Question

The undamped pendulum pivoted at point O shown in Figure E3.48 has a cylinder of mass m2 at its top that rotates without slipping on the interior of a cylinder. At the bottom end of the pendulum, a mass m1 is attached. The rod connecting the two masses is rigid and weightless. The system is in equilibrium at theta = 0. Determine an expression for the
period of oscillation of the system. Assume that m2L2 < m1L1. (Using Lagrange's method)

0 ム i/ m

Answer 1

This problem gives a simple problem for Lagrange method. This method is used to determine the governing equation for the system under conservative and nonconservative forces.

When the non-conservative forces are present then the right hand side will accommodate their expressions.

The obtained governing equations are uncoupled here which may not be the case all the time.

This problem gives a simple problem for Lagrange method. This method is used to determine the governing equation for the system under conservative and neoconservative forces. When the non-conservative forces are present then the right hand side will accommodate their expressions. The obtained governing equations are uncoupled here which may not be the case all the time. \r\n Lagrange equation for conservation forces, d/dt (Partialdifferential L/Partialdifferential x^._i) - Partialdifferential L/Partialdifferential x_i = 0 (I) where L = K - V K = kinetic energy of the system U = potential energy of the system K = 1/2 m, (l, theta^.)^2 + 1/2 m_2 (l_2 theta^.)^2 + 1/2 (1/2 m_2 r^2_2) phi^.2 (rotational K.E of mass M_2) U = m_1 g (l_1 - l_1 cos theta) + m_2 g ((l_1 + l_2)/2 - (l_1 + l_2/2) cos theta) L = K - U Partialdifferential L/Partialdifferential theta = m_1 g l_1 sin theta + m_2 g/2