More Problems

Problem 1:

Two point masses of mass m are joined by a rigid, weightless rod of length l, the center of which is constrained to move on a circle of radius a.  Set up the kinetic energy in generalized coordinates.

Solution:

Concepts, principles, relations that apply to the problem:
Generalized coordinates
Why do they apply?
We are asked to find generalized coordinates (independent quantities) that reduce the number of degrees of freedom of the system and express the kinetic energy of the system in terms of these generalized coordinates.
How do they apply?

The system has 3-degrees of freedom.
If we specify f, f', and q', the positions of both particles are specified.
T = kinetic energy of the system if all the mass were concentrated at the center of mass + kinetic energy of the motion about the center of mass.
Details of the calculation:
None

Problem 2:

A uniform ladder of mass M and length l slides without friction from wall or floor.

(a)  Set up the second order differential equation of motion, assuming the ladder remains in contact with the wall.
(b)  If the ladder is initially at rest at an angle a0 with the floor, at what angle, if any, will it break contact with the wall?

Image512.gif (1442 bytes)

 

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.

How do they apply?
(a)   As long as the ladder remains in contact with the wall, we have the constraints
x = (l/2)cosa and y = (l/2)sina for the coordinates of the CM.
The Lagrangian L is given by L = T - V.

T is the kinetic energy of the motion of the CM and the motion about the CM.




 is the equation of motion as long as there is contact with the wall.

Details of the calculations:
(b)  If the wall pushes on the ladder d2x/dt2 is greater than zero.  The ladder breaks contact with the wall when d2x/dt2 = 0.

To find d2x/dt2 as a function of a we need da/dt as a function of a.
From energy conservation we find da/dt.  T(a) + V(a) = V(a0).
.
Therefore
.
 .
d2x/dt2 is greater than zero as long as 3sina > 2sina0.  d2x/dt2 is smaller than zero if 3sina < 2sina0.  The board breaks contact with the wall when 3sina = 2sina0. 

Problem 3:

A particle of mass m moves in a force field of potential V(r,q,f).  Write:

(a) The Hamiltonian in spherical coordinates (r,q,f).

(b) Hamilton’s equations of motion in spherical coordinates.

Solution:

Concepts, principles, relations that apply to the problem:
The Lagrangian, the Hamiltonian, Hamilton's equations of motion
Why do they apply?
 L = T - U  is the Lagrangian of the system.  L is a function of the coordinates and the velocities.  The Hamiltonian H of a system is given by

.

H is a function of the generalized coordinates and momenta of the system.  The equations of motion can be obtained from Hamilton’s equations,

.

How do they apply?
(a)
 .
Details of the calculations:
(b)

Problem 4:

A particle of mass equal to 3 kg moves in the xy plane.  The potential energy of the particle as a function of position is given by  U = (36J/m2)xy - (48J/m2)x2.  The particle starts at time t = 0 from rest at the point with position vector (x, y) = (10m, 10m).
(a)  Set up the differential equations describing the motion and solve them to determine the position of the particle as a function of time.
(b)  Find the velocity as a function of time.

Solution:

Concepts, principles, relations that apply to the problem:
Lagrange's Equations:
Why do they apply?
The potential energy is given.  All forces except the forces of constraint are derivable from a potential.  The Lagrangian formalism is well suited for such a system.
How do they apply?
L = T - U = (1/2)m(2 + 2) - 36xy + 48x2  (SI units).
L/ = m,  L/ = m,  L/x = -36y + 96x,  L/y = -36x.
md2x/dt2 + 36y - 96x = 0,  md2y/dt2 + 36x = 0,  m = 3.
d2x/dt2 + 12y - 32x = 0,  d2y/dt2 + 12x = 0.
Details of the calculations:
(a)  Let x = Aexp(at),  y = A'exp(at).
Then a2A +12A' - 32A = 0,  a2A' + 12A = 0, a2A - 144A/a2 -32A = 0.
a4 - 32a2 - 144 = 0,  a2 = 16 ± 20.
a2 = 36, a = ±6 yields exponential solutions.
a2 = -4, a = ±2i yields sinusoidal solutions.
If a2 = 36, then A' = -1/3A.
If a2 = -4, then A' = 3A.
The most general solution is a linear superposition of all possible solutions.
x = a1exp(6t) + a2exp(-6t) + a3cos(2t) + a4sin(2t),
y = -(a1/3)exp(6t) - (a2/3)exp(-6t) + 3a3cos(2t) + 3a4sin(2t).
 = 6a1exp(6t) - 6a2exp(-6t) - 2a3sin(2t) + 2a4cos(2t),
 = -2a1exp(6t) + 2a2exp(-6t) - 6a3sin(2t) + 6a4cos(2t).

