Newton's Second Law for Rotation

What is the angular acceleration produced by applying a torque to an object? Newton's second law states that the angular acceleration is proportional to the net torque and inversely proportional to the moment of inertia.

Newton's Second Law: Στ = I α

Adjustable Moment of Inertia

Two heavy cylindrical masses are placed at opposite ends of a platform that rotates around its center. A torque is applied to the platform's axle by means of a string. The string passes over a pulley and a mass hangs from the other end of the string.

The system is released from rest, and the platform begins to rotate with a particular angular acceleration.

If the experiment is performed a second time with the masses moved closer to the center of the platform, what will happen?

  1. The angular acceleration will decrease.
  2. The angular acceleration will stay the same.
  3. The angular acceleration will increase.

According to Newton's Second Law, Στ = I α . The torque from the hanging mass is about the same in the two cases. Moving the masses closer to the center reduces the moment of inertia, which increases the angular acceleration.

Applying Newton's Second Law

Let's apply Newton's Second Law to our system of two cylindrical masses on the rotating platform. Determine the angular acceleration of the system in terms of:
M, the mass of each cylinder;
h, the distance from the rotation axis to each cylinder;
T, the tension applied by the string;
r, the radius of the axle.

We will assume that the platform itself has negligible mass, and that the radius of each cylinder is small compared to h.

As usual, start with a free-body diagram, and then apply Newton's Second Law for Rotation.

Στ = I α

The only torque we care about comes from the tension in the string. With torque given by τ = r × F , in this case the torque is:

τ = rT

There's a 90 degree angle between r and T.

Solving Newton's Second Law for α:

α = rT/I

For a cylinder rotating about its center-of-mass, where the rotation axis coincides with the axis of the cylinder, the moment of inertia is:

Icom = ½ MR2

Each of our cylinders rotates about a parallel axis (through the center of the platform), not about its center-of-mass. The parallel axis theorem tells us that each cylinder has a rotational inertia given by:

I1 = Icom + Mh2

I1 = ½ MR2 + Mh2

Assuming that R << h means we can approximate this as:

I1 = Mh2

The total moment of inertia is double this, because there are two cylinders:

I = 2Mh2

Our expression for the angular acceleration becomes:

α = rT/[ 2Mh2 ]

Numerical Example

For our demonstration apparatus, we have the following parameters:
mass of each cylinder is M = 2.5 kg
distance from axis to each cylinder is h = 7.5 cm
tension in the string is about F = 5 N
radius of the axle is r = 2.7 cm

α = 0.027 * 5 / (2 * 2.5 * 0.0752 )

α = 4.8 rad/s2

This predicts that the first full rotation of the system takes 1.6 s.

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:
ΣFy = May | ΣFy = may | Στ = I α
Mg - T1 = Ma | T2 - mg = ma | rT1 - rT2 = ½ mpr2α

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:

α = 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.

If mp = 200 g then a = g (10)/(510) = 0.19 m/s2