Semi-circle with small angle harmonic oscillations

Click For Summary
SUMMARY

The discussion focuses on deriving the angular frequency of small oscillations for a uniform semicircular disk of radius R using Lagrangian mechanics. The derived formula for angular frequency is ω = sqrt(8g / (R(9π - 16))). Key equations include the kinetic energy expression K = 0.5mv² + 0.5I_cω² and the potential energy U = mghcos(φ). Participants identified errors in the kinetic and potential energy calculations, emphasizing the importance of accurately defining the angle φ and the relationship between linear and angular velocities.

PREREQUISITES
  • Understanding of Lagrangian mechanics
  • Familiarity with kinetic and potential energy equations
  • Knowledge of the parallel axis theorem
  • Ability to define and manipulate angular variables
NEXT STEPS
  • Study the derivation of the Lagrangian for rolling objects
  • Learn about the parallel axis theorem in detail
  • Explore the relationship between linear velocity and angular velocity
  • Investigate the dynamics of small angle approximations in oscillatory motion
USEFUL FOR

Students and educators in physics, particularly those focusing on mechanics and oscillatory motion, as well as engineers working with rotational dynamics and energy conservation principles.

Liquidxlax
Messages
312
Reaction score
0

Homework Statement



Consider a uniform semicircular disk of radius R, which rolls without slipping on a horizontal surface. Recall that the kinetic energy of an object is the sum of the translational kinetic energy of the centre of mass (point C) and the rotational kinetic energy about the Centre of Mass.

Using Lagrangian methods, show that the angular frequency of small oscillations is

ω = sqrt([8g]/[R(9π -16)])

Homework Equations



.5mR2 = Ic + mh2 where h = 4R/3π

L = K - U

dL/dq = (d/dt)(dL/dq')

The Attempt at a Solution



First thing to do is find the center of mass of the object

so i solved for that to get (x,y) = (o, 4R/3π)

Then using that I solved for the moment which equals .5mR2

Using the parallel axis thheorem i found that

.5mR2 - mh2 = Ic

K = .5mv2 + .5Ic2

U = mghcos(∅) ( i think)

This has no R dependence so the Euler Lagrange only acts on theta and omega

problem is it's not work, so i obviously messed up on my kinetic and potential energies.
 
Physics news on Phys.org
How are you defining the angle φ? If it's the way I think you're doing it, the potential energy has the wrong sign.

Your expression for the kinetic energy is wrong, though I think it's just a typo. It should be
[tex]K=\frac{1}{2}mv^2 + \frac{1}{2}I_c \omega^2[/tex]You left out the ω. What is your expression for v in terms of ω?
 
Yeah i know i didnt transcribe it from my notes well sorry. I did manage to figure it out though. Not sure how to write out phi. Eaaier to draw it
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
Replies
13
Views
2K
  • · Replies 8 ·
Replies
8
Views
4K
Replies
4
Views
4K
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
2K
  • · Replies 18 ·
Replies
18
Views
3K
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K