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A thin uniform disc of mass $M$ and radius $R$ is rotating with angular velocity $\omega$ about an axis passing through its center and perpendicular to its plane. A thin ring of mass $m$ and radius $r$ is gently placed on the disc concentrically. The final angular velocity of the combined system is:

SUBHANI · Accepted Answer

## CORE INFORMATION\n✓ **Answer**: B\n## CONCEPT FRAMEWORK\n\t**Primary Concept**: Conservation of Angular Momentum\n\t**Related Topics**: Rotational motion, Moment of Inertia\n\t**Prerequisites**: Definition of angular momentum, moment of inertia for disc and ring\n## SOLUTION PATHWAY\n**Given Information**:\n\t* Mass of disc = M\n\t* Radius of disc = R\n\t* Initial angular velocity of disc = $\omega$\n\t* Mass of ring = m\n\t* Radius of ring = r\n\t* Ring is placed concentrically on the disc\n\t* No external torque is applied\n**Method & Steps**:\n\t1.  Calculate the initial angular momentum of the disc.\n\t\tReasoning: The angular momentum of a rotating object is given by the product of its moment of inertia and its angular velocity.\n\t\tFormula/Rule: $L = I\omega$\n\t2. Calculate the moment of inertia of the disc and the ring.\n\t\tWorking: Moment of inertia of a disc about its center is $I_{disc} = \frac{1}{2}MR^2$, and moment of inertia of a ring about its center is $I_{ring} = mr^2$\n\t\tKey Point: The ring is placed concentrically, so both objects rotate about the same axis.\n\t3. Calculate the total initial angular momentum of the system.\n\t\tWorking: Initially, only the disc is rotating.  Therefore, the initial angular momentum of the system is $L_i = I_{disc}\omega = \frac{1}{2}MR^2\omega$\n\t4. Calculate the total moment of inertia of the combined system.\n\t\tWorking: The total moment of inertia is the sum of the moment of inertia of the disc and the ring: $I_{total} = I_{disc} + I_{ring} = \frac{1}{2}MR^2 + mr^2$\n\t5. Apply conservation of angular momentum.\n\t\tReasoning: Since no external torque acts on the system, the total angular momentum is conserved.  $L_i = L_f$\n\t\tFormula/Rule: $L_i = I_i\omega_i = I_f\omega_f$\n\t6. Calculate the final angular velocity.\n\t\tWorking: $\frac{1}{2}MR^2\omega = (\frac{1}{2}MR^2 + mr^2)\omega_f$. Solving for $\omega_f$, we get $\omega_f = \frac{\frac{1}{2}MR^2\omega}{\frac{1}{2}MR^2 + mr^2} = \frac{MR^2\omega}{MR^2 + 2mr^2}$\n\t\tResult: $\omega_f = \frac{MR^2\omega}{MR^2 + 2mr^2}$\n\t\tVerification: The final angular velocity is less than the initial one, which is expected as the total moment of inertia has increased.\n ## OPTION ANALYSIS\n[A] :x: Incorrect. The denominator is missing a factor of 2 in front of $mr^2$.\n[B] ✓: Correct. The final angular velocity is derived correctly using conservation of angular momentum.\n[C] :x: Incorrect. The denominator incorrectly includes a 2 in front of $MR^2$.\n[D] :x: Incorrect. Both terms in the denominator have an incorrect factor of 2.\n ## LEARNING AIDS\n\t**Common Mistakes**:\n\t\t1.[Error 1]: Incorrectly using the moment of inertia of a solid cylinder instead of a disc. [How to avoid]: Ensure to use the moment of inertia for the specific shape, a disc in this case.\n\t\t2.[Error 2]: Forgetting to multiply the moment of inertia of the ring by $mr^2$. [How to avoid]: Remember the formula for the moment of inertia of a ring.\n\t**Quick Tips**:\n\t\t1.[Helpful shortcut]: When no external torque is present, angular momentum is always conserved. Remember to consider the moment of inertia of all objects in the system.\n\t\t2.[Memory aid]:  Remember that the moment of inertia increases when mass is moved farther from the axis of rotation.

SUBHANI · Answer

The correct Option is "(MR²ω) / (MR² + 2mr²)"

SUBHANI · Answer

(MR²ω) / (MR² + mr²)

SUBHANI · Answer

(MR²ω) / (2MR² + mr²)

SUBHANI · Answer

(MR²ω) / (2MR² + 2mr²)

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