A pendulum consists of point mass m on the end of a rigid massless rod of length

Question

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 theta (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; theta=theta_o, and theta'_o =0.

This time, M feels a repulsive force F = kXi from the force center at the origin, where k is a small positive constant. Use the same generalized coordinates THETA and X.

a) i. Write the Lagrangian for this system of two masses.
ii. Write the 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?

Explanation / Answer

To write the Lagrangeian all is identical with your first problem EXCEPT the potential energy of mass M:

U = -kX^2 / 2

The total Lagrangeian is

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

dL/dX = +k*X

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))

+k*x =(M+m)*(X_dot_dot)

(which is kind of Force = mass*generalized acceleration)

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)

g*sin(theta) = b*(theta_dot_dot)

b)

L = T- U does not depend on time t explicitly. Therefore H = Etot

H = T+U does not depend on time t explicitly. Therefore H is conserved

H= Etot, therefore Etot is conserved.

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