Question

A thin uniform circular disc of mass $M$ and radius $R$ is rotating in a horizantal plane about an axispassing through its centre and perpendicular to its plane with an angular velocity $\omega$ . Another disc ofsame dimension but of mass $M/4$ is placed gently on the first disc coaxially. The angular velocity of the system now is

• A. $4\omega /5$
• B. $2\omega /\sqrt{5}$
• C. $2\omega /5$
• D. $4\omega /\sqrt{5}$

Answer 1

The correct option is A $4\omega /5$

Initial angular momentum of one disc of mass M and radius R = $L=I\omega =\frac{1}{2}M{R}^2\omega$

when another disc of same dimension but mass $\frac{M}{4}$ is placed gently on it coaxially

Now tha angular velocity wll becomes ${\omega }^{\prime }$

Then angular momentum of the system=$L^{\prime }=(\frac{1}{2}M{R}^2+\frac{1}{2}\frac{M}{4}{R}^2){\omega }^{\prime }⟹L^{\prime }=\frac{5}{8}M{R}^2{\omega }^{\prime }$

As there is no external torque So angular momentum will be conserved
$\therefore L=L^{\prime }⟹\frac{1}{2}M{R}^2\omega =\frac{5}{8}M{R}^2{\omega }^{\prime }⟹{\omega }^{\prime }=\frac{4}{5}\omega$

Hence the final angular velocity of the system will be equal to $\frac{4}{5}\omega$