Velocity of vertices of a free rhombus

  • Thread starter guv
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  • #1
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Homework Statement:
Suppose we have a rhombus made of 4 point masses ##m## and massless rigid rod at length ##l##. It's placed on a horizontal frictionless table initially at rest. Let a corner be A and the angle at the corner ##\theta## initially.

Then a sudden impulse ##j## is applied symmetrically on the rhombus at the corner A such that A moves along the direction of the impulse. Find the velocities of all the corners.
Relevant Equations:
##v_{CM} = \frac{j}{4m}##
Let D be the opposite corner. In the CM frame, A moves towards CM, D moves towards CM as well. The other two corners (let them be B and C) moves away from the corner at ##v##. Then
##v_A \cos \theta/2 = v \sin \theta/2##

##v_{CM} = \frac{j}{4m}##

This is where it seems like the problem is under-constrained, it is not possible to determine individual velocities unless one velocity is specified. Let me know if I didn't explain this well enough or if I missed anything in solving this problem.
 

Answers and Replies

  • #2
ergospherical
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Just an idea, not sure if it's correct though. Just after the impulse, in the CM frame the point A initially moves at speed ##3j/4m## whilst all the other vertices initially move in the opposite direction at speed ##j/m##.

Since the momentum is zero in the CM frame, points B & C have equal vertical speed but can have no horizontal speed, whilst points A & D equal horizontal speed but can have no vertical speed, so tension forces in the rods act so as to quickly fulfil this condition.

But the energy of the system remains constant during this process, so the sum ##mv^2 + mv_A^2## equals the initial kinetic energy in terms of ##j## and ##m##. And this would provide the additional constraint.

Do you see any problems with this?
 
  • #3
haruspex
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It can't be underconstrained since one could set it up as an experiment.
I don't see how energy can be used since we don't know how much energy was injected.
You can simplify it by shifting to the CM frame, so you have equal and opposite impulses at A and D. Further, reduce it to a single rod with masses constrained to move along the X and Y axes.
Consider the impulse along the rod and write separate impulse equations for the two masses.
 
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  • #4
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I think the velocity component equation is a statement of impulse relationship as well.
 
  • #5
haruspex
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I think the velocity component equation is a statement of impulse relationship as well.
There is a relationship between the impulse components other than as derivable from the velocity relationship.
The rods are massless. Consider the impulsive torque on one.
 

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