Degrees of Freedom: Square & Triangular Lamina

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A square sheet moving in the XY plane has three degrees of freedom due to constraints between its particles. Each additional particle adds constraints, ultimately fixing the square's position while allowing movement. A triangular lamina exhibits similar behavior, also having three degrees of freedom under the same conditions. The discussion highlights the importance of generalized coordinates for calculating velocities, though the original question focuses on degrees of freedom rather than motion specifics. Overall, both shapes are constrained similarly in their movement within the XY plane.
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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