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Conservation of Angular Momentum in Divided Disc

Created by @nathanedwards

at November 1st 2023, 4:05:45 am.

AP_Physics_1-Questions

#moment-of-inertia

#conservation-of-angular-momentum

#angular-velocity

Question:

A disc of mass $m$ and radius $R$ is rotating with an angular velocity $\omega_1$ about its axis of rotation. The disc is now divided into two equal halves along a line perpendicular to its axis of rotation. One-half of the disc is then removed.

a) What is the new moment of inertia of the remaining portion of the disc about its axis of rotation? b) What is the new angular velocity of the remaining portion of the disc? c) Does the conservation of angular momentum hold in this scenario?

Assume the removed portion of the disc is instantly removed without any energy losses.

Answer:

a) To find the new moment of inertia of the remaining portion of the disc about its axis of rotation, we need to know the moment of inertia of the original disc and the moment of inertia of a half-disc.

The moment of inertia of a disc of mass $m$ and radius $R$ rotating about its axis of rotation is given by the formula:

I_{\text{disc}} = \frac{1}{2}mR^2

The moment of inertia of a half-disc can be found by considering it as a thin rod rotating about its end. The moment of inertia of a thin rod rotating about its end is given by the formula:

I_{\text{rod}} = \frac{1}{3}mR^2

Since the original disc is divided into two equal halves, the moment of inertia of each half will be:

I_{\text{half disc}} = \frac{1}{2} \times I_{\text{disc}} = \frac{1}{2} \times \frac{1}{2}mR^2 = \frac{1}{4}mR^2

Therefore, the new moment of inertia of the remaining portion of the disc about its axis of rotation is:

I_{\text{remaining}} = I_{\text{disc}} - I_{\text{half disc}} = \frac{1}{2}mR^2 - \frac{1}{4}mR^2 = \frac{1}{4}mR^2

b) To find the new angular velocity of the remaining portion of the disc, we can use the conservation of angular momentum. According to the conservation of angular momentum, the initial angular momentum of the system must be equal to the final angular momentum of the system.

Before the half-disc is removed, the initial angular momentum of the system can be calculated as:

L_{\text{initial}} = I_{\text{disc}} \times \omega_1

After the half-disc is removed, the final angular momentum of the system can be calculated as:

L_{\text{final}} = I_{\text{remaining}} \times \omega_{\text{final}}

Since angular momentum is conserved, we have:

L_{\text{initial}} = L_{\text{final}}

I_{\text{disc}} \times \omega_1 = I_{\text{remaining}} \times \omega_{\text{final}}

Substituting the expressions for $I_{\text{disc}}$ and $I_{\text{remaining}}$ , we get:

\frac{1}{2}mR^2 \times \omega_1 = \frac{1}{4}mR^2 \times \omega_{\text{final}}

Simplifying the equation, we find:

\omega_{\text{final}} = 2\omega_1

Therefore, the new angular velocity of the remaining portion of the disc is twice the initial angular velocity.

c) Yes, the conservation of angular momentum holds in this scenario. The total angular momentum of the system is conserved as there are no external torques acting on it. Initially, the angular momentum of the disc is balanced by the angular momentum of the removed half-disc, and after the removal, the remaining portion of the disc acquires the angular momentum of the removed portion, resulting in the conservation of angular momentum in the system.