Homework 3: # 2.13, 2.14
Michael Good
Sept 10, 2004
2.13 A heavy particle is placed at the top of a vertical hoop. Calculate the
reaction of the hoop on the particle by means of the Lagrange’s undetermined
multipliers and Lagrange’s equations. Find the height at which the particle falls
oﬀ.
Answer:
The Lagrangian is
L
=
T

V
⇒
L
=
1
2
m
( ˙
r
2
+
r
2
˙
θ
2
)

mgr
cos
θ
Where
r
is the distance the particle is away from the center of the hoop. The
particle will eventually fall oﬀ but while its on the hoop,
r
will equal the radius
of the hoop,
a
. This will be the constraint on the particle. Here when
θ
= 0, (at
the top of the hoop) potential energy is
mgr
, and when
θ
= 90
o
(at half of the
hoop) potential energy is zero. Using Lagrange’s equations with undetermined
multipliers,
∂L
∂q
j

d
dt
∂L
∂
˙
q
j
+
X
k
λ
∂f
k
∂q
j
= 0
with our equation of constraint,
f
=
r
=
a
as long as the particle is touching
the hoop. Solving for the motion:
d
dt
∂L
∂
˙
r
=
m
¨
r
∂L
∂r
=
mr
˙
θ
2

mg
cos
θ
λ
∂f
r
∂r
=
λ
*
1
thus

m
¨
r
+
mr
˙
θ
2

mg
cos
θ
+
λ
= 0
solving for the other equation of motion,
1
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dt
∂L
∂
˙
θ
=
mr
2
¨
θ
+ 2
mr
˙
r
˙
θ
∂L
∂θ
=
mgr
sin
θ
λ
∂f
θ
∂θ
=
λ
*
0
thus

mr
2
¨
θ

2
mr
˙
r
˙
θ
+
mgr
sin
θ
= 0
The equations of motion together are:

m
¨
r
+
mr
˙
θ
2

mg
cos
θ
+
λ
= 0

mr
2
¨
θ

2
mr
˙
r
˙
θ
+
mgr
sin
θ
= 0
To ﬁnd the height at which the particle drops oﬀ,
λ
can be found in terms
of
θ
. The force of constraint is
λ
and
λ
= 0 when the particle is no longer under
the inﬂuence of the force of the hoop. So ﬁnding
λ
in terms of
θ
and setting
λ
to zero will give us the magic angle that the particle falls oﬀ. With the angle
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 Spring '12
 atkin
 Equations, Energy, Kinetic Energy, Work, θ, Hoop

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