Newton's Second Law for Rotation for AP Physics C

By — McGraw-Hill Professional
Updated on Feb 11, 2011

Practice problems for these concepts can be found at:

Rotational Motion Practice Problems for AP Physics B & C

For linear motion, Newton says Fnet = ma; for rotational motion, the analog to Newton's second law is

where tnet is the τnet torque on an object. Perhaps the most common application of this equation involves pulleys with mass.

Draw a free-body diagram for each block, and use Fnet = ma. So we start that way.

The twist in this problem is the massive pulley, which causes two changes in the problemsolving approach:

  1. We have to draw a free-body diagram for the pulley as well as the blocks. Even though it doesn't move, it still requires torque to accelerate its spinning speed.
  2. We oversimplified things when we said, "One rope = one tension." The Physics C corollary to this rule says, "… unless the rope is interrupted by a mass." Because the pulley is massive, the tension in the rope may be different on each side of the pulley.

The Physics C corollary means that the free-body diagrams indicate T1 and T2, which must be treated as different variables. The free-body diagram for the pulley includes only these two tensions:

Now, we write Newton's second law for each block:

For the pulley, because it is rotating, we write Newton's second law for rotation. The torque provided by each rope is equal to the tension in the rope times the distance to the center of rotation; that is, the radius of the pulley. (We're not given this radius, so we'll just call it R for now and hope for the best.)

The acceleration of each block must be the same because they're connected by a rope; the linear acceleration of a point on the edge of the pulley must also be the same as that of the blocks. So, in the pulley equation, replace α by a/R. Check it out, all the R terms cancel! Thank goodness, too, because the radius of the pulley wasn't even given in the problem.

The pulley equation, thus, simplifies to

Now we're left with an algebra problem: three equations and three variables (T1, T2, and a). Solve using addition or substitution. Try adding the first two equations together—this gives a T1 - T2 term that meshes nicely with the third equation.

The acceleration turns out to be 5.6 m/s2. If you do the problem neglecting the mass of the pulley (try it!) you get 5.9 m/s2. This makes sense—the more massive the pulley, the harder it is for the system of masses to speed up.

Practice problems for these concepts can be found at:

Rotational Motion Practice Problems for AP Physics B & C

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