Problem: Two nearly equal masses hang from either side of a pulley (500 g, diameter = 35 cm) as shown. Find the acceleration of the system and the tensions in the cord. (Comment: This arrangement is known as "Atwood's Machine", and it was once used to determine the acceleration due to gravity g.)
Solution: Here are the free body
diagrams, except that the pulley's free body must also be considered:
Since I know that the right hand body accelerates downward, I chose that direction as positive; the left hand body accelerates upward and so that direction has been chosen as positive for that body. Since the cord does not slide on the pulley (that's why the pulley rotates), the tensions are different!
When this was done in module 5 using equal tensions, the equations were added and the T's subtracted out. Not this time! Now we have three unknowns and only two equations. So I need an equation for the pulley:
The acceleration is tangential (hence the "t" subscript). Since this is the same as the linear acceleration of the masses, I've dropped the subscript in the following line. Adding these three equations to eliminate the tensions, I get
Compare this to the acceleration of 0.0907 m/s2 obtained in module 5 (without considering the forces on the pulley). With some energy required to turn the pulley, we'd expect a smaller acceleration and that's what we got!
After substituting this acceleration into the equation containing T1, I got T1 = 5.93 N. Using either of the other two equations, I computed T2 = 5.95 N.
Comments: The frictional force between the cord and the pulley does not appear in the equations because it does not appear in the free body of the pulley/cord. It doesn't appear in the free body because we never cut the cord away from the pulley; I merely cut around the pulley and cord, cutting the cord twice to cause T1 and T2 to appear. If the final speed after falling a given distance were desired, I'd have to use the constant acceleration equations.