Unformatted text preview: ting it all together gives
t2 d2
dt2 ∂ L ˙ t2
β.
∂ x t1
¨
t1
t1
(313)
The boundary term involving β is zero, because β is assumed to vanish at the
˙
endpoints. But the boundary term involving β is not necessarily zero, because
the derivative is not assumed to be zero at the endpoints. The proposed result is
therefore not valid.
dS
=
da β ∂L
d
−
∂x
dt ∂L
∂x
˙ + ∂L
∂x
¨ dt + β ∂L
d
−
∂x
˙
dt t2 ∂L
∂x
¨ + 6.35. Constraint on a circle
Let Constraint on a curve
6.38. the constraining potential be V (r). Then the Lagrangian is 75
The true Lagrangian is L = (m/2 x22 + y 2 ) − V (η ), where η is the distance from
2)( ˙ ˙ ˙
1
L = m(r + r θ2 ) − V (r).
˙
(314)
the curve. The equations of 2
motion are
The equations of motion are
dV ∂η
∂η
mx = −
¨
·
=⇒ mx = F
¨
,
dη ∂ x
∂x
d
dV
d
2˙
˙2 ,
(mr) = −
˙
+ mrθ ∂η and
(mr θ) = 0.
dV
∂η
dt
dr
dt
my = −
¨
·
=⇒ my = F
¨
.
dη ∂ y
∂y
Using r = R =⇒ r = r = 0, the ﬁrst of these equations gives
˙¨
For a point on the curve, we ha...
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This note was uploaded on 01/19/2014 for the course PHYSICS 251 taught by Professor Brandenberger during the Fall '13 term at McGill.
 Fall '13
 Brandenberger
 mechanics, Centrifugal Force, Force

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