Uniform Rod Attached To Spring Motion Equation Problem

Click For Summary
SUMMARY

The discussion centers on deriving the equation of motion for a uniform rod of length L = 0.2 m and mass m = 0.2 kg, pivoted at one end and attached to a spring with a spring constant k = 3.0 N/m. The equation of motion is established as d²θ/dt² = (3k/m)sinθcosθ - (3g/2L)sinθ. Key concepts include the forces acting on the rod, specifically the spring force and gravitational force, which are resolved to derive the motion equation. The discussion emphasizes the importance of understanding both linear and angular motion principles.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with angular motion concepts, including torque and moment of inertia
  • Knowledge of harmonic motion and spring mechanics
  • Basic trigonometry for resolving forces
NEXT STEPS
  • Study the derivation of angular motion equations, focusing on T = Iα
  • Explore the principles of harmonic motion and the role of spring constants
  • Learn about the dynamics of pendulum systems and their equations of motion
  • Investigate the effects of gravitational forces on oscillatory systems
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the dynamics of oscillatory systems, particularly in the context of mechanics involving springs and rigid bodies.

TheDoctor078
Messages
1
Reaction score
0

Homework Statement



'Figure 2 shows a uniform rod of length L= 0.2m and mass m=0.2kg pivoted at one end. The other end is attached to a horizontal spring with spring constant k =3.0 N/m. The spring is neither stretched nor compressed when the rod is perfectly vertical. You can also assume that the force due to the spring is always horizontal.

a) Show that the equation of motion for the rod is:

\frac{d^2\theta}{dt^2}= \frac{3k}{m}\sin\theta\cos\theta - \frac{3g}{2L}\sin\theta

Homework Equations



F=-kx, <br /> F=ma, <br /> F=-mg\sin\theta<br />

The Attempt at a Solution



I have no real idea of how to tackle this problem, I presume we need to resolve the system horizontally in terms of the restoring forces needed by both parts, which in this case would be:

F=-kx-mg\sin\theta

After that, I have no idea how to tackle the problem, if someone could help point me in the right direction, it would be much appreciated as I'm getting a little bit stressed out at not being able to get the grips with this question...
 
Physics news on Phys.org
For linear motion we use F = ma.

Similarly for angular motion we use

T = I\alpha

where T = torque, I = moment of inertia and \alpha is the angular acceleration.
 

Similar threads

  • · Replies 21 ·
Replies
21
Views
5K
  • · Replies 48 ·
2
Replies
48
Views
9K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
3
Views
3K
Replies
2
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 10 ·
Replies
10
Views
4K