Since the Lagrangian depends explicitly on time, we know that the energy function is
not conserved:
dh
dt
= −
@L
@t
= −ml ˙ (¨X cos + ¨ Y sin ) − m˙R · ¨R − mg˙Y
We can also use this result to calculate the rate of change in the mechanical energy:
E = h + ml ˙ ( ˙X cos − ˙Y sin ) + mV 2
dE
dt
= dh
dt
+ d
dt
ml ˙ ( ˙X cos − ˙Y sin ) + mV 2
= −mg˙Y + ml¨( ˙X cos − ˙Y sin ) − ml ˙ 2( ˙X sin + ˙Y cos ) + m˙R · ¨R
which is of course the same expression we had obtained earlier.
Let us look at some particular cases:
• Small angle approximation: 1
The Lagrangian is
L =
1
2ml2 ˙ 2 + ml ˙ ( ˙X cos − ˙Y sin ) +
1
2m( ˙X 2 + ˙Y 2) + mgl cos + mgY
1
2ml2˙2 −
1
2mgl2 + ml ˙ ( ˙X − ˙Y ) +
1
2mV 2 + mgl
The equation of motion is
d
dt
@L
@˙
−
@L
@
= ml2 ¨ + ml(¨X − ¨ Y ) + mgl
0 = l¨ + ¨X + (g − ¨ Y )
The equation of motion does not necessarily with periodic solutions, unless ¨X = 0
and ¨ Y is constant, in which case the oscillations have a frequency !2 = (g − ¨ Y )/l.
The frequency increases if the motion is accelerated down (in the same direction than
gravity), or lower if the support is accelerated upwards.
The energy is
E = =
1
2ml2˙2 +
1
2mV 2 + ml ˙ ( ˙X cos − ˙Y sin ) − mgl cos − mgY
1
2ml2˙2 +
1
2mgl2 + ml ˙ ( ˙X − ˙Y ) +
1
2mV 2 − mgY − mgl
The rate of change in energy is
dE
dt
= −mg˙Y + ml¨( ˙X cos − ˙Y sin ) − ml ˙ 2( ˙X sin + ˙Y cos ) + m˙R · ¨R
l¨ ˙X − mg˙Y + m˙R · ¨R