The tension in the string as the yo-yo falls is
.
The tension in the string can be calculated by considering the forces and motions involved as the yo-yo falls.
First, let's determine the relationship between the linear speed
and the angular speed
of the yo-yo:

where r is the radius of the yo-yo.
Given that the center of the yo-yo moves down a distance y and the yo-yo turns through an angle y/r, we can relate linear speed to angular speed.

Now, let's consider the forces involved. The tension in the string (T) provides the force necessary to accelerate the yo-yo downward and also generates the torque for rotation.
The net force accelerating the yo-yo downward is the difference between the tension and the gravitational force:

where:
m = mass of each disk
g = acceleration due to gravity
a = acceleration of the yo-yo's center of mass
The factor of
accounts for the gravitational force acting on the three disks (two outer disks and the inner disk).
The linear acceleration of the yo-yo's center of mass (a) can be related to the linear speed
:

Using the chain rule:

Now, we know that
, so:

This is a differential equation, and to solve it, we'll separate variables and integrate:

Integrating both sides:

Where C is the constant of integration. Considering the initial condition that at t = 0, y = 0 (when the yo-yo is released), we find that C = 0.
So, we get:

Now, differentiating both sides with respect to
:

This means
.
Now, going back to our force equation:

Substitute
and rearrange:


