
This is what I think it should look like. I guess it's like a double pendulum with two rigid bodies.
The double pendulum sounds right to me. The moment of inertia of the disk about its pivot point P can be obtained using the parallel axis theorem. I think you would want to use two angles as your generalized coordinates, one for the rod and one for the disk.
Logarythmic
#5
Nov29-06, 09:40 AM
P: 285
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?
Logarythmic
#6
Nov29-06, 11:28 AM
P: 285
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?
OlderDan
#7
Nov29-06, 01:32 PM
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Quote Quote by Logarythmic View Post
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?
You will need translation and rotation for the disk- translation of the CM and rotation about the CM. I did not look at your equations of motion yet. Did you include more than what you have here?
Logarythmic
#8
Nov29-06, 02:04 PM
P: 285
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.
OlderDan
#9
Nov29-06, 07:31 PM
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Quote Quote by Logarythmic View Post
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.
With x and y the coordinates of the CM, then the I for the disc should be the I about the center of mass.
Logarythmic
#10
Nov30-06, 02:19 AM
P: 285
But if I'm using I about P, what should I include then?
OlderDan
#11
Nov30-06, 09:28 AM
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Quote Quote by Logarythmic View Post
But if I'm using I about P, what should I include then?
You could express the moment of inertia about the center of mass in terms of the moment of inertia about P by using the parallel axis theorem, but why would you want to? The CM is a special point in an assembly of particles or a rigid body that (among other things) permits the separation of kinetic energy into the translation of the CM term and the rotation about the CM term. No arbitrary point is so well behaved.

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.
Logarythmic
#12
Nov30-06, 09:32 AM
P: 285
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?
OlderDan
#13
Nov30-06, 12:32 PM
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Quote Quote by Logarythmic View Post
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?
Looks right to me. With Io being the moment of inertial of the rod about its end.

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.
Logarythmic
#14
Nov30-06, 01:28 PM
P: 285
That's how I've done it. =) Thanks for your help!
epdu
#15
Dec11-06, 08:32 PM
P: 1
which one is right?
Could you explain a little?

Quote Quote by Logarythmic View Post
That's how I've done it. =) Thanks for your help!


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