Q132 / 160 · TS EAMCET 2016 · Physics

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

  • Correct answer
    A
    $\frac{8}{9} w$
  • B
    $\frac{5}{9} w$
  • C
    $\frac{1}{3} w$
  • D
    $\frac{2}{9} w$

Solution

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, $$ \begin{aligned} & \omega_2=\omega \\ & \omega_2=\frac{8}{9} \omega_1=\frac{8}{9} \omega \end{aligned} $$

Hence, the correct answer is option: 'A'.

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