Question

Two discs of moment of inertia $I_1$ and $I_2$ about their respective axes which are passing through centre and perpendicular to surface, rotating with angular frequencies ${\omega }_1$ and ${\omega }_2$ respectively, are brought into contact face to face with their axes of rotation coincident. The angular frequency of the composite disc will be

• A. $\frac{I_1{\omega }_1-I_2{\omega }_2}{I_1-I_2}$
• B. $\frac{I_1{\omega }_1+I_2{\omega }_2}{I_1+I_2}$
• C. $\frac{I_2{\omega }_1+I_1{\omega }_2}{I_1+I_2}$
• D. $\frac{I_2{\omega }_1-I_1{\omega }_2}{I_1-I_2}$

Answer 1

The correct option is B $\frac{I_1{\omega }_1+I_2{\omega }_2}{I_1+I_2}$

Before the discs are placed in contact,
total initial angular momentum of the two discs, $L_i=I_1{\omega }_1+I_2{\omega }_2$

Since the two discs are brought into contact face to face (one on top of the other) and their axes of rotation coincide, the moment of inertia $I_c$ of the composite system will be equal to the sum of their individual moments of inertia, i.e.

$I_c=I_1+I_2$

If ${\omega }_c$ is the angular frequency of the composite system, the final angular momentum of the system,

$L_f=I_c{\omega }_c=(I_1+I_2){\omega }_c$

Since no external torque acts on the system, so we can apply conservation of angular momentum.

By the law of conservation of angular momentum,

$\therefore L_f=L_i$

$(I_1+I_2){\omega }_c=I_1{\omega }_1+I_2{\omega }_2$

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

Hence, option $\left(\text{B}\right)$ is correct.