The Question is from neet-physics rotational-motion , and it was created by the author SUBHANI

Question

A disc of mass M and radius R is rotating about its axis with angular velocity $\omega$. If a small portion of mass m is detached from the rim of the disc, the new angular velocity of the remaining disc will be:

SUBHANI · Accepted Answer

## CORE INFORMATION
✓ **Answer**: A
## CONCEPT FRAMEWORK
	**Primary Concept**: Conservation of Angular Momentum
	**Related Topics**: Rotational Motion, Moment of Inertia
	**Prerequisites**: Angular momentum, Moment of inertia of a disc
## SOLUTION PATHWAY
**Given Information**:
	* Mass of the disc: $M$
	* Radius of the disc: $R$
	* Initial angular velocity: $\omega$
	* Mass detached: $m$
**Method & Steps**:
	1. **Initial Angular Momentum ($L_i$)**: The initial moment of inertia of the disc is $I_i = \frac{1}{2}MR^2$.  The initial angular momentum is $L_i = I_i \omega = \frac{1}{2}MR^2\omega$.
		Reasoning: The angular momentum of a rotating object is the product of its moment of inertia and angular velocity.
		Formula/Rule: $L = I\omega$
	2. **Final Angular Momentum ($L_f$)**: When a small mass $m$ is detached from the rim, the remaining mass is $(M-m)$. Assuming the detached mass is small and doesn't significantly alter the radius, the new moment of inertia is $I_f = \frac{1}{2}(M-m)R^2$. Let the new angular velocity be $\omega'$.  Then, $L_f = I_f \omega' = \frac{1}{2}(M-m)R^2\omega'$.
		Working: We calculate the moment of inertia of the remaining disc and express its angular momentum.
		Key Point:  We assume the removal of mass $m$ doesn't change the radius of the disc significantly, which is a valid approximation for a small $m$.
	3. **Conservation of Angular Momentum**: Since no external torque acts on the system, the angular momentum is conserved. Therefore, $L_i = L_f$.
		Result: $\frac{1}{2}MR^2\omega = \frac{1}{2}(M-m)R^2\omega'$
		Verification: Solving for $\omega'$, we get $\omega' = \frac{M\omega}{M-m}$.
## OPTION ANALYSIS
[A] ✓: $\frac{M\omega}{M-m}$: This is the correct answer derived from the conservation of angular momentum.
[B] :x: $\frac{M\omega}{M+m}$: This option incorrectly adds the detached mass in the denominator.
[C] :x: $\frac{(M-m)\omega}{M}$: This option incorrectly places the mass terms in the numerator.
[D] :x: $\frac{(M+m)\omega}{M}$: This option incorrectly adds the detached mass in the numerator.
## LEARNING AIDS
	**Common Mistakes**:
		1. Incorrectly assuming the moment of inertia remains constant: The moment of inertia changes when mass is removed from the rotating body.
			How to avoid: Always recalculate the moment of inertia after a change in mass distribution.
		2. Not applying conservation of angular momentum:  The absence of external torque implies conservation of angular momentum.
			How to avoid: Recognize that in the absence of external torques, angular momentum is conserved.
	**Quick Tips**:
		1. Remember the formula for angular momentum: $L = I\omega$.
		2. Conservation of Angular Momentum is analogous to Conservation of Linear Momentum in linear motion.

SUBHANI · Answer

The correct Option is "$\frac{Mω}{M-m}$"

SUBHANI · Answer

$\frac{Mω}{M+m}$

SUBHANI · Answer

$\frac{(M-m)ω}{M}$

SUBHANI · Answer

$\frac{(M+m)ω}{M}$

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