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.

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.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.