Question

Two discs of moments of Inertia $I_1$ and $I_2$ about their respective axes (normal to the disc and. passing through the center), and rotating with angular speeds ${\omega }_1$, and ${\omega }_2$ are brought Into contact face to face with their axes of rotation coincident.

(A) What is the angular speed of the two-disc system?

(B) Show that the kinetic energy of the combined system is less than the sum of the initial kinetic energies of the two discs. How do you account for this loss In energy? Take ${\omega }_1\ne {\omega }_2$.

Answer 1

Step 1: Given data

Two discs of moments of inertia $I_1$ and $I_2$ about their respective axes rotating with angular speed ${\omega }_1$ and ${\omega }_2$, where ${\omega }_1\ne {\omega }_2$.

Step 2: Calculation of the angular speed of the two-disc system

Total initial angular momentum($L_i$) is given by, $L_i=I_1{\omega }_1+I_2{\omega }_2$.

The total moment of inertia about the axis when the discs are joined together is $I=I_1+I_2$.

Suppose the angular speed of the two-disc system is $\omega$,So the final angular momentum($L_f$) becomes,

$\begin{array}{rcl}{L}_f& =& I\omega \\ & & \\ & =& (I_1+I_2)\omega ........\left(2\right)\end{array}$

Now applying the conservation of angular momentum,i.e. $L_i=L_f$, we can write

$\begin{array}{rcl}{I}_1{\omega }_1+I_2{\omega }_2& =& (I_1+I_2)\omega \end{array}$

So,

$\begin{array}{rcl}& & \omega =\frac{I_1{\omega }_1+I_2{\omega }_2}{(I_1+I_2)}\end{array}$

Step 3: Showing that the kinetic energy of the combined system is less than the sum of the initial kinetic energies of the two discs

The total initial kinetic energy($E_i$) of the system is,

$E_i=\frac{1}{2}{I}_1{\omega }_1^1+\frac{1}{2}{I}_2{\omega }_2^2$

And final kinetic energy($E_f$) of the system is,

$\begin{array}{rcl}{E}_i& =& \frac{1}{2}(I_1+I_2){\omega }^2\\ & =& \frac{1}{2}(I_1+I_2){\left(\frac{I_1{\omega }_1+I_2{\omega }_2}{(I_1+I_2)}\right)}^2\\ & =& \frac{1}{2}\frac{{(I_1{\omega }_1+I_2{\omega }_2)}^2}{(I_1+I_2)}\end{array}$

Now the energy difference is given by,

$E_i-E_f=\frac{I_1{I}_2{({\omega }_1-{\omega }_2)}^2}{2(I_1+I_2)}>0$

Hence, it is clear that the kinetic energy of the combined system($E_f$) is less than the sum of the initial kinetic energies of the two discs($E_i$), and this loss of kinetic energy is due to the frictional force that comes into play when the two discs come in contact with each other.