Pendulum with Moving Support A pendulum consists of point mass m on the end of a

Question

Pendulum with Moving Support

A pendulum consists of point mass m on the end of a rigid massless rod of length b. The other end of the rod is attached to point mass M that is free to move along a frictionless, horizontal surface. The origin is on the surface (see diagram), and we set U(0) = 0. As generalized coordinates: use ? (the angle between the pendulum bar and the vertical) and X (the distance from M to the origin). At t = 0: M is at the origin (X = 0)moving at vo;?=?o, and ?'o =0.

a) i. Express x and y (the coordinates of m) in terms of the generalized coordinates and constants. Write the Lagrangian for this system of two masses.

ii. Write the two Euler-Lagrange equations for this system. Simplify, but DO NOT SOLVE!

b) Without further calculations: answer each question YES or NO, and support your answer.

i. Is ETOT = H?

ii. Is H conserved?

iii. Is ETOT conserved?

c) Assume ? is limited to small oscillations.
i. Use first order approximations for sin? and cos? to find expression for ?(t) and

X(t). Use the initial conditions to eliminate all undetermined constants. At the end, write out the expression for X and ?, and give the period of motion for the oscillation of the pendulum.

ii. Describe the general motion of the two masses (without the small angle approximation).

iii. Now, describe all special cases for the motion that depend on the values of the initial conditions.

Explanation / Answer

a i)

X and theta are generalized coordinates

x = X +b*sin(theta)

y =-b*cos(theta)

a ii)

The lagrangeian is

L = T- U   (=kinetic energy - potential energy)

We note

dX/dt = X_dot  

d(theta)/dt = theta_dot

Ek(M) = M*(X_dot)^2 / 2

Ek(m) = Ek (translation) + Ek(rotation) = m(X_dot)^2 / 2 + m*(theta_dot)^2*b^2 / 2

because:

-for translation m has the same translational speed as M

-for rotation m has speed v = omega*b = (theta_dot)*b

Potetial energy is only for m

U = -m*g*b*cos(theta)

L = (M+m)*(X_dot)^2 /2 + m*b^2*(theta_dot)^2 /2 - m*g*b*cos(theta)

dL/dX =0   (X does not appear explicitly, but only X_dot)

dL/d(X_dot) =(M+m)*(X_dot)

d/dt (dL/d(X_dot)) = (M+m)*d/dt (X_dot) = (M+m)*(X_dot_dot)

First Euler Lagrange equation is

dL/dX = d/dt (dL/d(X_dot))

0 = (m+M)*(X_dot_dot)

dL/d(theta) = m*g*b*sin(theta)

dL/d(theta_dot) = m*b^2*(theta_dot)

d/dt (dL/d(theta_dot)) =m*b^2*d/dt (theta_dot) = m*b^2*(theta_dot_dot)

Second Euler Lagrange equation is

dL/d(theta) = d/dt (dL/d(theta_dot))

m*g*b*sin(theta) = m*b^2*(theta_dot_dot)

b) Hamiltonian H=T+U and total energy Etot are the same if Lagrangeian does not depend EXPLICITLY on time t.

In our case it is true, l does not depend on t.

Thus,YES H and Etot are the same.

YES, H is conserved since it does not depend EXPLICITLY on time t

YES, Etot is conserved since it is equal to H

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