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Physics 325 Homework 9 – Solutions – Fall 2010
1)
A hoop of mass
m
and radius
R
rolls without slipping down an inclined plane of mass
M
, that makes an angle
θ
with the horizontal. Find the Lagrange equations and the
integrals of the motion if the plane can slide without friction along a horizontal
surface. [
Hint
: Recall that the kinetic energy of rotation can be written
as
T
rot
=
1
2
I
!
2
, with
I
the moment of inertia. For a hoop,
I
=
mR
2
.]
Solution
: We set up a coordinate system as is shown in the figure:
We place the origin of the
x

y
coordinate system fixed on the surface that holds the
inclined plane. The position of the incline will be measured as
x
1
,
y
1
( )
and the
position of the hoop will be measured as
x
2
,
y
2
( )
. We will measure the position of the
hoop along the plane as
s
measured from the bottom of the plane. Note we have
defined the height and base of the incline as
h
and
d
respectively for convenience,
they will not show up in the result.
We begin by noting that five variables are required to specify the motion of the
hoop, four spatial dimensions (
x
1
,
y
1
,
x
2
,
y
2
) and one angular dimension (
φ
) to specify
the orientation of the hoop. We have three constraints:
•
The inclined plane remain on the table:
y
2
=0.
•
The hoop moves on the surface of the incline plane. This relates the x2 and y2
coordinates or, alternatively, relates
x
2
and
y
2
to
s
. This can be written as:
sin
=
y
2
/
s
.
(0.1)
•
The rotational motion of the hoop is related to the translational motion. This
constraint can be written as:
s
=
R
"
!
s
=
R
!
.
(0.2)
Five coordinates with three constraints gives two degrees of freedom. We choose
coordinates
x
1
and
s
to specify our system.
The potential energy function can be written as:
U y
( )
=
mgy
2
=
mgs
sin
(0.3)
R
g
hoop,
mass
m
I
=
mR
2
h
d
incline mass M
y
x
s
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View Full Documentwhere we have defined the potential function so that
U
(0)=0. We have additionally
written the potential energy in terms of
s
based upon equation (0.1).
The kinetic energy of the inclined plane is:
T
plane
=
1
2
M
!
x
1
2
(0.4)
The kinetic energy of the hoop is going to have a contributions from the
translational motion of the hoop along the inclined plane, the translational motion of
the inclined plane along the surface, and a contribution from rotational motion:
T
hoop
=
1
2
m
!
v
i
!
v
+
1
2
I
"
!
2
(0.5)
We can write the velocity of the hoop as the sum of the motion along the inclined
plane plus the motion of the inclined plane:
!
v
=
"
s
ˆ
s
+
"
x
1
ˆ
i
=
!
"
s
cos
"
+
"
x
1
( )
ˆ
i
+
"
s
sin
ˆ
j
(0.6)
Note: as the hoop rolls down the incline (
x
2
increasing)
s
is decreasing. We now
write the kinetic energy of the hoop as:
T
hoop
=
1
2
m
!
!
s
cos
+
!
x
1
( )
2
+
!
s
2
sin
2
#
$
%
’
(
+
1
2
I
!
)
2
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 Fall '08
 Staff
 Physics, mechanics, Mass, Work

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