Why Does the Disk Have Two Moments of Inertia?

  • Thread starter Thread starter don_anon25
  • Start date Start date
  • Tags Tags
    Lagrange
don_anon25
Messages
35
Reaction score
0
Lagrange Problem redux -- super urgent...

See the attachment to help you visualize this.
A rod of length L and mass m is povoted at the origin and swings in the vertical plane. The other end of the rod is attached/pivoted to the center of a thin disk of mass m and radius r.
OK, I know that the rod can swing and that the disc can swing and rotate.
So for my kinetic energy, I should have a moment of inertia for the rod. According to the text, the disk should have two moments of inertia. Why and what are they?
I suppose my expression for Kinetic Energy should be (Moment of inertia for rod)*(d theta/ dt)^2 + (Moment of inertia for disk)*(d phi/dt)^2 +??
For potential energy, I should have PE = mgh. But h = L/2 (1-cos theta). So PE = 1/2 mgL (1-cos theta). Or should my mass be 2m, since I have both the mass of the rod and the mass of the disk?
If someone could just help with the setup, I can do the Euler-Lagrange equations from there.
Disregard my earlier post... I was totally off-base with my discussion...
Please help soon! I am really struggling with this problem...
 

Attachments

Physics news on Phys.org
The other contribution to the kinetic energy comes from the fact that the disk's center of mass can move. As for the potential energy, you need to find the position of the center of mass. Also, perhaps I'm missing something, but isn't the rotation of the disk about its center of mass completely decoupled from the rest of the motion?
 
Thanks so much for the help...but I need some further clarification.
You said that the contribution to KE comes from the fact that the disk's center of mass can move. How do I express this mathematically as a term in my kinetic energy expression? Is what I have for Kinetic energy thus far entirely wrong?

Secondly, how do I find the position of the center of mass? And then what do I do with it?

We did nothing like this in class...

Is my expression for potential energy entirely wrong as well?

Should this problem have constraints as well (i.e. need to use Lagrangian multipliers?)
 
What you have so far is two of the three pieces of the kinetic energy. You should have three terms, one term corresponding to the rotation of the rod about its pivot (this should involve theta dot), another term corresponds to spinning of the disk about its center of mass (this should involve ph dot), the third term which you are missing comes from the fact that the whole disk swings with the rod i.e. the center of mass of the disk moves. What is the kinetic energy of this center of mass motion? Hint: think about the kinetic energy of a particle constrained to move in a circle.

I can't see your attachement, but it looks like you have the part corresponding to the potential energy of the rod right (as long as you've defined your angle consistently). You are however missing the potential energy of the disk. Where is the center of mass of the disk located (the disk is uniform so this boils to where the center is located)?

As for constraints, you have already solved the constraints by introducing a reduced number of generalized coordinates. You only have to use Lagrange multipliers when some of your coordinates are not independent of each other i.e. when you haven't solved the equations of constraint.

Let me ask again, are you sure the phi motion is supposed to be completely independent of the theta motion?
 
Thanks for being so prompt and helpful in your response!
So, constrained to move in a circle...that sounds like polar coordinates!
So the third term I am missing is the expression of kinetic energy for the disc in polar coordinates?

So I should find the center of mass of the disk, and that will aid me in finding my second term for potential energy, correct?

And no, I am not entirely convinced that theta and phi are independent on one another, but I am unsure as to how to relate them...

Thank goodness I don't have to use Lagrangian multipliers :)
 
##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
Hello everyone, I’m considering a point charge q that oscillates harmonically about the origin along the z-axis, e.g. $$z_{q}(t)= A\sin(wt)$$ In a strongly simplified / quasi-instantaneous approximation I ignore retardation and take the electric field at the position ##r=(x,y,z)## simply to be the “Coulomb field at the charge’s instantaneous position”: $$E(r,t)=\frac{q}{4\pi\varepsilon_{0}}\frac{r-r_{q}(t)}{||r-r_{q}(t)||^{3}}$$ with $$r_{q}(t)=(0,0,z_{q}(t))$$ (I’m aware this isn’t...
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
Back
Top