A wedge of mass M rests on a horizontal frictionless surface. A point mass m is placed on the wedge, whose surface is also frictionless. Find the horizontal acceleration a of the wedge.

Solution:

- Concepts:

Lagrangian Mechanics - 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:

L = T – U.

T = T_{wedge}+ T_{m}= ½M(dx_{2}/dt)^{2}+ ½m[(dx_{1}/dt)^{2}+ (dy_{1}/dt)^{2}]. U = mgy_{1}.

Constraints: x_{1}= x_{2}+ z cosα, y_{1}= z sinα.

L = ½m[(dx_{2}/dt)^{2}+ (dz/dt)^{2}+ 2(dx_{2}/dt)(dz/dt) cosα] + ½M(dx_{2}/dt)^{2}– mgz sinα.

The equations of motion are obtained from d/dt(∂L/∂(dq/dt)) - ∂L/∂q = 0.

for q_{1}= x_{2}and q_{2}= z.

(i) (m + M)(d^{2}x_{2}/dt^{2}) + m(d^{2}z/dt^{2})^{ }cosα = 0. The horizontal position of the CM does not change.

m(d^{2}z/dt^{2})^{ }+ m(d^{2}x_{2}/dt^{2})cosα + mg sinα = 0.

a = d^{2}x_{2}/dt^{2}= mg cosα sinα/(M + m - m cos^{2}α) = mg cosα sinα/(M + m sin^{2}α).

(ii) You can also work from Newton's laws. The equations of motion then are

-Nsinα = m(d^{2}x_{1}/dt^{2}) = -M(d^{2}x_{2}/dt^{2}), -mg + Ncosα = m(d^{2}y_{1}/dt^{2}).

x_{1}= x_{2}+ z cosα, y_{1}= z sinα.

Solve for d^{2}x_{2}/dt^{2}.

A wedge of mass M = 4.5 kg sits on a horizontal surface. Another mass m = 2.3 kg sits on the sloping side of the wedge. The incline is at an angle θ = 31.7 degrees with respect to the horizontal, and g = 9.8 m/s. All surfaces are frictionless. The mass m is released from rest on mass M, which is also initially at rest. Use Lagrangian mechanics to determine the vector values of the accelerations of both M and m once the mass is released.

Solution:

- Concepts:

Lagrangian Mechanics - Reasoning:

We are asked to use Lagrangian Mechanics to solve the problem. - Details of the calculation:

L = T – U.

T = T_{wedge}+ T_{m}= ½M(dx_{2}/dt)^{2}+ ½m[(dx_{1}/dt)^{2}+ (dy_{1}/dt)^{2}]. U = mgy_{1}.

Constraints: x_{1}= x_{2}+ z cosθ, y_{1}= z sinθ.

L = ½ m[(dx_{2}/dt)^{2}+ (dz/dt)^{2}+ 2(dx_{2}/dt)(dz/dt) cosθ] + ½M(dx_{2}/dt)^{2}– mgz sinθ.

The equations of motion are obtained from d/dt(∂L/∂(dq/dt)) - ∂L/∂q = 0, for q = x_{2}and q = z.

(m + M)(d^{2}x_{2}/dt^{2}) + m(d^{2}z/dt^{2})^{ }cosθ = 0,

m(d^{2}z/dt^{2})^{ }+ m(d^{2}x_{2}/dt^{2}) cosθ + mg sinθ = 0.

d^{2}x_{2}/dt^{2}= (mg cosθ sinθ)/(M + m - m cos^{2}θ).

Inserting the numerical values we have:

d^{2}x_{2}/dt^{2}= 1.9624 m/s^{2},

d^{2}z/dt^{2}= -6.82 m/s^{2},

d^{2}x_{1}/dt^{2}= -3.85 m/s^{2},

d^{2}y_{1}/dt^{2}= -3.58 m/s^{2}.

The system, shown in the diagram, consists of a wedge placed on a horizontal surface and a block connected to a string that passes over a massless pulley attached to the wedge. The other end of the string is attached to the wall so that the string is horizontal between the wall and the pulley, and parallel to the inclined surface between the pulley and the block. The wedge and the block have equal masses m. The angle θ is given. When the system is released from rest the block begins to slide along the inclined surface of the wedge. Neglecting all friction, find the acceleration of the wedge.

Solution:

- Concepts:

Lagrangian mechanic - Reasoning:

The Lagrangian method often simplifies the treatment of systems with holonomic constraints. - Details of the calculation:

Let the string have length L. Let x be the coordinate of the point where the string touches the pulley.

Then coordinates of the point where the string attaches to the block are

x' = x + (L – x)cosθ, y' = h - (L – x)sinθ, where h is the height of the point where the string attaches to the wall.

