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Double Atwood Machine

Main Concept

The double Atwood machine consists of an Atwood machine with one of the original masses replaced by a second Atwood machine. In total there are three masses, m1, m2, m3,suspended via massless, frictionless, inextensible cords connected to massless, frictionless pulleys. For convenience, it is assumed that the pulley system and cords have no mass or friction. A straightforward application of Newton's laws can be tedious as it requires the solution of three equations. Instead, the Lagrangian approach to deriving the equations of motion using generalized coordinates is used, which leads to only two equations to be solved.


Lagrangian Derivation


Suppose the double Atwood machine is composed of three masses, m1, m2, m3 connected by two chords of length L and L' respectively through two ideal pulleys. The system has two degrees of freedom, since the height of m3 can be found from the height of m2 and the second pulley. The most convenient set of generalized coordinates to describe the motion are y1: the distance from m1 to the the top pulley, and y2: the distance from m2 to the second pulley. Given these coordinates, the distance from m2 to the top pulley is Ly1 +y2, and the distance from m3 to the top pulley is L+L'  y1 y2.

The kinetic energy of the system is then:




where the dot denotes a time derivative. By setting the potential energy to zero at the height of the top pulley, the total potential energy of the system is:


U = m1g y1m2gLy1+y2m3gLy1+L'  y2.


 The Lagrangian is then L = T  U  or:


L = 12m1y.12+12m2y1.+y2.2+12m3y1.y2.2+m1g y1+m2gLy1+y2+m3gLy1+L'y2.


The equations of motion then follow from evaluating the Euler-Lagrange equations:


tLy.1Ly1 = 0,

tLy.2Ly2 = 0.


Differentiating and simplifying these equations give the following equations of motion for the double Atwood machine:


m2y1..+m2y1.. y2..+m3gm1m2m3=0,



Rearranging terms gives the accelerations of each block as:

y1..=gm1m2 + m3 4 m2m3m1m2+m3 + 4 m2m3,

 y2..=gm1m2  3 m3+ 4 m2m3m1m2+m3 + 4 m2m3,

 y3..=gm1m3  3 m2+ 4 m2m3m1m2+m3 + 4 m2m3. 

where we used that y1..=y2..+y3..2 and solved for y3.., the position of the third mass.

Left Mass, m1

Centre Mass , m2

Right mass, m3



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