At t = 0 we have = 0, = 0, 3a1 - 3a2 + a4 = 0, -a1 + a2 + 3a4 = 0.
At t = 0 we have x = 10, y = 10, a1 + a2 + a3 = 10, -a1 - a2 + 9a3 = 30.
Four equations for four unknowns yield a1 = 3, a2 = 3, a3 = 4, a4 = 0.
x(t) = 3exp(6t) + 3exp(-6t) + 4cos(2t) = 6cosh(6t) + 4cos(2t),
y(t) = -exp(6t) - exp(-6t) + 12cos(2t) = -2cosh(6t) + 12 cos(2t).
Below are contour and a surface plots of the potential for
-12.5 < x < 12.5,
-12.5 < y < 12.5.


(b)  = 18exp(6t) - 18exp(-6t) - 8sin(2t) = 36 sinh(6t) - 8sin(2t).
 = -6exp(6t) + 6exp(-6t) - 24sin(2t) = -12sinh(6t) - 24sin(2t).

Problem 5:

Two masses m1 and m2 are connected by a massless string that runs over a frictionless pulley.  The length of the string, l, somehow increases at a constant rate, i.e., l(t) = l0 + l1 t.  Use the method of Lagrange multipliers to determine the tension of the string at time t.

Solution:

Concepts, principles, relations that apply to the problem:
Lagrange's Equations, ,  Lagrange multipliers
, 
Why do they apply?
The problem requires us to use the method of Lagrange multipliers. 
How do they apply?
 Let the generalized coordinate xi denote the distance along the string from the top of the pulley to the mass mi.
Then L = m112/2 + m222/2 + m1gx1 + m2gx2.
The equation of constraint is x1 + x2 = l0 + l1t, dx1 + dx2 - l1dt = 0.  There is only one equation of constraint.  We write l(ax1 dx1 + ax2 dx2 + atdt)  = 0.
ax1 = 1,  ax2 = 1,  at = -l1.
Details of the calculations:
The equations of motion are
m1d2x1/dt2 - m1g = l,
m2d2x2/dt2 - m2g = l.
From dx1 + dx2 - l1dt = 0 we obtain
d2x1/dt2 + d2x2/dt2 = 0.
Solving these 3 equations we obtain
m1d2x1/dt2 - m2d2x2/dt2 = (m1 - m2)g .
m1d2x1/dt2 + m2d2x1/dt2 = (m1 - m2)g .
d2x1/dt2 = (m1 - m2)g/(m1 + m2).
l = m1(m1 - m2)g/(m1 + m2) - m1g = -2m1m2g/(m1 + m2).
l is the force of constraint acting on m1, its magnitude is the tension in the string, the - sign indicates that it is acting upward on m1.

Problem 6:

A block of mass m moves on a frictionless surface with a displacement x under the influence of a spring force
F = -kx,  where k is the spring constant.
(a)  Calculate the potential energy U(x), and and the Lagrangian.
(b)  Find the Lagrange's equations of motion.
(c)  Find the Hamiltonian.
(d)  Find the Hamilton's equations of motion.

Solution:

Concepts, principles, relations that apply to the problem:
Lagrange's Equations:
,   
Why do they apply?
F is a conservative force.  It is derivable from a potential.  We can use the Lagrangian and Hamiltonian formalism.
How do they apply?
(a)  Hooke's law:  U = (1/2)kx2,  L = T - U = (1/2)m(dx/dt)2 - (1/2)kx2.
(We have one generalized coordinate.)
(b)  L/ = m,  L/x = kx,  d2x/dt2 = -(k/m)x.
(c)  Px = L/ = m.
m2 - L = (1/2)m2 + (1/2)kx2 = Px2/(2m) + (1/2)kx2.
H = Px2/(2m) + (1/2)kx2.
(d)  = H/Px = Px/m,  = -H/x = -kx.
Details of the calculations:
None

Problem 7:

A particle of mass m is rigidly attached to a uniform hoop of radius a and mass m.  The combination is released from rest on a frictionless horizontal surface with the line joining the mass m to the center of the hoop initially horizontal.  The plane of the hoop remains vertical throughout the motion.

