A yo-yo is constructed of three disks: two outer disks of mass $\mathit{M}$, radius $\mathit{R}$, and thickness $\mathit{d}$, and an inner disk (around which the string is wrapped) of mass , radius , and thickness $\mathit{d}$. The yo-yo is suspended from the ceiling and then released with the string vertical (figure).

Calculate the tension in the string as the yo-yo falls. Note that when the center of the yo-yo moves down a distance $\mathit{y}$, the yo-yo turns through an angle$\mathit{y}\mathbf{/}\mathit{r}\mathbf{,}$which in turn means that the angular speed$\mathit{\omega }$is equal to ${\mathit{v}}_{\mathbf{C}\mathbf{M}}\mathbf{/}\mathit{r}$. The moment of inertia of a uniform disk is$\frac{\mathbf{1}}{\mathbf{2}}\mathit{M}{\mathit{R}}^{\mathbf{2}}$.

Tension can be defined as an action-reaction pair of forces acting at each end of the said elements. While considering a rope, the tension force is felt by every section of the rope in both the directions, apart from the endpoints. The endpoints experience tension on one side and the force from the weight attached.

The torque$\left(\tau \right)$acting on the yo-yo due to the tension $\left(T\right)$on the string is defined as the product of the tension and the radius $\left(r\right)$of the disk , and it Is given by the equation.

$\tau =Tr$ …(i)

If $I$be the moment of inertia and $\text{'}\alpha \text{'}$is the angular acceleration of the disk, then the torque acting on the yo-yo defined as the

$\tau =I\alpha$ …(ii)

The situation of the problem is shown in the below figure.

Mass of the yo-yo is the sum of the masses of the two identical disks$\left(M\right)$and mass of the inner disk$\left(m\right)$.

$\begin{array}{rcl}{M}_{yo-yo}& =& M+M+m\\ & =& 2M+m\\ & & \end{array}$

The gravitational force acting on the yo-yo $\left(M_{yo-yo}\right)$in the downward direction is the product of the mass of the yo-yo and acceleration due to gravity$\left(g\right)$. Hence, it is given by the equation

$\begin{array}{rcl}{F}_g& =& M_{yo-yo}g\\ & =& \left(2M+m\right)g\\ & & \end{array}$

Using Newton second law of motion, the net force acting on the yo-yo is

$\begin{array}{rcl}{F}_{net}& =& M_{yo-yo}a\\ & =& \left(2M+m\right)a\\ & & \end{array}$

$\begin{array}{rcl}{F}_{net}& =& M_{yo-yo}a\\ & =& \left(2M+m\right)a\\ & & \end{array}$

Here is the acceleration of the yo-yo.

The yo-yo moves down a distance . Hence, the net force acting on yo-yo is

$F_{net}=F_g-T$

$\left(2M+m\right)a=\left(2M+m\right)g-T$ …(iii)

$T=\left(2M+m\right)\left(g-a\right)$

The tension on the string of the yo-yo is calculated by using the equations (i) and (ii).

$Tr=I\alpha$

$T=\frac{I\alpha }{r}$ …(iv)

The angular acceleration of the yo-yo in terms of the linear acceleration $\left(a\right)$is

$\alpha =\frac{a}{r}$

Now, substituting the value in the equation (iv) ,we get

$T=\frac{I\left(\frac{a}{r}\right)}{r}$ …(v)

$T=\frac{Ia}{r^2}$

Using the equation (iii) and, we can eliminate$T$and obtain an expression for linear acceleration$\text{'}a\text{'}$

$\left(2M+m\right)\left(g-a\right)=\frac{Ia}{r^2}$

$\left(2M+m\right)g=\left(\left(2M+m\right)+\frac{I}{r^2}\right)a$

$a=\frac{\left(2M+m\right)g}{\left(\left(2M+m\right)+\frac{I}{r^2}\right)}$

Now substituting this value in the equation$\left(5\right)$, we get the tension on the string

$\begin{array}{rcl}T& =& \frac{\left(\frac{\left(2M+m\right)g}{\left(\left(2M+m\right)+\frac{I}{r^2}\right)}\right)}{r^2}\\ & =& \left(\left(2M+m\right)g\right)\left(\frac{\frac{I}{r^2}}{\left(2M+m\right)+\frac{I}{r^2}}\right)\\ & =& \left(\left(2M+m\right)g\right)\left(\frac{\frac{I}{r^2}}{\frac{\left(2M+m\right)r^2}{I}+I}\right)\\ & & \end{array}$

The moment of inertia of the yo-yo is the sum of the moment of inertia of the two identical disks and inner disk.

$I=2\left(\frac{1}{2}M{R}^2\right)+\frac{1}{2}m{r}^2$

$I=M{R}^2+\frac{1}{2}m{r}^2$

Now, substituting this value in the above equation, we get the tension in the yo-yo string

$T=\left(\left(2M+m\right)g\right)\left(\frac{\frac{I}{\left(2M+m\right)r^2}}{M{R}^2+\frac{1}{2}m{r}^2}+I\right)$