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

Let's apply Newton's Second Law to a system of two cylindrical masses on a 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

=

r T 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:

I_com = 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 = I_com + 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

=

r T 2 M h2

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

=

4.8 rad/s2

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