A mechanical system known as an Atwood machine consists of three weights of
mass m_{1}, m_{2}, and m_{3}, respectively, connected by a light
(massless) inextensible cords of length l - πR and l' - πR, respectively,
which pass over identical pulleys with radius R, mass M, and moment of inertia I. Find the acceleration of m_{1}.
Let m_{1} = m, m_{2} = m, m_{3} = 0.25 m, M = 0.5 m and
I = 0.25 mR^{2}.

Solution:

- Concepts:

Lagrange's equations - Reasoning:

All forces except the forces of constraint are derivable from a potential. The Lagrangian formalism is well suited for such a system. - Details of the calculation:

T = ½[m_{1}(dx/dt)^{2}+ M(dx/dt)^{2}+ (I/R^{2})(dx/dt)^{2}+ m_{2}(d(x - x')/dt)^{2}+ m_{3}(d(x + x')/dt)^{2}+ (I/R^{2})(dx'/dt)^{2}]

= ½[(m_{1}+ m_{2}+ m_{3}+ M + I/R^{2})(dx/dt)^{2}+ (I/R^{2}+ m_{3}+ m_{2})(dx'/dt)^{2}+ 2(m_{3}- m_{2})(dx/dt)(dx'/dt).

U = [-m_{1}x - M(l - x) - m_{2}(l - x + x') - m_{3 }(l - x + l' - x')]g = [(-m_{1}+ M + m_{2}+ m_{3})x + (m_{3}- m_{2})x']g + constant.

L = T - U.

∂L/∂v_{x}= (m_{1}+ m_{2}+ m_{3}+ M + I/R^{2})(dx/dt) + (m_{3}- m_{2})(dx'/dt).

∂L/∂v_{x'}= (I/R^{2}+ m_{3}+ m_{2})(dx'/dt) + (m_{3}- m_{2})(dx/dt).

∂L/∂x = -(-m_{1}+ M + m_{2}+ m_{3})g, dL/dx' = -(m_{3}- m_{2})g.

(m_{1}+ m_{2}+ m_{3}+ M + I/R^{2})d^{2}x/dt^{2}+ (m_{3}- m_{2})d^{2}x'/dt^{2}+ (-m_{1}+ M + m_{2}+ m_{3})g = 0.

(I/R^{2}+ m_{3}+ m_{2})d^{2}x'/dt^{2}+ (m_{3}- m_{2})d^{2}x/dt^{2}+ (m_{3}- m_{2})g = 0.

We have two equations for two unknowns, d^{2}x/dt^{2}, and d^{2}x'/dt^{2}.

Plug in numbers:

3 d^{2}x/dt^{2}- 0.75 d^{2}x'/dt^{2}+ 0.75 g = 0.

1.5 d^{2}x'/dt^{2}- 0.75 d^{2}x/dt^{2}- 0.75 g = 0.

d^{2}x'/dt^{2}= 0.5 d^{2}x/dt^{2}+ 0.5 g.

(3 - 0.375)d^{2}x/dt^{2}+ (0.75 - 0.375) g = 0.

d^{2}x/dt^{2}= -0.143 g.

m_{1 }accelerates upward with a = 0.143 g.

A bead, of mass m, slides without friction on a wire that is in the shape of
a cycloid with equations

x = a(2θ + sin2θ),

y = a(1 - cos2θ),

- π/2 ≤ θ
≤ π/2.

A
uniform gravitational field **g** points in the negative y-direction.**
**(a) Find the Lagrangian and the second order differential equation of
motion for the coordinate θ.

(b) The bead moves on a trajectory s with elements of arc length ds.

Integrate ds = (dx

(c) Rewrite the equation of motion, switching from the coordinate θ to the coordinate s and solve it. Describe the motion.

Solution:

- Concepts:

Lagrangian Mechanics - Reasoning:

We are asked to obtain Lagrange's equation of motion, using two different generalized coordinates. - Details of the calculation:

(a) L = T - U. T = ½m(v_{x}^{2}+ v_{y}^{2}), v_{x}= dx/dt = 2a(dθ/dt)(1 + cos2θ),

v_{y}= dy/dt = 2a(dθ/dt)sin2θ.

T = 2ma^{2}(dθ/dt)^{2}[(1 + cos2θ)^{2}+ sin^{2}2θ] = 4ma^{2}(dθ/dt)^{2}(1 + cos2θ)

= 8ma^{2}(dθ/dt)^{2}cos^{2}θ.

U = mgy = mga(1 - cos2θ) = 2mga sin^{2}θ.

L = 8ma^{2}(dθ/dt)^{2}cos^{2}θ - 2mga sin^{2}θ.

Lagrange's equations: d/dt(∂L/∂(dq_{i}/dt)) - ∂L/∂q_{i}= 0.

We have only one generalized coordinate.

∂L/∂(dθ/dt) = 16ma^{2}(dθ/dt)cos^{2}θ.

d/dt(∂L/∂(dθ/dt)) = 16ma^{2}d^{2}θ/dt^{2}cos^{2}θ - 32ma^{2}(dθ/dt)^{2}cosθ sinθ.

∂L/∂θ = -16ma^{2}(dθ/dt)^{2}cosθ sinθ - 4mga cosθ sinθ.

Equation of motion:

16ma^{2}(d^{2}θ/dt^{2})cos^{2}θ - 16ma^{2}(dθ/dt)^{2}cosθ sinθ + 4mga cosθ sinθ = 0.

d^{2}θ/dt^{2}cosθ - (dθ/dt)^{2}sinθ + (g/(4a)) sinθ = 0.(b) ds = 4a

^{ }cosθ dθ. s(θ) = ∫_{0}^{θ}ds = ∫_{0}^{θ}4a^{ }cosθ'dθ' = 4a^{ }sinθ.(c) ds/dt = 4a

^{ }cosθ (dθ/dt). d^{2}s/dt^{2}= 4a^{ }cosθ d^{2}θ/dt^{2}- 4a^{ }sinθ (dθ/dt)^{2}.

Therefore d^{2}θ/dt^{2}cosθ - (dθ/dt)^{2 }sinθ = (4a)^{-1}d^{2}s/dt^{2},

and (g/(4a))sinθ = (g/(4a)^{2})s.The equation of motion in terms of the coordinate s is d

^{2}s/dt^{2}= -(g/(4a))s.

This is Hooke's law with the solution s(t) = s_{0 }cos(ωt + φ).ω = (g/(4a))

^{½}, s_{0}and φ depend on the initial conditions.

The bead executes simple harmonic motion in s.

A particle of mass m is described by the Lagrangian function

L = ½m[(dx/dt)^{2} + dy/dt)^{2} + (dz/dt)^{2}] +
½ωl_{3},

where l_{3} is the z-component of the angular momentum and ω is a
constant angular frequency.

(a) Find the equations of motion, write them in terms of the variables (x + iy)
and z and solve them.

(b) Construct the Hamiltonian function and find the kinematic and canonical
momenta.

Show that the particle has only kinetic energy and that the latter is
conserved.

Solution:

- Concepts:

Lagrange's Equations: d/dt(∂L/∂(dq_{i}/dt)) - ∂L/∂q_{i}= 0; Canonical momenta: ∂L/∂(dq_{i}/dt) = p_{i};

The Hamiltonian function: H(q, p, t) = ∑_{i}(dq_{i}/dt)p_{i}- L - Reasoning:

The Lagrangian of the system is given, we are asked to solve the equations of motion and to construct the Hamiltonian function.

L = ½m[(dx/dt)^{2}+ dy/dt)^{2}+ (dz/dt)^{2}] + ½ωl_{3}, l_{3}= m(r × v)_{z}= m(x(dy/dt) - y(dx/dt)).

L = ½m[(dx/dt)^{2}+ dy/dt)^{2}+ (dz/dt)^{2}] + ½mω(x(dy/dt) - y(dx/dt)).

The generalized coordinates q_{i}are the Cartesian coordinates. The coordinate z is cyclic. - Details of the calculation:

(a) ∂L/∂(dz/dt) = ∂L/∂v_{z}= mv_{z}= p_{z}= constant, since z is cyclic.

From Lagrange's equations we have

∂L/∂(dx/dt) = mdx/dt - ½mωy, d/dt(∂L/∂(dx/dt)) = md^{2}x/dt^{2}- ½mωdy/dt,

∂L/∂x = ½mω(dy/dt),

d^{2}x/dt^{2}- ωdy/dt = 0.

∂L/∂(dy/dt) = mdy/dt + ½mωx, d/dt(∂L/∂(dy/dt)) = md^{2}y/dt^{2}+ ½mωdy/dt,

∂L/∂y = -½mω(dx/dt),

d^{2}y/dt^{2}+ ωdx/dt = 0.

Combining these two equations we can write

d^{2}x/dt^{2}+ id^{2}y/dt^{2}+ ω(idx/dt - dy/dt) = 0.

d^{2}x/dt^{2}+ id^{2}y/dt^{2}+ Iω(dx/dt + idy/dt) = 0.

Let c = x + iy. Then d^{2}c/dt^{2}+ iωdc/dt = 0.

c = Bexp(-iωt) + C.B and C are complex constants determined by the initial conditions.

Let B = Aexp(-iΦ), A = real, C = 0, then c = Aexp(-i(ωt + Φ)).

Solve for x and y as function of time by isolating the real and the imaginary parts.

x = Acos(ωt + Φ), y = -Asin(ωt + Φ).

For z we have z = z(0) + p_{z}t/m.

(b) The canonical momenta are

p_{z}= ∂L/∂(dz/dt) = mdz/dt, p_{x}= ∂L/∂(dx/dt) = mdx/dt - ½mωy,

p_{y}= ∂L/∂(dy/dt) = mdy/dt + ½mωx.

The kinematic momenta are

mdz/dt = p_{z}, mdx/dt = p_{x}+ ½mωy, mdy/dt = p_{y}- ½mωx.Note: The kinematic momenta are not equal to the canonical momenta.

The Hamiltonian is a function of the generalized coordinates and momenta.

H = p_{z}dz/dt + p_{x}dx/dt + p_{y}dy/dt - L

= ½[p_{z}^{2}/m + (p_{x}+ ½mωy)^{2}/m + ( p_{y}- ½mωx)^{2}/m] = H(p,q) = E.

H does not explicitly depend on time.

∂H/∂t = 0, E is conserved.

In terms of the kinematic momenta we have:

E = [(p_{x_mech})^{2 }+ (p_{y_mech})^{2 }+ (p_{z_mech})^{2}]/2m = T.

Think about: Which physical situation is described by this Lagrangian?

For a symmetrical prism (one in which the apex
angle lies at the top of an isosceles triangle), the total deviation angle
ϕ of a light ray is minimized when the ray
inside the prism travels parallel to the prism's base.

Assume that a beam of light passes through a glass
equilateral prism with refractive index 1.5. The prism is in air and is mounted
on a rotation stage, as shown in the figure. When the prism is rotated, the
angle by which the beam
is deviated changes. What is the minimum angle ϕ by which the beam is deflected?

Solution:

- Concepts:

Snell's law, geometry - Reasoning:

sinθ = n sinθ_{glass}.

At the angle of minimum deviation, the refraction at each interface is the same, so the beam inside the prism is parallel to the base of the prism. At both interfaces, the internal angle between the beam and the normal is thus 30^{o}. - Details of the calculation

By Snell's Law, the external angle between the beam and the normal is thus θ,

where 1.0*sinθ =1.5*sin30^{o}.

θ = sin^{-1}(1.5*sin30^{o}) = sin^{-1}(0.75) = 48.6^{o}.

From geometry:

The angle of minimum deviation is ϕ = 2θ -2*30^{o}.

ϕ = 2 sin^{-1}(0.75) - 60^{o}= 37.2^{o}.

A point particle with mass m is restricted to move on the inside surface of a
horizontal ring. The radius of the ring increases steadily with time as R
= R_{0} + R't.

At t = 0 the speed of the particle is v_{0}.

(a) Are there any constants of motion?

(b) Let E(t_{1}) be the energy of the particle at the time the
radius of the ring is 2R_{0} and E_{0} its energy at t = 0.

Find the ration E(t_{1})/E_{0}.

Solution:

- Concepts:

Lagrange's equations - Reasoning:

There are no net forces except the forces of constraint. To find the constants of motion we find the Lagrangian and look for cyclic coordinates.

If the generalized coordinates and momenta we choose do not depend on time and the Lagrangian depends explicitly on time, then the energy is not conserved. - Details of the calculation:

(a) Let x and y be the Cartesian coordinates in the plane of the ring and r and φ be the polar coordinates.

r = R_{0 }+ R't is the equation of constraint.

The constraint is holonomic and time dependent.

x = (R_{0}+ R't)cosφ, y = (R_{0}+ R't)sinφ.

T = ½m(v_{x}^{2}+ v_{y}^{2}) = ½m((dφ/dt)^{2}(R_{0}+ R't)^{2}+ R'^{2}).

L = T, we have only one generalized coordinate, φ, which does not explicitly depend on time.

∂L/∂φ = 0, φ is cyclic.

∂L/∂(dφ/dt) = m(dφ/dt)(R_{0}+ R't)^{2}= constant, since (d/dt)(∂L/∂(dφ/dt)) = 0.

The angular momentum M = m(dφ/dt)(R_{0}+ R't)^{2}about the vertical axis is a constant of motion.

The energy is NOT a constant of motion.

(b) E = M^{2}/(2m(R_{0}+ R't)^{2}) + ½mR'^{2}.

E(t_{1})/E_{0}= (M^{2}/(8mR_{0}^{2}) + ½mR'^{2})/(M^{2}/(2mR_{0}^{2}) + ½mR'^{2})

= (M^{2}+ 4m^{2}R'^{2}R_{0}^{2})/ (4M^{2}+ 4m^{2}R'^{2}R_{0}^{2}).

If R'^{2}<< M^{2}/(2m^{2}R_{0}^{2}), then E(t_{1})/E_{0}= ¼.

The energy of the particle decreases, the particle does positive work.