solutions_HW4 - y 1 = a (cos ) + l 2 (cos ) x 2 =-a (sin )...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Homework 4 Classical Mechanics (Phys701) Due: October 28, 2011 Problem 1-14 Two points of mass m are joined by a rigid weightless rod of length l , the center of which is constrained to move on a circle of radius a . Express the kinetic energy in generalized coordinates. Write down the Lagrange equations. Solution: The position of the masses are determined by two angles. Angle ϕ fixes the postion of the center of the rod on the circle, and angle θ gives the orientation of the rod. The coordinates of the first mass are x 1 = a cos ϕ + l 2 cos θ y 1 = a sin ϕ + l 2 sin θ The coordinates of the second mass are x 1 = a cos ϕ - l 2 cos θ y 1 = a sin ϕ - l 2 sin θ Differentiating with respect to time one gets ˙ x 1 = - a (sin ϕ ) ˙ ϕ - l 2 (sin θ )
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: y 1 = a (cos ) + l 2 (cos ) x 2 =-a (sin ) + l 2 (sin ) y 2 = a (cos ) -l 2 (cos ) The kinetic energy T = m ( x 2 1 + y 2 1 + x 2 2 + y 2 2 ) / 2 becomes T = m ( a 2 2 + l 2 4 2 ) while the potential energy is equal to zero. Lagrange equations read = 0 , = 0 or = 1 = const = 2 = const Note that this independence of and motions happens only because we considered the particles of identical masses. Otherwise the energy T would have had a term proportional to . 1...
View Full Document

This note was uploaded on 12/13/2011 for the course PHYS 701 taught by Professor Bazaliy during the Fall '11 term at South Carolina.

Ask a homework question - tutors are online