Question

A mass mi falls under its own weight while connected by a string to a frictionless pulley of mass m2 and radius R, whose center is fixed. Use two generalized coordinated, y (the length of the hanging string) and 6 (the angle through (a) Assuming that the string does not slip relative to the pulley, find the constraint governing 1. y and θ, and express this in the form of a constraint equation gy.9,-0. (You can assume (b) Write the Lagrangian in terms of the generalized coordinates y and 6. Note: You will (c) write out the Euler-Lagrange equations for y and θ. Don't forget the Lagrange (d) Find the magnitude of the acceleration of mi that mi is initially at the same level as the center of the pulley.) need the moment of inertia of a disc-shaped pulley, which is I-m2R2/2.) undetermined multipliers! (e) Find the tension in the rope directly from the Euler-Lagrange equations

Answer 1

Given that the given A mass m_1 falls under its own weigh connected by a string to a friction pully of mass m_2 and radius R, whose center is fixed here, use two generation coordinates, y and theta where y(the length of the hanging string) and theta (the angle through which the fully rotates) (a) assuming that the string does not slip relative to the fully, then the constraints of motion is y = R theta hence, the equation of constraint is g(y, theta) = y - R theta = 0 (b) to write the lagrangian in terms of the generalized coordinates y and theta here the moment of inertia of a disc-shaped polly, which is G = m_2 R^2/2 the lagrangian is = kE - PE where, KE rightarrow (kinetic energy): PE(potential energy) now, we take starting point as o potential energy therefore pe = = -mgy therefore kE = 1/2 m_2 y^2 + 1/2 I theta^2 hence the equation of lagrangian is L = 1/2 m_1 y^2 + 1/2 I theta^2 + mgy therefore I = m_2 R^2/2 therefore L = 1/2 m_1 y^2 + 1/2 times 1/2 m_2 R^2 theta^2 + mgy (C)