Phys Final (Astronomy&Mechanics)

[mooeypoo]

Hellooo all you astronomy/mechanics wizzes.

 

I had my final final today, and I am hoping I did well. I felt okay, but it's a confusing subject so.. well, I'd appreciate actually seeing if I made it. It's not too much about the grade (class avg is in the 40s so far), it's more about making sense of it and makign sure I understood things right. Either I will have to take this course again (meh) or I continue to more difficult subjects, in both cases I should understand this material well.

 

So.. on we go:

 

Problem 1

Consider a simple pendulum where the bob of mass m is attached to the support by a thin spring (taken to be massless) of spring constant k. The equilibrium length of the spring is [math]l_{0}[/math]. The pendulum can oscillate in a single vertical plane. The spring can strethc and the angle [math]\theta[/math] can change. Obtain the lagrangian and the equations of motion of the system (You do not have to solve the equations of motion).

 

I feel so stupid.. I just noticed. I wrote in the notebook [math]\frac{1}{2}k(l-l_{0})[/math] ... but I didn't add that to the lagrangian!!!! AAA!!! <sigh> anyhoo, here's my answer:

 

[math]L = 1/2 m (\dot{l}^2+l^2\dot{\theta}^2) + mglcos\theta[/math]

 

Obviously my equations of motion are wrong after this point since my lagrangian is lacking the spring but anyhoo, I'm hoping for partial credit, and for - at least - me understanding THAT part:

 

[math]\frac{d}{dt} (\frac{\partial L}{\partial \dot{\theta}}) =\frac{\partial L}{\partial \theta}[/math]

 

AND

 

[math]\frac{d}{dt} (\frac{\partial L}{\partial \dot{l}}) =\frac{\partial L}{\partial l}[/math]

 

Which result in:

[math]\ddot{\theta}=-gsin\theta[/math]

[math]\ddot{l}=\dot{\theta}l+gcos\theta[/math]

 

I know I forgot the spring, but is my method right after the setting up of the lagrangian?

 

I don't have my notes for problem 2, so there's nothing much to report here.. :\

 

Problem 3

A disk of mass M and radius R can roll on a horizontal plane along the x-axis (Ignore motion along the other directions). It is attracted to a fixed point P with a force F=-ks where s is the distance of the center of the disk from P (see figure). Obtain the Lagrangian and the equations of motion. The disk can oscillate (in a rolling motion) as shown. Obtain the frequency of this oscillation. (Moment of Inertia of a disk for rotations around its axis is 1/2 MR^2. HINT: Obtain the potential energy corresponding to F and then use the geometry to relate s to x).

 

 

Okay, [math]F=-\nabla V[/math] so:

 

[math]V=-\int ks = -\frac{1}{2}ks^2[/math]

 

and I decided to represent everything in terms of x and xdot, (where [math]s=\frac{x}{cos^2 \theta }[/math] and an added [math]\frac{1}{2} I \dot{\phi }^2 [/math] for the rotating disk, and a relationship of arclengths [math]R\phi = x[/math]) so:

 

[math]L=\frac{3}{4}m\dot{x}^2-mgR-\frac{1}{2}k\frac{x^2}{cos^2\theta}[/math]

 

and

[math]\frac{d}{dt} (\frac{\partial L}{\partial \dot{x}}) =\frac{\partial L}{\partial x}[/math]

 

[math]\ddot{x}=\frac{2kx}{3mcos^2\theta}[/math]

 

Question 4

A globular cluster of stars has a spherically symmetric (independent of angles) distribution of mass which falls off with distance r' from the center, up to radius R. The density can be approximated by:

[math]\rho (r') = \rho_{0}(1-\frac{r'}{R})[/math] for r'<R

[math]\rho (r') = 0[/math] for r'>R

 

where [math]\rho_{0}[/math] is a constant. Obtain the gravitational potential at a distance r from the center, where r<R (the point under consideration is inside the mass distribution. You can express the result in terms of [math]\rho_{0}[/math]

So..

 

[math]V(r')=-G \int_{0}^{r} \frac{\rho(r')}{|r-r'|} d\tau = -G \int_{0}^{r} \frac{\rho_{0}(1-\frac{r'}{R})}{r}r^2dr sin\theta d\theta d\phi = -4\pi G\rho_{0} r^2(\frac{1}{2}-\frac{r}{3R})[/math]

Edited December 23, 2008 by mooeypoo