# Lagrangian of a rotating disc with mass attached

1. Apr 21, 2014

### albega

1. The problem statement, all variables and given/known data
A uniform disk of mass 2M, radius R, is mounted on a frictionless horizontal pivot through its principal axis. The disk has an additional point-mass, M, fixed to a point on its circumference.

(a) Give the Lagrangian for this system.

(b) Find the frequency of small oscillations of the weight about its lowest point.

The disk is now placed on its rim on a at surface and set in motion such that it can roll without slipping.
(c) Show that the period of small oscillations has increased by a factor of √3/2 relative to the answer obtained in part (b).

2. Relevant equations
L=T-V
EL equation
T=0.5Iω2 for rotation about a principal axis.

3. The attempt at a solution
I have completed a) and b) to obtain an angular frequency of oscillation of √(g/2R) about equilibrium (although this may not be right).

Now I'm having a lot of trouble with c). This must be with determining the KE of the system. I define the angle θ between the line from the centre vertically down to the point of contact with the ground and the line from the centre to the mass.

The PE U=-mgRcosθ as before.

The KE has two contributions. One due to rotation and translation of the disc, and one due to rotation and translation of the mass.

For the disc, T=MR2(dθ/dt)2 - I think this is ok as the angle the disc rolls through matches with the angle the mass moves through and the pure rolling condition holds.

For the mass, the problems start. I though it would be best to write down it's position, differentiate and find (dx/dt)2+(dy/dt)2 from this to get it's total KE. So [x,y]=[Rsinθ+R(dθ/dt)t,-Rcosθ]. [dx/dt,dy/dt]=[R(dθ/dt)cosθ+R(dθ/dt)2t+R(dθ/dt),R(dθ/dt)sinθ]. Squaring the components, adding and multiply by M/2 gives the KE of the mass.

Now this gets very messy, and two attempts to work through the algebra haven't led to anything that cancels nicely. So I believe my KE is wrong (more specifically my [x,y]) so if anybody could point out why I would be grateful, thankyou.

2. Apr 21, 2014

### collinsmark

Are you sure the given factor is not [STRIKE]3/√2[/STRIKE] [sorry another edit] 2/√3, instead of the √3/2 figure stated above?

[Edit: By that I mean in part (c) of the original problem statement.]

Last edited: Apr 21, 2014
3. Apr 21, 2014

### collinsmark

My question above still holds,

But in the mean time, I believe the trick to this problem is to realize that you are looking for the oscillations when the point mass is in the lowest point.

This will allow you to use the $\cos \theta \approx 1 - \frac{\theta^2}{2}$ approximation when calculating the potential energy, as you already must have done in part (b). [This assumes that you've set up your equations such that $\theta$ is with respect to the vertical, and the potential energy is zero when the point mass is at its lowest point.]

Then realizing, that in part (c), where the system is rolling without slipping, the point mass has zero kinetic energy at this location. All the kinetic energy (both rotational and translational) is in the disk.

Last edited: Apr 21, 2014
4. Apr 22, 2014

### fayled

The point mass would have kinetic energy no? It would still have to move if the disc moved...

Last edited: Apr 22, 2014
5. Apr 22, 2014

### physiks

I think you should write the position as (x,y)=R(1-sinθ,Rcosθ), i.e not using R(dθ/dt)t.

Even using that, I'm getting the same period as for part b) so I can't help any further...

Last edited: Apr 22, 2014
6. Apr 22, 2014

### collinsmark

Not when using the small angle approximation. Again, we're only interested in the small oscillations when the point mass is at its lowest point. At this location, when the system is allowed to roll on a surface without slipping, the point mass' velocity is negligible, and thus can be neglected. One still needs to use the point mass for the potential energy though; that's not neglected.