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A thin wiform circular disc of mass $M$ and radius $R$ is rotating in a horizontal plane about an axis passing through its centre and perpendicular to its plane with an angular velocity $\omega$. Another disc of same thickness and radius but of mass $\frac{1}{8} M$ is placed gently on the first disc coaxially. The angular velocity of the system is now
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Verified Answer
The correct answer is:
$\frac{8}{9} w$
Given, mass of first circular disc $\left(m_1\right)=M$
Radius of first circular disc $\left(r_1\right)=R$
Mass of second circular disc $\left(m_2\right)=\frac{1}{8} M$
For first disc,
$$
\begin{aligned}
& I_1=\frac{M R^2}{2} \\
& \Rightarrow \quad L_1=\frac{M R^2 \omega_1}{2} \\
&
\end{aligned}
$$
For first and second disc system,
$$
I_2=\frac{M R^2}{2}+\frac{M}{8} \cdot \frac{R^2}{2}
$$
or,
$$
\begin{aligned}
& I_2=\frac{9 M R^2}{8 \times 2} \\
& L_2=\frac{9 M R^2}{8 \times 2} \omega_2
\end{aligned}
$$
Now from conservation of angular momentum,
$$
\begin{aligned}
L_1 & =L_2 \\
\Rightarrow \quad \frac{M R^2 \omega_1}{2} & =\frac{9 M R^2}{8 \times 2} \omega_2 \Rightarrow \omega_1=\frac{9}{8} \omega_2
\end{aligned}
$$
Given,
$$
\begin{aligned}
& \omega_2=\omega \\
& \omega_2=\frac{8}{9} \omega_1=\frac{8}{9} \omega
\end{aligned}
$$
Radius of first circular disc $\left(r_1\right)=R$
Mass of second circular disc $\left(m_2\right)=\frac{1}{8} M$
For first disc,
$$
\begin{aligned}
& I_1=\frac{M R^2}{2} \\
& \Rightarrow \quad L_1=\frac{M R^2 \omega_1}{2} \\
&
\end{aligned}
$$
For first and second disc system,
$$
I_2=\frac{M R^2}{2}+\frac{M}{8} \cdot \frac{R^2}{2}
$$
or,
$$
\begin{aligned}
& I_2=\frac{9 M R^2}{8 \times 2} \\
& L_2=\frac{9 M R^2}{8 \times 2} \omega_2
\end{aligned}
$$
Now from conservation of angular momentum,
$$
\begin{aligned}
L_1 & =L_2 \\
\Rightarrow \quad \frac{M R^2 \omega_1}{2} & =\frac{9 M R^2}{8 \times 2} \omega_2 \Rightarrow \omega_1=\frac{9}{8} \omega_2
\end{aligned}
$$
Given,
$$
\begin{aligned}
& \omega_2=\omega \\
& \omega_2=\frac{8}{9} \omega_1=\frac{8}{9} \omega
\end{aligned}
$$
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