Question

Two discs of moments of inertia $I_1$ and $I_2$ about their respective axes (normal to the disc and passing through the centre), 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

(a) Total initial angular momentum=$I_1{\omega }_1+I_2{\omega }_2$

When the discs are joined together, total moment of inertia about the axis becomes=$I_1+I_2$

Let the angular speed of the two-disc system be $\omega$.

Then from conservation of angular momentum=$I_1{\omega }_1+I_2{\omega }_2=(I_1+I_2)\omega$

Thus $\omega =\frac{I_1{\omega }_1+I_2{\omega }_2}{I_1+I_2}$

(b) Total initial kinetic energy of the system=$E_i=\frac{1}{2}{I}_1{\omega }_1^2+\frac{1}{2}{I}_2{\omega }_2^2$

Final kinetic energy of the system=$\frac{1}{2}(I_1+I_2){\omega }^2$

$=\frac{1}{2}(I_1{\omega }_1+I_2{\omega }_2{)}^2/(I_1+I_2)$

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

The loss of KE can be attributed to the frictional force that comes into play when the two discs come in contact with each other.