Question

Reference: JEE Adv 2016 Phys

Two thin Circular discs of mass m and 4 m having radii of a and 2a respectively are rigidly fixed by a mass less rigid rod of length L = √24 a through their centers. This assembly is laid on a firm flat surface. And set rolling without slipping on the surface so that the angular speed about the axis of the rod is ω. The angular momentum of the entire assembly about the point ‘O’ is L (vector).

A) The center of mass of the assembly rotates about the z-axis with an angular speed of ω/5
B) Magnitude of angular momentum of the assembly about its center of mass is 17 m a2/2.

Answer 1

See the diagram.

Moment of Inertia of disc of mass m about the axis OCD:  I1 = 1/2 m a²

MOI of disc of mass 4m about axis OCD = I2 = 1/2 (4m) (2a)² = 8 m a²
 
Total MOI of the assembly about OCD = I1 + I2 = I = 17/2 m a²

Angular momentum of the assembly about COM E (or axis of rotation)
  = L = I ω = 17/2 * m a² ω

======

Angular velocity vector of the assembly along OCD = $\vec{\omega}$

Cos θ = L / √[L²+a²] = √24 / 5 
sin θ =  a / (5 a) = 1/5

Component of $\vec{\omega}$ along z axis = ω Sin θ = ω / 5 upwards