The energy is
E =
1
2ml2˙2 +
1
2mV 2 + ml ˙ ( ˙X cos − ˙Y sin ) − mgl cos − mgY
=
1
2ml2˙2 − mgl cos +
1
2mY 2
0
2 cos2
t + mlY0
˙ sin cos
t − mgY0 cos
t
and the power absorbed is
dE
dt
= −mg˙Y + ml¨( ˙X cos − ˙Y sin ) − ml ˙ 2( ˙X sin + ˙Y cos ) + m˙R · ¨R
= mgY0
sin
t + mlY0
¨sin
t sin + mlY0
˙2 sin
t cos + mY 2
0
2 sin
t cos
t
= mgY0
sin
t +
1
2mY 2
0
2 sin 2
t + mlY0
sin
t(¨sin + ˙2 cos )
• Support point in uniform circular motion
The support will move with X(t) = Rsin
t, Y (t) = Rcos
t.
L =
1
2ml2 ˙ 2 + ml ˙ ( ˙X (t) cos − ˙Y (t) sin ) +
1
2m( ˙X (t)2 + ˙Y (t)2) + mgl cos + mgY (t)
=
1
2ml2˙2 + mgl cos + mlR
˙ cos(
t − ) +
1
2mR2
2 + mgRcos
t
Lagrange’s equation is
0 = d
dt
@L
@˙
−
@L
@
= ml2 ¨ + mgl sin − mlR
2 sin(
t − )
= ml2 ¨ + ml(g + R
2 cos
t) sin − mlR
2 cos sin
t
For small oscillations, we see that there is an oscillating contribution to the restoring
gravitational force, and an oscillatory driving term.
The energy is
E =
1
2ml2˙2 − mgl cos + mlR˙ sin(
t − ) +
1
2mR2
2 − mgRcos
t
and the power absorbed by the system is
dE
dt
= −mg˙Y + ml¨( ˙X cos − ˙Y sin ) − ml ˙ 2( ˙X sin + ˙Y cos ) + m˙R · ¨R
= mgR
sin
t + mlR¨
cos(
t − ) + mlR˙2
sin(
t − )