(a)  Find the angular velocity of the hoop as a function of the angle q.
(b)  When q = p/2, what is the force exerted on the hoop by the surface?

 

Solution:

Concepts, principles, relations that apply to the problem:
 Lagrange's Equations: , Newton's second law:  F = dp/dt
Centripetal acceleration: a = v2/r, cyclic coordinates
Why do they apply?
Solving Lagrange's equations we can find dq/dt.  Given dq/dt as a function of q, we can find the acceleration at q = p/2.
How do they apply?
Let x be the x-coordinate of the center of the hoop, and x' and y' be the coordinates of the mass m.  We have
x' = x + acosq,  dx'/dt = dx/dt -asinq(dq/dt),
y' = -asinq,  dy'/dt = -acosq(dq/dt).
T = (1/2)ma2(dq/dt)2 + (1/2)m(dx/dt)2 + (1/2)m[(dx'/dt)2 + (dy'/dt)2]
= ma2(dq/dt)2 + m(dx/dt)2 - m a sinq (dx/dt)(dq/dt).
U = -mga sinq.
T + U = 0, energy is conserved.
There is another conservation law.  The coordinate x is cyclic, it does not appear in L.
.
Insert this into T + U = 0 to eliminate dx/dt.
ma2(dq/dt)2 + (1/4)ma2sin2q(dq/dt)2 - (1/2)ma2sin2q(dq/dt)2 = mga sinq.
(dq/dt)2 = (g/a)sinq/[1 - (1/4)sin2q].
Details of the calculations:
(a)  At q = p/2 (dq/dt)2 = (g/a)/[1 - (1/4)], dq/dt = (4g/(3a))1/2.
(b)  At q = p/2 the CM of the system is accelerating towards the center of the loop.
a = jvCM2/(a/2) = j(a/2)2(dq/dt)2/(a/2) = j(a/2)(dq/dt)2.
The total force exerted by the surface on the system therefore is
F = (2mg + 2m(a/2)(dq/dt)2)j.
(dq/dt)2 = (g/a)/[1 - (1/4)] = 4g/(3a).
F = (10mg/3)j.

Problem 8:

A simple pendulum of length b with a bob of mass m is attached to a massless support moving horizontally with constant acceleration a. 
(a) Compute the Lagrangian function. 
(b) Write down Lagrange’s equation of motion. 
(c)  Simplify and compare the equation of motion to that for a simple pendulum with a fixed (motionless) support.

Solution:

Concepts, principles, relations that apply to the problem:
Lagrange's equations
Why do they apply?
We are asked to obtain Lagrange's equation of motion.
How do they apply?
(a)  L = T – U.
T = ˝m((dx/t)2 +(dy/dt)2).
x = ˝at2 + bsinq,  y = -bcosq.
dx/dt = at + bcosq ,  dy/dt = bsinq .
T = ˝m(b2 2 + a2t2 + 2at bcosq ).
U = mgy = -mg bcosq.
L = ˝m(b2 2 + a2t2 + 2at bcosq ) + mg bcosq.
(b)  L/  = mb2  + mat bcosq,  d/dt(L/ ) = mb2 - mat bsinq  + ma bcosq, 
L/¶q = -matbsinq  - mg bsinq.
b + g sinq + a cosq = 0.
(c)  sin(q + q0) = sinq cosq0 + cosq sinq0.
Let g sinq = c sinq cosq0, and a cosq = c cosq sinq0.
Then g  = c cosq0, and a = c sinq0,  c = (g2 + a2)1/2,  tanq0 = a/g.
Define f = q + q0.  Then the equation of motion becomes
bdf2/dt2 = -c sinf.  This equation becomes bdq2/dt2 = -g sinq when a = 0.
For a simple pendulum we have bdq2/dt2 = -g sinq.
Details of the calculation:
 None