Two discs are rotating about their axes, normal to the discs and passing through the centres of the discs. Dis

Question

Two discs are rotating about their axes, normal to the discs and passing through the centres of the discs. Disc $D_1$ has $2	ext{ kg}$ mass and $0.2	ext{ m}$ radius and initial angular velocity of $50	ext{ rad s}^{-1}$. Disc $D_2$ has $4	ext{ kg}$ mass, $0.1	ext{ m}$ radius and initial angular velocity of $200	ext{ rad s}^{-1}$. The two discs are brought in contact face to face, with their axes of rotation coincident. The final angular velocity (in $	ext{rad s}^{-1}$) of the system is

Accepted Answer

Let the moments of inertia of the two discs be $I_1$ and $I_2$. For a uniform disc, $I = \frac{1}{2}MR^2$.
$I_1 = \frac{1}{2} M_1 R_1^2 = \frac{1}{2} 	imes 2 	imes (0.2)^2 = 0.04	ext{ kg m}^2$
$I_2 = \frac{1}{2} M_2 R_2^2 = \frac{1}{2} 	imes 4 	imes (0.1)^2 = 0.02	ext{ kg m}^2$
Initial angular momentum of the system, $L_i = I_1\omega_1 + I_2\omega_2$
$L_i = (0.04 	imes 50) + (0.02 	imes 200) = 2 + 4 = 6	ext{ kg m}^2	ext{ s}^{-1}$
When the two discs are brought in contact, they will rotate together with a common final angular velocity $\omega$ due to internal friction between them. There is no net external torque on the system along the axis of rotation. 
By conservation of angular momentum:
$L_i = L_f$
$6 = (I_1 + I_2)\omega$
$6 = (0.04 + 0.02)\omega$
$6 = 0.06\omega$
$\omega = \frac{6}{0.06} = 100	ext{ rad s}^{-1}$

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