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 $\vec{L}$ (vector).

State whether the following are correct and find their values.

A) The magnitude of angular momentum of center of mass of the assemble about the point O is 81 m a² ω

B) The magnitude of the z-component of $\vec{L}$ is 55 m a² ω

Answer 1

Given values above are not correct.

See the diagram.

The center of mass of the assembly is E.
   CE = [0*m + 4m* L] / (5 m)    = 4L/5
  OE = 9 L/5 = 9√24 a /5 
  
  cos θ =  L / (5 a) = √24 a / 5

  When the smaller disc rolls on the floor A goes in a circle of radius 5a. Total circumference 2π(5a).  It takes the smaller disc to rotate 10πa / (2πa) = 5 rotations for it to complete one revolution around O on the ground. 

 So the angular velocity of the COM about point O = ω1 = ω / 5
  
 Angular momentum of COM 'E' about O = L2 
      [ formula  R X p = m r X v = M  R² ω..   as R and v are ⊥
  L2  = 5m * [9 L/5]² * ω1
       = 81 m L² ω / 25
       = 81 * 24 m a² ω / 25
            The direction is parallel to DB downwards.

Moment of inertia of disc m about axis OCD = 1/2 m a²
MOI of disc 4m about axis OCD = 1/2 (4m) (2a)² = 8 m a²
Total MOI = I = 17 m a² / 2

Angular momentum of the assembly about axis OCD = L1
   L1 = I ω = 17 m a² ω / 2     
        direction is along OCD upwards.

Z component of the angular momentum about O : 
    = L1 Sinθ - L2 Cosθ
    = ma²ω * [ 17/2 * 1/5 -  81*24/25 * √24/5 ]