A Merry-go-round, made of a ring-like platform of radius R and mass M, is revolving with angular speed ω . A person of mass M is standing on it. At one instant, the person jumps off the round, radially away from the centre of the round (as seen from the round). The speed of the round afterwards is

The correct answer is Option 2: ω

Let us analyze the system using the conservation of angular momentum.
Initially, the Merry-go-round is modeled as a ring-like platform of radius ^R and mass ^M.
The moment of inertia of the ring about its central axis of rotation is:
$I_{\text{ring}}=M{R}^2$

A person of mass ^M is standing on it. Assuming the person is standing at the outer edge (distance ^R from the center), the person's moment of inertia is:
$I_{\text{person}}=M{R}^2$
The total initial moment of inertia of the system (ring + person) is:
$I_i=I_{\text{ring}}+I_{\text{person}}=M{R}^2+M{R}^2=2M{R}^2$
The system is revolving with an initial angular speed $\omega$. Therefore, the initial angular momentum of the system is:
$L_i=I_i\omega =2M{R}^2\omega$

Now, the person jumps off the round, radially away from the centre of the round (as seen from the rotating frame of the round).
When a person jumps radially outward relative to the platform, the force exerted by the person on the platform (and vice-versa) is purely radial.
A radial force passes through the axis of rotation, which means it produces zero external torque about the axis of rotation:
$\tau =0$
Therefore, the angular momentum of the system about the central axis must be conserved.

Furthermore, let's examine the velocity of the person at the instant of the jump.
As seen from the ground, the velocity of the person is the vector sum of:
1. The tangential velocity due to the rotation: $v_t=R\omega$ (tangential direction)
2. The relative radial jump velocity: $u_r$ (radial direction)

Since the jump is strictly in the radial direction, the person's component of velocity in the tangential direction remains unchanged during the jump, which is $R\omega$.
This means the person carries away an angular momentum about the axis equal to:
$L_{\text{person, final}}=M{v}_{t}R=M\left(R\omega \right)R=M{R}^2\omega$

Let the new angular speed of the ring after the jump be ${\omega }^{\text{'}}$. The final angular momentum of the ring is:
$L_{\text{ring, final}}=I_{\text{ring}}{\omega }^{\text{'}}=M{R}^2{\omega }^{\text{'}}$

By conservation of angular momentum for the entire system:
$L_i=L_{\text{ring, final}}+L_{\text{person, final}}$
Substitute the values into the equation:
$2M{R}^2\omega =M{R}^2{\omega }^{\text{'}}+M{R}^2\omega$

Divide the entire equation by $M{R}^2$:
$2\omega ={\omega }^{\text{'}}+\omega$
Subtracting $\omega$ from both sides, we get:
${\omega }^{\text{'}}=\omega$

Thus, the angular speed of the Merry-go-round remains unchanged and is equal to $\omega$.