CM & Relativity-Slides-Week 4 (1)
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Transcript of CM & Relativity-Slides-Week 4 (1)
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7/30/2019 CM & Relativity-Slides-Week 4 (1)
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Lagrange, 1736-1831
What Lagrange did?
---- Formulated and interpreted D Alemberts principle ingeneralized coordinate.
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APPLICATIONS OF LAGRANGEs EQUATION
---- (1) --- (2)
For any given physical system
(Not necessarily a mechanical system):
1. Step 1: Locate generalized coordinate/s.2. Step 2: Express K. E. and P. E in terms of g.c.
3. Step 3: Determine L = T-V
4. Step 4: Use equation (1) if there is a force or (2) if force free.
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Example 4.1: (By Lagrangess equation)
Q. Determine equation of motion of an idea Atwoods machine
Answer (hints)
Let g. C. Be x (position of m1).
Follow the steps (1-3).
Use equation (2)
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Example 4.2: (By Lagrangess equation)
Q. A] Analyze variation of charge with time in LC circuit. (Assumption:
Charge does not flow at time t = 0).
Answer (Hints)
Take q as g.c.
Write expression of K. E. And P. E. In
Terms of L and C.
Use equation 2.
B. Show that S. H. O. motion is a mechanical analog of the above.
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If there is a dissipation in the system.
Examples:
(i) Ball falls under gravity in a liquid.
(ii) A resistance in LR circuit.
(iii) Damped harmonic oscillator.
MechanicalAnalog
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Example 4.3: (Particle in an Electromagnetic field)
Q. Generalized potential or velocity dependent potentialis applicableTo a very important force field, namely electromagnetic forces of moving
charges. From x component of the Lorentz forceF = q [E + (v x B)
show that, the potential energy is
U = q - qA.v [Where, = (x,y,z,t) and A = A (x,y,z,t) are scalar and vectorpotential respectively]
What is the Lagrangian of the system?
Example 4.4: (Generalized kinetic energy)
Q. A] Derive an expression for generalized kinetic energy. Show that if transformation
Equations do not contain time explicitly, then K. E. Is always a homogeneous,
quadratic form of generalized velocities.
B] Show an example in case of time dependent transformation.
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Example 5 (On degrees of freedom)
Q. A] To describe a rigid body in space, one must know three non collinearpoints on it. The constraint condition is that the distance between two
points is constant. Calculate number of degrees of freedom.
B] If now one point is made fixed in space, what is the degree of freedom?
Ans. (Hints)
Write down equations of distances for three points and use constraint
Conditions [Constraint equations].
No. of d.o.f. = 9 -3 = 6
The motion of a rigid body can always be understood as a translation of any of its points relative
to an inertial system and a rotation of the body about this point. -- M. Chasles
Now, if one point is fixed, d. o. f. = 6-3=3