Assignment 4, solutions

Problem 1:

Two spheres are of the same radius R and mass M, but one is solid and the other is a hollow shell (of negligible thickness).  Both spheres roll (without sliding) down a ramp of incline q.
(a)  Which sphere will have the greater acceleration down the ramp?
(b)  Determine the Lagrangian for the motion of the sphere and derive the equation of motion for both cases.

Solution:

Concepts, principles, relations that apply to the problem:
Lagrange's Equations,
Why do they apply?
All forces except the forces of constraint are derivable from a potential.  The Lagrangian formalism is well suited for such a system.  We are asked to derive Lagrange's equation of motion.
How do they apply?
(a)  We have a conservative system.  Potential energy is converted into rotational and translational kinetic energy ½Iw2 + ½Mv2 with v = wR.  The moment of inertia of the solid sphere I = (2/5)MR2 is smaller than the moment of inertia of the thin spherical shell I = (2/3)MR2, so the solid sphere will have the greater acceleration down the ramp.
(b)  Set up a coordinate system:

L = T – V.  T = ½(I/R2 + M)(dx/dt)2,  V = Mgxsinq.
L = ½(I/R2 + M)(dx/dt)2 - Mgxsinq.
Equation of motion: (d/dt)(L/v) – L/x = 0, with v = dx/dt.
(I/R2 + M)d2x/dt2 + Mgsinq = 0.
a =  d2x/dt2 = (M/(I/R2 + M))gsinq.
For the solid sphere:  a = (5/7) gsinq.
For the spherical shell:  a = (3/5) gsinq.

Details of the calculation:
None

Problem 2:

Obtain Lagrange's equations of motion for a spherical pendulum (a mass point suspended by a rigid, weightless rod).

Solution:

Concepts, principles, relations that apply to the problem:
 Lagrange's Equations: ; generalized coordinates
Why do they apply?
We are asked to obtain Lagrange's equation of motion.
How do they apply?

The system has only two degrees of freedom.  Choose the generalized coordinates to be q and f.  The constraints are holonomic, the applied forces are conservative.

,  and  q1 = q,  q2  = f.
L = T – U.
 .
U = mg r cosq.

.

.

In this problem  f is cyclic,  is a constant of motion.

Details of the calculations:
Equations of motion:


Note:
 Because f is cyclic, we already have a first integral.  We can insert df/dt = constant/sin2q into the expression for d2q/dt2.

Problem 3:

The Lagrangian for a simple spring is given by

Find the Hamiltonian and the equations of motion using the Hamiltonian formulation.  Identify any conserved quantities.

Solution:

Concepts, principles, relations that apply to the problem:
The Lagrangian and the Hamiltonian, Hamilton's equations of motion
Why do they apply?
We are asked to find the Lagrangian and the Hamiltonian and Hamilton's equations of motion.
How do they apply?
H = p2/(2m) + ½ kx2.  H is a function of x and p, while the Lagrangian is a function of x and v.
Equations of motion: dx/dt = H/p,  dp/dt = -H/x.
dx/dt = p/m,  dp/dt = -kx.  We have two first-order differential equations which can be combined to yield the second order differential equation d2x/dt2 = -(k/m)x.
The coordinates and the Hamiltonian do not contain time explicitly, so H = E, and E = constant.  The energy E is a conserved quantity.
Details of the calculation:
 None

Problem 4:

A particle of mass m moves in one dimension under the influence of a force

F(x,t) = -k x exp(-t/t)

where k and t are positive constants. (Note carefully the dependence on x to the first power).
(a) Compute the Lagrangian function.
(b) Use Lagrange’s equation to determine the equation of motion explicitly.
(c) Compute the Hamiltonian function in terms of the generalized coordinate and generalized momentum. (Show clearly how you get this.)
(d) Determine Hamilton’s equations of motion explicitly for this particular problem (not just general formulae).
(e) Does the Hamiltonian equal the total energy?
(f) Is the total energy of the mass conserved?
(g) What is it about the force F which supports your answer to part f?

