The center of a disk of radius $ r $ and mass $ m $ is attached to a spring of spring constant $ k $, inside a ring of radius $ R>r $ as shown in the figure. The other end of the spring is attached on the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can only roll along the inside periphery of the ring, without slipping. The spring can only be stretched or compressed along the periphery of the ring, following Hooke’s law. In equilibrium, the disk is at the bottom of the ring. Assuming small displacement of the disc, the time period of oscillation of center of mass of the disk is written as $ T = \frac{2\pi}{\omega} $. The correct expression for $ \omega $ is ( $ g $ is the acceleration due to gravity):
A wheel of mass \( 4M \) and radius \( R \) is made of a thin uniform distribution of mass \( 3M \) at the rim and a point mass \( M \) at the center. The spokes of the wheel are massless. The center of mass of the wheel is connected to a horizontal massless rod of length \( 2R \), with one end fixed at \( O \), as shown in the figure. The wheel rolls without slipping on horizontal ground with angular speed \( \Omega \). If \( \vec{L} \) is the total angular momentum of the wheel about \( O \), then the magnitude \( \left| \frac{d\vec{L}}{dt} \right| = N(MR^2 \Omega^2) \). The value of \( N \) (in integer) is: