Q103 / 180 · NEET 2013 (Karnataka) · Physics

The discs are rotating about their axes, normal to the discs and passing through the centres of the discs. Disc $D_1$ has $2 \mathrm{~kg}$ mass and $0.2 \mathrm{~m}$ radius and initial angular velocity of $50 \mathrm{rad} \mathrm{s}^{-1}$. Disc $\mathrm{D}_2$ has $4 \mathrm{~kg}$ mass, $0.1 \mathrm{~m}$ radius and initial angular velocity of $200 \mathrm{rad} \mathrm{s}^{-1}$. The two discs are brought in contact face to face, with their axes of rotation coincident. The final angular velocity (in rad $s^{-1}$ ) of the system is:

  • A
    60
  • Correct answer
    B
    100
  • C
    120
  • D
    40

Solution

M.J. of $\mathrm{D}_1$ is $\mathrm{I}_1=\frac{\mathrm{MR}^2}{2}=\frac{2(0.2)^2}{2}$ $=0.04 \mathrm{~kg} \mathrm{~m}^2$ initial angular velocity of $D_1$ is $\omega_1=50 \mathrm{rad} \mathrm{S}^{-1}$ M.J of $\mathrm{D}_2$ is $\mathrm{I}_2=\frac{4 \times(0.1)^2}{2}=0.02 \mathrm{~kg} m_2$ initial angular velocity of $\mathrm{D}_2$ is $\omega_2=$ 200 and $\mathrm{S}^{-1}$. According to the law of conservation of angular moment $$ \begin{gathered} L_1=L_2 \\ \therefore \mathrm{I}_1 \omega_1+\mathrm{I}_2 \omega_2=\left(\mathrm{I}_1+\mathrm{I}_2\right) \\ \Rightarrow \omega=\frac{l_1 w_1+l_2 w_2}{\mathrm{I}_1+1} \\ \therefore \omega=\frac{(0.04)^2 \times 50+(0.02)^2 \times 200}{0.04+0.02} \\ =\frac{2+4}{0.06}=100 \mathrm{rads}^{-1} \end{gathered} $$

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

Relevant Questions