Lagrangian rolling cylinders + small oscillations

In summary, the point mass is inside a hollow cylinder of radius R, mass M and moment of inertia I = MR^2. The cylinder rolls without slipping. The velocity of the point mass is x2' = x'(1-cos(x/R)) y2' = x'(sin(x/R)) and the frequency of small oscillations about the stable equilibrium state is w^2 = v^2/R^2.
  • #1
phys
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Homework Statement


A point mass m is fixed inside a hollow cylinder of radius R, mass M and moment of inertia I = MR^2. The cylinder rolls without slipping

i) express the position (x2, y2) of the point mass in terms of the cylinders centre x. Choose x = 0 to be when the point mass is at the bottom.
Show the velocity of the point mass is:

x2' = x'(1-cos(x/R))
y2' = x'(sin(x/R))

ii) find the lagrangian for the generalised co-ordinate x
and write down the Euler lagrange equation to obtain the equation of motion for x (DO NOT SOLVE)

iii) find the frequency of small oscillations about the stable equilibrium state

Homework Equations


L = T-V

The Attempt at a Solution


i) can find this just by using the geometry of the situation and then differentiating
ii) I *think* the lagrangian is:
L = (x'^2)(M+m)(1-cos(x/R)) + mgRcos(x/R)

using the Euler lagrange equation I think the equation of motion is:

2x''(M+m)[1-cos(x/R)] - (x')^2((M+m)/R)(sin(x/R) + mgsin(x/R) = 0

iii) this is where I am stuck...
stable equilibrium is at x = 0
if you use sin x = x then the e.o.m reduces to:

2x''(M+m)(1-cos(x/R)) + mgx/R = 0
I have neglected the middle term since x/R^2 is negligible...

I was expecting to get an equation of the form x'' + w^2 x = 0 were w is the frequency ... however when expanding the cos you don't get this...

Have I done something really wrong?
 

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  • #2
phys said:
ii) I *think* the lagrangian is:
L = (x'^2)(M+m)(1-cos(x/R)) + mgRcos(x/R)
Please show your work. It does not look correct to me.

phys said:
if you use sin x = x then the e.o.m reduces to:
This is a correct assumption for linearising the equation near the stable point. However, if ##\sin(x) = \mathcal O(x)##, then ##\cos(x) = 1 + \mathcal O(x^2)##.
 
  • #3
Thanks ... for the lagrangian
i thought the KE would be rotational KE of the cylinder which is 0.5Iw^2 plus the KE of the mass = 0.5m(x2'^2+y2'^2)
w^2 = v^2/R^2 since no slip
I have attached my working ... am I doing this really wrong?
 

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  • #4
You say the cylinder rolls without slipping, but what does it roll upon? Is it on a horizontal surface, a set of rollers, across a hill, or what? The kinematics makes all kinds of difference here.
 
  • #5
It is rolling on a flat horizontal surface
 
  • #6
A couple of things that I noticed in your derivation of L:

You used the same symbol V for the speed of the point mass and also for the speed of the center of the cylinder. But these speeds are different.

You took the kinetic energy of the cylinder to be (1/2)Icω2 where Ic is the moment of inertia of the cylinder about its central axis. But the cylinder is not just rotating about this axis. It is also in translational motion. (Or, you can think of the cylinder as instantaneously rotating about the point of contact of the cylinder with the ground.)
 
  • #7
Ah yes! knew i'd made a stupid mistake somewhere - thanks for pointing out!
Just to check... the translational velocity of the C of M is v'
and therefore the rotational velocity w is x'/R?
thanks so much
 
  • #8
phys said:
Just to check... the translational velocity of the C of M is v'
and therefore the rotational velocity w is x'/R?
I'm a little confused with the notation. In most of your work, a prime appears to denote a time derivative. So, v' would be an acceleration.

If x denotes the position of the CoM, then x'/R would be the rotational velocity (or, angular velocity) of the cylinder.
 
  • #9
Ah yes sorry .. I meant x'
Thanks
 

1. What is the concept of Lagrangian rolling cylinders?

Lagrangian rolling cylinders refer to a system in which a cylinder is rolling on a surface while also experiencing small oscillations. This system can be described using the Lagrangian mechanics, which is a mathematical framework for analyzing the dynamics of complex mechanical systems.

2. How are Lagrangian rolling cylinders different from regular rolling cylinders?

Regular rolling cylinders only consider the motion of the cylinder as it rolls on a surface, while Lagrangian rolling cylinders take into account additional oscillations or vibrations that may occur. This makes the system more complex, but also allows for a more accurate analysis of its dynamics.

3. What are small oscillations in the context of Lagrangian rolling cylinders?

Small oscillations refer to the slight back-and-forth motions that a rolling cylinder may experience while in motion. These oscillations are usually small enough to be neglected in regular rolling cylinder systems, but are important to consider in Lagrangian rolling cylinders.

4. How is the Lagrangian of a rolling cylinder with small oscillations calculated?

The Lagrangian of a rolling cylinder with small oscillations is calculated by considering the kinetic and potential energies of the system. The kinetic energy includes the rotational and translational motion of the cylinder, while the potential energy takes into account the effects of gravity and any other external forces acting on the system.

5. What can the analysis of Lagrangian rolling cylinders with small oscillations tell us?

The analysis of this system can provide insights into the motion and stability of rolling cylinders, as well as help in designing and optimizing mechanical systems that involve rolling cylinders. It also helps in understanding the principles of Lagrangian mechanics and their application to real-world scenarios.

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