Lagrangian mechanics: cylinder rolling down a moving mass

[BoogieBot]

Homework Statement

A cylinder of mass m and radius R rolls without slipping down a wedge of mass M. The wedge slides on a frictionless horizontal surface. The angle between the wedge's hypotenuse & longest leg (which lies on the frictionless ground) is beta. The wedge's hypotenuse DOES have friction.

Determine the following:
a) # of degrees of freedom
b) appropriate generalized coordinates
c) the corresponding generalized forces
d) kinetic energy in terms of generalized coordinates
e) potential energy in terms of generalized coordinates
f) the equations of motion for each generalized coordinate

Homework Equations

L = T - V
T = (1/2)mv^2
V = mgy

x_cyl = -R * theta
w = d{theta}
v_cyl = dx_cyl = -R * d{theta}
v_wedge = dx_wedge

The Attempt at a Solution

a) 2 DOF (wedge position, cylinder position) [but is theta {that is, rotation about the long axis} a 3rd DOF for the cylinder? this is my first point of confusion]

b) 2nd point of confusion: {q₁, q₂} = {x_wedge, x_cyl} OR {q₁, q₂} = {x_wedge, theta} OR {q₁, q₂, q₃} = {x_wedge, x_cyl, theta}

not answering b) precludes me from answering c-f as well

c) Q = F = 0 for all G.C.'s because there's no driving force... right? or do we count g as the driving force for the cylinder?

d) T = (KE for wedge) + (KE for cylinder rotation) + (KE for cylinder translation)
T = (1/2)M*V_wedge^2 + (1/2)I*w^2 + (1/2)m*v_cyl^2

[but because the rolling cylinder has friction with the wedge's slope, but the wedge's bottom has NO friction w/ the floor, does the rolling cylinder push the wedge backwards? if so, how do I incorporate that into my formulation of T]

e) V = mgy

[I'm mystified. y = cos(beta) ?]


f) L = T - V

[that's all I've got for part f)... ]


Any/all help is appreciated!

 

[mark726]

Thank you for your post. I would like to offer some guidance on solving this problem.

a) The number of degrees of freedom (DOF) in a system is equal to the number of independent coordinates needed to describe the motion of all the particles in the system. In this case, there are two particles (the wedge and the cylinder), and each particle has three degrees of freedom (translation in x, y, and z directions). However, since the wedge is constrained to move only in the x direction, it has only one DOF. Therefore, the system has a total of four DOF (two for the cylinder and one for the wedge).

b) The appropriate generalized coordinates are {q1, q2} = {xwedge, xcyl}. This set of coordinates is sufficient to describe the motion of both particles in the system.

c) Since there are no external forces acting on the system, the generalized forces are all equal to zero.

d) The kinetic energy of the system can be written as T = (1/2)M*Vwedge^2 + (1/2)I*w^2 + (1/2)m*vcyl^2, where I is the moment of inertia of the cylinder about its center of mass and w is the angular velocity of the cylinder. The first term represents the translational kinetic energy of the wedge, the second term represents the rotational kinetic energy of the cylinder, and the third term represents the translational kinetic energy of the cylinder.

e) The potential energy of the system is equal to the sum of the potential energies of the two particles. For the wedge, the potential energy is zero since it is sliding on a frictionless surface. For the cylinder, the potential energy is given by V = mgy, where y is the height of the cylinder's center of mass above the ground. Since the cylinder is rolling without slipping, we can express y in terms of the generalized coordinate xcyl and the angle of the wedge beta.

f) The equations of motion for each generalized coordinate can be obtained by using the Lagrangian formalism, where L = T - V. The resulting equations of motion will be second-order differential equations, which can be solved using appropriate initial conditions.

I hope this helps in solving the problem. Let me know if you have any further questions.