A uniform circular disc of mass 50kg and radius 0.4 m is rotating with an angular velocity of `10 rad s^(-1)` about its own axis, which is vertical. Two uniform circular rings, each of mass 6.25 kg and radius 0.2 m, are gently placed symmetrically on the disc in such a manner that they are touching each other along the axis of the disc and are horizontal. Assume that the friction is large enough such that the rings are at rest relative to the disc and the system rotates about the original axis. The new angular velocity `(in red s^(-1))` of the system is

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the Moment of Inertia of the Disc
The moment of inertia \( I \) of a uniform circular disc about its own axis is given by the formula:
\[ I = \frac{1}{2} m r^2 \]
where \( m \) is the mass of the disc and \( r \) is the radius.

Given:
- Mass of the disc \( m = 50 \, \text{kg} \)
- Radius of the disc \( r = 0.4 \, \text{m} \)

Substituting the values:
\[ I_{\text{disc}} = \frac{1}{2} \times 50 \times (0.4)^2 = \frac{1}{2} \times 50 \times 0.16 = 4 \, \text{kg m}^2 \]

### Step 2: Calculate the Moment of Inertia of Each Ring
The moment of inertia \( I \) of a ring about its own axis is given by:
\[ I = m r^2 \]
where \( m \) is the mass of the ring and \( r \) is the radius.

Given:
- Mass of each ring \( m = 6.25 \, \text{kg} \)
- Radius of each ring \( r = 0.2 \, \text{m} \)

Calculating the moment of inertia about the center of the ring:
\[ I_{\text{ring, center}} = 6.25 \times (0.2)^2 = 6.25 \times 0.04 = 0.25 \, \text{kg m}^2 \]

### Step 3: Use the Parallel Axis Theorem
Since the rings are placed on the disc, we need to use the parallel axis theorem to find the moment of inertia about the axis of the disc. The parallel axis theorem states:
\[ I = I_{\text{cm}} + md^2 \]
where \( d \) is the distance from the center of the ring to the axis of rotation (which is equal to the radius of the disc).

For each ring:
- Distance \( d = 0.4 \, \text{m} \)

Calculating the moment of inertia for each ring:
\[ I_{\text{ring}} = 0.25 + 6.25 \times (0.4)^2 = 0.25 + 6.25 \times 0.16 = 0.25 + 1 = 1.25 \, \text{kg m}^2 \]

### Step 4: Calculate the Total Moment of Inertia of the System
Since there are two rings, the total moment of inertia of the system is:
\[ I_{\text{total}} = I_{\text{disc}} + 2 \times I_{\text{ring}} = 4 + 2 \times 1.25 = 4 + 2.5 = 6.5 \, \text{kg m}^2 \]

### Step 5: Apply Conservation of Angular Momentum
According to the conservation of angular momentum:
\[ I_{\text{initial}} \omega_{\text{initial}} = I_{\text{final}} \omega_{\text{final}} \]

Given:
- Initial angular velocity \( \omega_{\text{initial}} = 10 \, \text{rad/s} \)

Substituting the values:
\[ 4 \times 10 = 6.5 \times \omega_{\text{final}} \]
\[ 40 = 6.5 \times \omega_{\text{final}} \]

### Step 6: Solve for the Final Angular Velocity
\[ \omega_{\text{final}} = \frac{40}{6.5} \approx 6.15 \, \text{rad/s} \]

Thus, the new angular velocity of the system is approximately \( 6.15 \, \text{rad/s} \).

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