Solution:
Using conservation of angular momentum: \(I_1\omega = (I_1 + I_2)\omega_f\). Since \(I_1 = \frac{1}{2}MR^2\) and \(I_2 = \frac{1}{2}M(R/2)^2 = \frac{1}{8}MR^2\), we get \(\omega_f = \frac{1/2}{1/2+1/8}\omega = \frac{4}{5}\omega\).
A uniform disc of radius R rotates about an axis through its centre and perpendicular to its plane with angular velocity \(\omega\). A stationary disc of the same mass but half the radius is placed on it axially. The final angular velocity of the system is
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