A thin circular ring of mass M and radius R is rotating about its axis with a constant angular velocity ω. Four objects each of mass m, are kept gently to the opposite ends of two perpendicular diameters of the ring. The angular velocity of the ring will be

The correct option is Mω/(M + 4m).

To find the new angular velocity of the ring, we can use the law of conservation of angular momentum. Since the four masses are placed gently onto the rotating ring, no external torque acts on the system about its axis of rotation. Consequently, the total angular momentum of the system remains constant.

Step 1: Calculate the initial angular momentum
The initial moment of inertia of the thin circular ring of mass $M$ and radius $R$ about its central axis is:
$I_i=M{R}^2$
Given that the initial angular velocity of the ring is $\omega$, the initial angular momentum ($L_i$) is:
$L_i=I_i\omega =M{R}^2\omega$

Step 2: Calculate the final moment of inertia of the system
Four objects, each of mass $m$, are placed at the opposite ends of two perpendicular diameters. Since they lie on the ring, their distance from the axis of rotation is equal to the radius of the ring, $R$.
The final moment of inertia ($I_f$) of the combined system (ring + four masses) is:
$I_f=I_{\text{ring}}+4I_{\text{object}}$
$I_f=M{R}^2+4m{R}^2=(M+4m)R^2$

Step 3: Apply conservation of angular momentum
Let ${\omega }^{\prime }$ be the final angular velocity of the ring.
$L_i=L_f$
$I_i\omega =I_f{\omega }^{\prime }$
Substitute the values of $I_i$ and $I_f$:
$M{R}^2\omega =(M+4m)R^2{\omega }^{\prime }$

Dividing both sides by $R^2$ yields:
$M\omega =(M+4m){\omega }^{\prime }$
Solving for ${\omega }^{\prime }$:
${\omega }^{\prime }=\frac{M\omega }{M+4m}$