Question

Two equal masses, m, are joined by a massless string of length L that passes through a hole in a frictionless horizontal table. First mass slides on table while the second hangs below the table and moves up and down in a vertical line.

a.) Assuming the string remains taut, write down the Lagrangian for the system in terms of the polar coordinates of the mass on the table.

(r, 0)

b.) Find the two Lagrangian equations of motion and interpret the $\phi$ equation in terms of the angular momentum .

c.) Express $\dot{\phi}$ in terms of and eliminate   $\dot{\phi}$ from the r equation. Now use the r equation to find the value $r=r_0$ at which the first mass can move in a circular path. Interpret your answer in Newtonian terms.

d.) Suppose the first mass is moving in a circular path and is given a small radius nudge. Write $r(t) = r_0 +\epsilon(t)$ , and rewrite the r equation in terms of $\epsilon(t)$ dropping all powers of $\epsilon(t)$ higher than linear. Show that the circular path is stable and oscillates about $r_0$

e.) Find the frequency of these oscillations.

Answer 1

Given that, the following given figure m_1 = m_2 = m length of string = L (a) Kinetic energy of the mass m_1, K = 1/2 M (r^2 + v)^2 Kinetic energy of m_2 K_2 = 1/2 m (- dy/dt)^2 = 1/2 mr^2 Potential energy of m_1 v_1 = 0 Potential energy m_2 v_2 = -mg (L - r) Lagrangian in L = K_1 + K_2 - u_1 - u_2 L = 1/2 m (r^2 + r^2 v^2 + r^2 + mg(L - r) (b) Lagrangian equation for phi d/dt (Partial differential L/ Partial differential theta) - Partial differential L/ Partial differential phi = 0, d/dt (mr^2 phi) = 0 Angular momentum of system l = mt^2 phi dL/dt = 0 �l� is constant. \r\n (c) if v = l/mr^2 Lagrangian equation for V d/dt (Partial differential L/ Partial differential r) - (Partial differential L/ Partial differential r) = 0 d/dt (2 mr) - (- mg - mv^2) = 0 2mr = mr v^2 = mr