Need help problem 34 boxed problem use lagranges method Dete

Need help problem 34 boxed problem use lagrange\'s method

Determine the equations of motion in terms of x and phi. Assume small angles and that the wheel rolls without slip. The mass of the thin homogeneous large disk of radius 2R is 2m. The mass of the thin homogeneous inner disk of radius R is m. The rod of length 2R is massless and rigid. The two pulleys are massless. Use Lagrange\'s Method. Answer [11/2m 2mR 2mR 4mR^2] {double dot x double dot phi} + [c 0 0 0] {dot x dot phi} + [13 k + 2(mg/R) - 4kr -4kr 4kR^2 + 2mgR] {x phi} = {-F(t) + M(t)/R -2RF(t)}

Solution

...Points on a sphere of radius R are determined by two angular coordinates, an az-
imuthal angle and a polar angle :
r = xˆi + y
ˆj + zkˆ = R(sin cos
ˆi + sin sin
ˆj + cos kˆ)
When moving on the sphere, the differential arc length ds is
ds2 = dx2 + dy2 + dz2
= R
2
((cos cos d sin sin d)
2 + (cos sin d + sin cos d)
2 + ( sin d)
2
)
= R
2
(d2 + sin2 d2
)
The distance on the sphere between two points is then
l =
Z
ds = R
Z q
d2 + sin2 d2 = R
Z
ds
d
d 2
+ sin2 = R
Z
df(, 0
)
We can use a variational principle for finding the path with minimum length between
two point. The path is described by a function (), and the (differential) equation for
can be obtained from the Euler-Lagrange equation using f(, 0
) = p
sin2 + 02.
Back to the variational principle: the equation for is
0 =
d
d
f
0

f
0
=
d
d

0
f


sin cos
f
=

00
f


0f
0
f
2

sin cos
f
=

00
f


0
f
2

0
sin cos +
0
00
f

sin 0 = (
00 sin cos )f
2
02
(
00 + sin cos )
0 = (
00 sin cos )(
02 + sin2 )
02
(
00 + sin cos )
=
00 sin2 2
02
sin cos sin3 cos
This looks like a complicated equation to solve! It’s always useful if we know the
solution before we obtain it, admittedly not the most common case, but true in this
case. We know that the shortest path between points in the sphere are great circles.
Great circles are the intersection between the sphere and a plane. If the unit vector
normal to the plane as nˆ = aˆi+b
ˆj+ckˆ, the points in the great circle are those points
in the sphere that satisfy nˆ · r = 0 = R(sin (a cos + b sin ) + c cos ), or those
points with coordinates , satisfying
cos
sin
= A cos + B sin
with A2 + B2 < 1. If we define a function q() = cos ()/ sin(()), we are looking
for an equation of the form d
2
q/d2
= q. If q = 1/ tan , then q
0 =
0/ sin2 ,
and q
00 =
00/ sin2 + 2
02
cos / sin3 . Lagrange’s equation can then be written
as
0 = q
00 sin4 sin3 cos
q
00 = cos / sin = q
which is the equation we were looking for, with a general solution
q =
cos
sin
= A cos + B sin
which we know describes points on a great circle.
2. Exercise 2-14: A hoop rolling on a cylinder
We can find out the angle at which the hoop falls from the cylinder by obtaining an
expression for the normal force on the hoop as a function of the position of the hoop:
the hoop will fall off when the normal force vanishes.
We set up a coordinate system with the origin at the center of the cylinder, and
describe the center of mass of the hoop with polar coordinates r, , and an angular
coordinate for the rotation about the hoop’s axis, as shown in the figure.
The kinetic energy is
T =
1
2
mv2 +
1
2
I2 =
1
2
m( r
2 + r
2
2
) + 1
2
ma22
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