The kinetic energy of the system is T = ½ mv_{x}^{2}+ ½ m(v_{x'}^{2}+ v_{y'}^{2}) = ½ m v_{x}^{2}(3 - 2cosθ).

The potential energy is U = m g x sinθ + constant.

L = ½ m v_{x}^{2}(3 - 2cosθ) - m g x sinθ

∂L/∂v_{x}= mv_{x}(3 - 2cosθ), d/dt(∂L/∂v_{x}) = ma_{x}(3 + 2cosθ), ∂L/∂x = - m g sinθ.

a_{x}= - g sinθ/(3 - 2cosθ).

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 x_{0} = 0, dx/dt|_{0} = 0.

(b) Determine the alternative Lagrangian L(x, dx/dt, θ, dθ/dt) and
the holonomic constraint f(x, θ) = 0 that must accompany it. Derive the
associated three equations of motion for the two unknown dynamical variables x
and θ, and the undetermined Lagrange multiplier λ. 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:

Lagrange's Equations, Lagrange multipliers - Reasoning:

In part (a) we use the constraint of rolling to eliminate the coordinate θ.

In part (b) we are asked to use the method of Lagrange multipliers. Then the equations of motion are be obtained from

d/dt(∂L/∂(dq_{k}/dt)) - ∂L/∂q_{k}= ∑_{l}λ_{l}a_{lk}, Σ_{k}a_{lk }dq_{k }+ a_{lt }dt = 0. - Details of the calculation:

(a) L = T - U.

T = (m/2)v^{2}+ (I/2)ω^{2}. v = dx/dt, ω = dθ/dt = v/r, I = mr^{2}, U = -mgxsinΦ.

L = m(dx/dt)^{2}+ mgxsinΦ.

∂L/∂v = 2mdx/dt, (d/dt)∂L/∂v = 2md^{2}x/dt^{2}, ∂L/∂x = mgsinΦ.

d^{2}x/dt^{2}= (g/2)sinΦ. a_{x}= (g/2)sinΦ, v_{x}= (gt/2)sinΦ, x = (gt^{2}/4)sinΦ.

(b) L = (m/2)(dx/dt)^{2}+ (mr^{2}/2)(dθ/dt)^{2}+ mgxsinΦ.

Constraint: dθ = dx/r, dx - rdθ = 0.

We have two coordinates, x and θ, and one equation of constraint, dx - rdθ = 0, which must be cast into the form λ_{1}∑a_{1k }dq_{k }= 0, λ_{1}(a_{1x }dx + a_{1θ}dθ)_{ }= 0. Therefore a_{1x }= 1, a_{1θ }= -r.

∂L/∂v = mdx/dt, (d/dt)∂L/∂v = md^{2}x/dt^{2}, ∂L/∂x = mgsinΦ.

∂L/∂ω = mr^{2}dθ/dt, (d/dt)∂L/∂ω = mr^{2}d^{2}θ/dt^{2}, ∂L/∂θ = 0.

md^{2}x/dt^{2}- mgsinΦ = λ, mr^{2}d^{2}θ/dt^{2}= -rλ, d^{2}x/dt^{2}= rd^{2}θ/dt^{2}.

3 equations, 3 unknowns.

d^{2}x/dt^{2}= (g/2)sinΦ, d^{2}θ/dt^{2}= (g/(2r))sinΦ, λ = -(mg/2)sinΦ.

Here λ = F_{x}, the static frictional force of constraint between the hoop and the incline.

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 θ.

(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:

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. We are asked to derive Lagrange's equation of motion. - Details of the calculation:

(a) We have a conservative system. Potential energy is converted into rotational and translational kinetic energy ½Iω^{2}+ ½Mv^{2}with v = ωR. The moment of inertia of the solid sphere I = (2/5)MR^{2}is smaller than the moment of inertia of the thin spherical shell I = (2/3)MR^{2}, so the solid sphere will have the greater linear acceleration down the ramp.

(b) Set up a coordinate system:

L = T – U. T = ½(I/R

^{2}+ M)(dx/dt)^{2}, U = Mgxsinθ.

L = ½(I/R^{2}+ M)(dx/dt)^{2}- Mgxsinθ.

Equation of motion: (d/dt)(∂L/∂v) – ∂L/∂x = 0, with v = dx/dt.

(I/R^{2}+ M)d^{2}x/dt^{2}+ Mgsinθ = 0.

a = d^{2}x/dt^{2}= (M/(I/R^{2}+ M))gsinθ.

For the solid sphere: a = (5/7) gsinθ.

For the spherical shell: a = (3/5) gsinθ.