A thin circular ring of mass M and radius R is rotating about its axis with a constant angular velocity ω. Two objects each of mass ‘m’ are attached gently to the opposite ends of a diameter of the ring. The ring will now rotate with an angular velocity –

CONCEPT:

• The angular momentum of a particle rotating about an axis is defined as the moment of the linear momentum of the particle about that axis.
• It is measured as the product of linear momentum and the perpendicular distance of its line of action from the axis of rotation.
• It is analogous to linear momentum and its SI unit is kg m2/s.

• The relation between the angular momentum and moment of inertia is given by

L = Iω

Where I = moment of inertia, L = angular momentum, and ω = angular velocity.

• Moment of inertia of a circular ring about an axis passing through its center and perpendicular to its plane is

Where M = mass of the ring and R = radius of the ring

EXPLANATION:

Given – M1 = M, M2 = 2m, R1 = R2 = R (let), initial velocity (ω1) = ω and final velocity (ω2) = ω’

We have a thin circular ring of mass M and radius r and it is rotated at constant angular velocity ω

• The initial moment of inertia is of the circular ring is

The initial momentum of the circular ring is

     ---------- (1)

• When another ring of the same mass is put on the 1st rotating ring, then the final moment of inertia is

Final momentum of the circular ring is

     ------------ (2)

• Since there is no external torque acting on this system, this means that angular momentum of the system will be conserved
• Thus by conservation of angular momentum

⇒ L = L'

⇒  MR²ω = (M + 2m) R2ω’ 

 Hence  option 2 is correct among all.