Solution:

Concepts, principles, relations that apply to the problem:
Lagrangian and Hamiltonian mechanics
Why do they apply?
We are asked to find the Lagrangian and Hamiltonian for the system.
How do they apply?
(a)  F(x,t) = - U(x,t)/x.   U(x,t) = k(x2/2)exp(-t/t).
L = T – U = ½ m -  k(x2/2)exp(-t/t).
(b)  
L(x,t)/x = -kxexp(-t/t).
md2x/dt2 = -kx exp(-t/t).
(c)  H(x, p, t) = p – L.
p = L/= m.   H = p2/(2m) + k(x2/2)exp(-t/t).
(d)   = H/p = p/m.    = -H/x = -kxexp(-t/t).
= /m = -(k/m)xexp(-t/t).
(e)  Yes.  If the generalized coordinates do not explicitly depend on time, then H = T + U = E, the total energy of the system.
(f)  No.  If H does explicitly depend on time, H = H(t), then H is not a constant of motion.
(g)  The force F is a function of position and time.  It is derivable from a time dependent potential energy function.  F = -k(t)x, it is a Hooke’s law type force with a spring constant that decreases exponentially.  Assume a particle has kinetic energy E0 = ½mv02 as it moves through the origin at t = 0 and comes to a stop at position x.  It then accelerates back towards the origin.  Since the magnitude of the force decreases with time, the magnitude of the positive work done by F on the return trip is smaller than the magnitude of the negative work done by F on the trip from the origin to x.  The potential energy at the origin is zero, the kinetic energy of the particle has decreased, the particle has lost energy.
Details of the calculation:
 None

Problem 5:

A point mass m slides without friction along a wire bent into a vertical circle of radius a.  The wire rotates with constant angular velocity W about the vertical diameter, and the apparatus is placed in a uniform gravitational field g parallel to the axis of rotation.
(a)  Construct the Lagrangian for the point mass using the angle q (angular displacement measured from the downward vertical) as its generalized coordinate and find its equation of motion.
(b)  What are the equilibrium values for q?  For each equilibrium value, what is the condition for the equilibrium to exist?
(c)  For each equilibrium value, is the equilibrium stable or unstable against small displacements along the wire?  What is the oscillation frequency or growth rate of such perturbations?

Solution:

Concepts, principles, relations that apply to the problem:
Lagrange’s equations
Why do they apply?
Lagrange’s equations are the equations of motion.
How do they apply?

(a)  L = T – U.
T = ½m(a2(dq/dt)2 + a2sin2q W2),  U = -mgacosq.
L =  ½m(a2(dq/dt)2 + a2sin2q W2) + mgacosq.
Here r and W are constants.

,

 .

 d2q/dt2 = sinq cosq W2 – (g/a) sinq.

(b)  d2q/dt2|qequ = 0 for a stationary point.
The angles q = 0, q = p, or cosq  = g/(aW2) define stationary points.
We need g/(aW2) < 1 for the point q = cos-1(g/aW2) to exist.
For a stable point we need a restoring force.  Let a = qequ + d.
For small d we need d2d/dt2 = -cd, with c a positive number.
d2d/dt2 = sin(qequ + d)cos(qequ + d) W2 – (g/a) sin(qequ + d)
= d(sinq cosq W2 – (g/a) sinq)/dq|equ*d  (series expansion)
= (cos2qequ W2 - sin2qequ W2 - (g/r)cosqequ)d.
We need cos2qequ W2 - sin2qequ W2 - (g/a)cosqequ to be negative.

Details of the calculation:
For the stationary point qequ = 0:
cos2qequ W2 - sin2qequ W2 - (g/a)cosqequ = W2 – (g/a).
If g/a > W2  then this point is stable, otherwise it is unstable.
Frequency of small oscillations if g/a > W2: w = ((g/a) – W2)1/2.
Growth rate if g/a < W2: l = (W2 – (g/a))1/2.

For the stationary point qqequ = p:
cos2qqequ W2 - sin2qequ W2 - (g/a)cosqequ = W2 + (g/a).
This point is always unstable.
Growth rate if: l = (W2 + (g/a))1/2.

For the stationary point qqequ = cos-1(g/(aW2)):
cos2qequ W2 - sin2qequ W2 - (g/a)cosqequ = g2/(a2W2) - W2 < 0 since g/(aW2) < 1 for the point to exist.
This point is always stable if it exists.
Frequency of small oscillations if point exists: w = (W2 – g2/(a2W2))1/2.

Problem 6:

A small object with mass m moves on a smooth, friction-free horizontal surface.  It is attached to a peg at the origin by an ideal massless spring with spring constant k and equilibrium length r0.  At time t = 0, the mass is set in motion in an arbitrary direction from point (r,q).

