1. Constant of motion of a freely falling particle
A particle of mass m is falling from rest under gravity with uniform acceleration g.
After time t from starting, let the position (from the point of start) and the speed of
the particle be z and v, respectively. Show that an infinitesimal translation in space will
correspond to a symmetry transformation of the Lagrangian. Using Noether's theorem,
find the corresponding constant of motion.
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Answer:
If we consider first 2t seconds, distance travelled is x
1
+x
2
Hence, v
2
−u
2
=2as
Since u=0, a=g and s=x
1
+x
2
,
v
2
=2g(x
1
+x
2
)
v=
2g(x
1
+x
2
)
Now, v=u+at gives us
t=
a
v
=
g
2g(x
1
+x
2
)
Thus, t=
2g
(x
1
+x
2
)
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