A pendulum consists of point mass m on the end of a rigid massless rod of length
ID: 3897633 • Letter: A
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.
Assume mass M is slowly increasing, according to M = Mo + kt, where k is a small positive constant and t is time in seconds. 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(amiltonian) conserved?
iii. Is ETOT conserved?
Explanation / Answer
To write the Lagrangeian all is identical with your first problem EXCEPT the kinetic energy of mass M
Ek(M) = (M0+k*t)*(X_dot)^2 /2
the total Lagrangeian is
L = (M+m+kt)*(X_dot)^2 /2 + m*b^2*(theta_dot)^2 /2 - m*g*b*cos(theta)
Observation: this lagrangeian DOES depend on the time explicitly
dL/dX = 0
dL/d(X_dot) =(M+m+kt)*(X_dot)
d/dt (dL/d(X_dot)) = d/dt [(M+m+kt)*(X_dot)] = k*(X_dot) +(M+m+kt)*(X_dot_dot)
First Euler Lagrange equation is
dL/dX = d/dt (dL/d(X_dot))
0 = k*(X_dot) +(M+m+kt)*(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))
g*sin(theta) = b*(theta_dot_dot)
b)
i) L = T-U does depend on time t explictly. Therefore H is different from Etot.
ii) H = T+U does depend on the time t explicitly. Therefore H is not conserved.
iii) There are no external forces that act on the system except the conservative force of gravity on mass m. Therefore the total energy Etot is conserved.