(a)  Find the Lagrangian L for the system, then
(b)  calculate the generalized momenta pj .
(c)  Construct the Hamiltonian function, H(pj, qj, t);
(d)  then work out the equations of motion dpj/dt and dqj/dt. 
(e)  Are any of the variables cyclic, thereby giving especially simple equations of motion?  If so, integrate the equation(s) and interpret your results physically.
(f)  Consider the special case that r = constant.  Deduce the condition(s) that allow this case and discuss how this occurs physically.

Solution:

Concepts, principles, relations that apply to the problem:
Lagrangian and Hamiltonian mechanics
Why do they apply?
We are asked to find the Lagrangian and Hamiltonian for the system.
How do they apply?
Assume the spring can rotate about the origin, but not bend.
This is a two-dimensional problem.
(a)  L = T - U.
T = ½m(vx2 + vy2) = ½m(r2 + ),  U = ½k(r-r0)2.
L = ½m(r2 + ) - ½k(r-r0)2.
(b)  L/ = pr,   L/= pf,   pr = m,  pf  =  mr2.
(c)   H = pr2/(2m)  + pf2/(2mr2) + ½k(r-r0)2, since  .
(d)  pr/t = -H/r = -k(r-r0) + pf2/(mr3), pf/t = -H/f  = 0.
r/t = H/pr = pr/m, ¶f/t = H/pf  = pf/(mr2).
(e)  The coordinate f is cyclic.
If the Lagrangian does not contain a given coordinate qj then the coordinate is said to be cyclic and the corresponding conjugate momentum pj is conserved.
pf = constant = M,  = M/(mr2).
The Lagrangian and the coordinates do not contain time explicitly, therefore E = T + U is a constant of motion.
(f)  We want  and  to be zero.
m  = -k(r-r0) + M2/(mr3) = 0,  k(r-r0) = M2/(mr3).
k(r-r0)  = mr =  mv2/r.
Rotation with uniform angular velocity, spring is stretched providing the centripetal force.
Details of the calculation:
None

Problem 7:

Consider a hoop of mass m and radius r rolling without slipping down an incline.
(a)  Determine the Lagrangian L(x, dx/dt) of this one-degree-of-freedom system.  Derive from it the Lagrange equation and its solution for initial condition x0 = 0, dx/dt|0 = 0.
(b)  Determine the alternative Lagrangian L(x, dx/dt, q, dq/dt)  and the holonomic constraint f(x, q) = 0 that must accompany it.  Derive the associated three equations of motion for the two unknown dynamical variables x, q and the undetermined Lagrange multiplier l.  Solve these equations for the same initial conditions as in (a) and determine the static frictional force of constraint between the hoop and the incline.

Solution:

Concepts, principles, relations that apply to the problem:
Lagrange's equations, , Lagrange multipliers
Why do they apply?
In part (a) we use the constraint of rolling to eliminate the coordinate q.
In part (b) we are asked to use the method of Lagrange multipliers.  Then the equations of motion are be obtained from

and    .

How do they apply?
(a)  L = T - U.
T = (m/2)v2 + (I/2)w2.  v = dx/dt,  w = dq/dt = v/r, I = mr2, U = -mgxsinf.
L = m(dx/dt)2 + mgxsinf.
L/v = 2mdx/dt,  (d/dt)L/v = 2md2x/dt2, L/x = mgsinf.
d2x/dt2 = (g/2)sinf.  ax = (g/2)sinf, vx = (gt/2)sinf, x = (gt2/4)sinf.
Details of the calculation:
(b)  L = (m/2)(dx/dt)2 + (mr2/2)(dq/dt)2 + mgxsinf.
Constraint: dq = dx/r,  dx - rdq  = 0.
We have two coordinates, x and q, and one equation of constraint, dx - rdq  = 0, which must be cast into the form l1S a1k dqk = 0,  l1(a1x dx + a1q dq) = 0.  Therefore a1x = 1, a1q  =  -r.
L/v = mdx/dt,  (d/dt)L/v = md2x/dt2, L/x = mgsinf.
L/¶w = mr2dq/dt,  (d/dt)L/¶w = mr2d2q/dt2, L/¶q = 0.
md2x/dt2 - mgsinf = l,  mr2d2q/dt2 = -rl,  d2x/dt2 = rd2q/dt2.
3 equations, 3 unknowns.
d2x/dt2 = (g/2)sinf, d2q/dt2 = (g/(2r))sinf,  l =-(mg/2)sinf.
Here l = Fx, the static frictional force of constraint between the hoop and the incline.