Question

Two discs of moments of inertia $I_{1}$ and $I_{2}$ about their respective axes, 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 $A$

• A. $\frac{I_{1} \omega_{1}+I_{2} \omega_{2}}{I_{1}+I_{2}}$
• B. $\frac{I_{2} \omega_{1}+I_{1} \omega_{2}}{I_{1}+I_{2}}$
• C. $\frac{I_{1} \omega_{1}+I_{2} \omega_{2}}{I_{1}-I_{2}}$
• D. $\frac{I_{2} \omega_{1}+I_{1} \omega_{2}}{I_{1}-I_{2}}$

Answer 1

Answer: (A)

Total initial angular momentum of the two discs is $L_{i}$ $=I_{1} \omega_{1}+I_{2} \omega_{2}$
When 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 of the system is equal to the sum of their individual moments of inertia.
i.e. $I=I_{1}+I_{2}$
Let $\omega$ be the final angular speed of the system.
The final angular momentum of the system is
$L_{f}=I \omega=\left(I_{1}+I_{2}\right) \omega$
As no external torque acts on the system, therefore according to law of conservation of angular momentum, we get
$L_{i}=L_{f} $
$ I_{1} \omega_{1}+I_{2} \omega_{2}=\left(I_{1}+I_{2}\right) \omega$
$ \omega=\frac{I_{1} \omega_{1}+I_{2} \omega_{2}}{I_{1}+I_{2}}$