#### Atwood's machine re-visited

Atwood's machine is a device where two masses, M and m, are connected by a string passing over a pulley. Assume that M > m. The pulley is a solid disk of mass mp and radius r.

What is the acceleration of the two masses?

Start with three free-body diagrams, one for each mass and one for the pulley.

When we did this problem before we assumed the pulley was frictionless and massless, so the tension is the same everywhere in the string. Now we'll account for the pulley's mass, so the tensions will be different.

Think about what the system will do. If the system is released from rest, the heavy mass will accelerate down, the lighter one will accelerate up, and the pulley will have a clockwise angular acceleration.

Align the coordinate systems with the acceleration. Each object has its own coordinate system, but they must be consistent.
Take +y down for mass M.
Take +y up for mass m.
Take clockwise to be positive for the pulley.

Recognize that the masses have the same acceleration, a.

Apply Newton's second law for each mass.
 For mass M: | For mass m: | For the pulley: SFy = May | SFy = may | St = I a Mg - T1 = Ma | T2 - mg = ma | rT1 - rT2 = ½ mpr2a

The two masses give two equations in three unknowns. The pulley gets us the third equation we need. Assuming the string does not slip while in contact with the pulley, the acceleration of the string, a, is related to the angular acceleration of the pulley by:
a =
 a r

The pulley equation then becomes:

T1 - T2 = ½ mp a

Combining the three equations to eliminate the two tensions gives:

(Mg - Ma) - (mg + ma) = ½ mp a

Mg - mg = Ma + ma + ½ mp a
a =
 g (M - m) M + m + ½ mp

This is a smaller acceleration than the a = g (M - m) / (M + m) we got when we neglected the pulley, which makes sense.

Numerical example: M = 210 g and m = 200 g.
a =
 g * 10 410
= 0.24 m/s2 if mp = 0.
a =
 g * 10 510
= 0.19 m/s2 if mp = 200 g.