分析力学第三次作业解答

We get the same results from the Newton’s second law
6.1 Cartesian or cylindrical coordinates (from Greiner p349). A mass point m shall
move in a cylindrically symmetric potential . Determine the Hamiltonian and the
= 1 (r 2 + r 2 sin 2 θ φ 2 ) − mgr cosθ + λ (θ − α ) 2 + r 2 θ L 2
From equation (2) we get
2 + mgr sin α λ = mr 2 sin α cos α φ
Since
ˆ ) = − rN φ ˆ ˆ × ( − Nθ τ N = r × N = rr
Therefore, Lets use Newton’s laws to arrive at this result
ˆ + r sin α sin φ y ˆ + r cos α z r = r sin α cos φ x ˆ + sin α sin φ y ˆ + cos α z =r [sin α cos φ x ˆ] r
Lagrange’s equation for
is
After a litter algebra, this becomes
By enforcing the constraint,
, this becomes
By drawing a free body diagram, you can convince yourself that is the normal force which the cylinder exerts upon the block, which will vanish when the block leaves the surface of the cylinder. The equations for are like
will be the constraint force in
So, we get
λ = N z = N sin α ⇒ N =
From Newton’s second law:
m( z + g) sin α m( z + g) sin α
N z − mg = maz = m z⇒N =
(a) Find the Lagrangian using generalized spherical coordinates. (b) Identify the cyclic coordinate(s) and associated conserved quantities.
(c) Suppose the particle has very large energy E and fixed angular momentum, . What are the approximate maximum and minimum possible values of the distance from the vertex, ? (d) Find the angular speed for motion having constant and (e) Using the Lagrange method of undetermined multipliers to find the normal force on the particle, and verify your answer using Newton’s second law. . Express your answer in terms of
The coordinate (x,y,z) has a relationship with the coordinate (x’,y’,z’) in the rotating system
Then
So
Thus
(b)
Then
Then
6.2 Small Particle in Bowl (Stony Brook) A small particle of mass m slides without
, we must have:
Or The angular speed is given by
Therefore,
(e) We will write the constraint that the mass moves on the conical surface in two ways; both shall give the constraint force, 1) Spherical polar coordinate
Also, L has no explicit time dependence. Therefore, energy is conserved
(c) Plot
and
are given by
. For E 1, ra 1, rb 1 as
So
as
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(d) For circular motion at
5.1 Block on a cylinder II. Let's return to the block on a cylinder problem of Problem Set 5. A
block of mass m starts at rest at the top of a cylinder of mass M and radius R which is placed on the floor, as shown in the figure. At time t = 0 the block is given a gentle tap and it begins sliding down the surface of the cylinder. Ignore the friction between the block and the cylinder and the cylinder and the floor. Use the method of Lagrange multipliers to nd the normal force which the cylinder exerts upon the block. Determine the angle (as a function of the mass ratio . ) at
which the block loses contact with the cylinder. Check your result in the limit
Solution: Start by introducing the generalized coordinates
, such that
. At this point we haven’t enforced the constraint, which is , since we want to determine the force of constraint and find the angle at which it vanishes. In terms of these coordinates, the Lagrangian is
ˆ ˆ + rφφ r =r
ˆ + rφ ˆ − rφ sin αφ φ 2 sin α [cos φ x ˆ + 2r ˆ + sin φ y ˆ] = r r φ r ˆ ˆ = cos α cos φ x ˆ + cos α sin φ y ˆ − sin α k Now θ
canonical equations of motion with respect to a coordinate system that rotates with constant angular velocity about the symmetry axis: (a) in Cartesian coordinates, and (b) in cylindrical coordinates. Solution: (a)
Where
. We also have
Using the equations above
Solving the equation for
Expressing the result completely in terms of
; after lots of algebra,
To find the critical angle
friction on the inside of a hemispherical bowl, of radius R, that has its axis parallel to the gravitational field g. Use the polar angle and the azimuthal angle to describe the location
5.3. Particle moving along inverted cone. A particle of mass
with the gravitational acceleration, , and the cone angle is .
slides without friction on the surface of a stationary inverted come in a uniform gravitational field. The cone axis is aligned
Solution: (a) In spherical coordinates:
Imposing the constraint
(b) L does not have explicit dependence on is conserved, and is cyclic
. Therefore, the associated generalized momentum
at which the block loses contact with the cylinder, set
; this
results in the following cubic equation for
For
, we recover the result so that
which we had obtained for the immobile cylinder. If is . This
of the particle (which is to be treated has a point particle). (a) Write the Lagrangian for the motion. (b) Determine formulas for the generalized momenta (c) Write the Hamiltonian for the motion. (d) Develop Hamilton’s equations for the motion. (e) Combine the equations so as to produce one second order differential equation for as a function of time. (f) If and independent of time, calculate the velocity (magnitude and direction). and .
Hence, 2) Cartesian coordinates
i
= 1 m( x 2 + y 2+ z L 2 ) − mgz + λ ( z − cot α x 2 + y 2 ) 2
Where we have written the Cartesian coordinates as
Note that g has dimensions of length, and written in this form, the Z-direction
, the physically meaningful solution of
makes sense----if the cylinder is massless, it will immediately shoot to the left once we tap the block, and the block will lose contact with the cylinder.