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 a 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 (T 1 , T 2 , and a ). Solve
using addition or substitution. Try adding the first two equations together—this gives a T 1 − T 2 term that
meshes nicely with the third equation.
The acceleration turns out to be 5.6 m/s^2 . If you do the problem neglecting the mass of the pulley (tryit!) you get 5.9 m/s^2 . This makes sense—the more massive the pulley, the harder it is for the system of
masses to speed up.
Rotational Kinetic Energy
The pulley in the last example problem had kinetic energy—it was moving, after all—but it didn’t have
linear kinetic energy, because the velocity of its center of mass was zero. When an object is rotating, its
rotational kinetic energy is found by the following equation:
Notice that this equation is very similar to the equation for linear kinetic energy. But, because we’re
dealing with rotation here, we use rotational inertia in place of mass and angular velocity in place of
linear velocity.
If an object is moving linearly at the same time that it’s rotating, its total kinetic energy equals the sum
of the linear KE and the rotational KE.
Let’s put this equation into practice. Try this example problem.