System of Particles and Rotational Motion - Result Question 30
30. Two discs are rotating about their axes, normal to the discs and passing through the centres of the discs. Disc $D_1$ has $2 kg$ mass and $0.2 m$ radius and initial angular velocity of $50 rad s^{-1}$. Disc $D_2$ has $4 kg$ mass, $0.1 m$ radius and initial angular velocity of $200 rad s^{-1}$. 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 [NEET Kar. 2013]
(a) 40
(b) 60
(c) 100
(d) 120
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Answer:
Correct Answer: 30. (c)
Solution:
(c) Given: $m_1=2 kg$
$m_2=4 kg$
$r_1=0.2 m$
$r_2=0.1 m$
$\omega_1=50 rad s^{-1}$
$\omega_2=200 rad s^{-1}$
As, angular momentum,
$I_1 \omega_1+I_2 \omega_2=(I_1+I_2) \omega_1$
$\therefore \omega_f=\frac{I_1 \omega_1+I_2 \omega_2}{I_1+I_2}=\frac{\frac{1}{2} m_1 r_1^{2} \omega_1+\frac{1}{2} m_2 r_2^{2} \omega_2}{\frac{1}{2} m_1 r_1^{2}+\frac{1}{2} m_2 r_2^{2}}$
By putting the value of $m_1, m_2, r_1, r_2$ and solving we get $=100 rad s^{-1}$
