Degrees of Freedom: Square & Triangular Lamina

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SUMMARY

The discussion focuses on the degrees of freedom for a square sheet and a triangular lamina moving in the XY plane. It is established that both shapes possess three degrees of freedom due to constraints imposed by fixed distances between particles. The analysis includes a comparison to a simple pendulum, which has one degree of freedom defined by the angle of oscillation. The conversation emphasizes the importance of generalized coordinates in determining motion and velocity.

PREREQUISITES
  • Understanding of degrees of freedom in mechanics
  • Familiarity with generalized coordinates
  • Basic knowledge of particle constraints in physics
  • Concept of motion in two-dimensional space
NEXT STEPS
  • Research the application of generalized coordinates in classical mechanics
  • Study the degrees of freedom in rigid body motion
  • Learn about constraints and their effects on particle systems
  • Explore the dynamics of simple pendulums and their degrees of freedom
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This discussion is beneficial for physics students, mechanical engineers, and anyone interested in the principles of motion and constraints in two-dimensional systems.

Jhansi1990@gma
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What are the number of degrees of freedom of
1)a square sheet moving in XY plane
2)a triangular lamina moving freely in XY plane
 
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This looks like homework: What did you find out so far?
 
A square is composed of many particles with the constraint that distance between every particle is the fixed.so such a square is moving in XY plane.

if i consider two particles they have 4-1 =3degree of freedom(one is subtracted due to constraint that distance between particles are fixed).if i consider third particle it is defined by two co-ordinates and two constraints and therefore no degree of freedom...the same for fourth fifth and so on...

so my answer is three

and i don't find any difference in this respect with a triangular lamina...



I do not know the answer to the question...please comment on this and say if you have any other opinion.
 
Those 3 degrees of freedom just fix the current position of the square. What about its movement?
 
if i get generalized co-ordinates i can calculate co-ordinate velocities from it
 
Is that related to the original question?
What do you mean with "get"?
 
The question does not ask that .But i told a general principle.For example a simple pendulum which is oscillating in a plane.since its moving along the arc of a circle(distance from orgin is fixed) it has only one degree of freedom...theta...which is the angle dat the string makes with vertical...so if i know theta as a function of time...i can differentiate "theta" to find generalised co-ordinate velocity
 

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