Question

Initial angular velocity of a circular disc of mass $M$ is ${\omega }_1$. Then two small spheres of mass $m$ are attached gently to two diametrically opposite points on the edge of the disc. What is the final angular velocity of the disc?

• A. $\left(\frac{M+m}{M}\right){\omega }_1$
• B. $\left(\frac{M+m}{m}\right){\omega }_1$
• C. $\left(\frac{M}{M+4m}\right){\omega }_1$
• D. $\left(\frac{M}{M+2m}\right){\omega }_1$

Answer 1

The correct option is C $\left(\frac{M}{M+4m}\right){\omega }_1$

As the mass is attached gently, there is no external torque, and thereby the angular momentum of the system will be consereved.

The initial angular momentum $L_i=I_{disc}.{\omega }_1$

$I_{disc}$ is the moment of inertia of the circular disc.

The moment of inertia of each of the two spheres is,

$I_s=m{R}^2$

If the final angular velocity of the system is $\omega$, then-

Initial angular momentum $=$ Final angular momentum

$I_{disc}.{\omega }_1=\left(I_{disc}+I_s+I_s\right).\omega$

$\Rightarrow \left(\frac{M{R}^2}{2}\right){\omega }_1=\left(\frac{M{R}^2}{2}+2m{R}^2\right)\omega$

$\Rightarrow M{\omega }_1=\left(M+4m\right)\omega$

$\Rightarrow \omega =\left(\frac{M}{M+4m}\right){\omega }_1$