A disc of mass 1 kg and radius 0.1 m is rotating with angular velocity `"20 rad s"^(-1)`. What is angular velocity, if a mass of 0.25 kg is put on periphery of the disc?

Options

A. `"24 rad s"^(-1)`
B. `"36 rad s"^(-1)`
C. `"19 rad s"^(-1)`
D. `"26 rad s"^(-1)`

Answer

To solve the problem of finding the new angular velocity of a disc after a mass is added to its periphery, we will use the principle of conservation of angular momentum. Here’s a step-by-step solution:

### Step 1: Understand the System
We have a disc with:
- Mass \( M = 1 \, \text{kg} \)
- Radius \( R = 0.1 \, \text{m} \)
- Initial angular velocity \( \omega_i = 20 \, \text{rad/s} \)

A mass \( m = 0.25 \, \text{kg} \) is placed on the periphery of the disc.

### Step 2: Calculate the Initial Angular Momentum
The moment of inertia \( I \) of the disc is given by the formula:
\[ I = \frac{1}{2} M R^2 \]
Substituting the values:
\[ I = \frac{1}{2} \times 1 \, \text{kg} \times (0.1 \, \text{m})^2 = \frac{1}{2} \times 1 \times 0.01 = 0.005 \, \text{kg m}^2 \]

The initial angular momentum \( L_i \) is:
\[ L_i = I \omega_i = 0.005 \, \text{kg m}^2 \times 20 \, \text{rad/s} = 0.1 \, \text{kg m}^2/s \]

### Step 3: Calculate the Final Moment of Inertia
When the mass \( m \) is added to the periphery of the disc, the new moment of inertia \( I_f \) is:
\[ I_f = I + m R^2 \]
Substituting the values:
\[ I_f = 0.005 \, \text{kg m}^2 + 0.25 \, \text{kg} \times (0.1 \, \text{m})^2 = 0.005 + 0.25 \times 0.01 = 0.005 + 0.0025 = 0.0075 \, \text{kg m}^2 \]

### Step 4: Apply Conservation of Angular Momentum
According to the conservation of angular momentum:
\[ L_i = L_f \]
Where \( L_f = I_f \omega_f \). Thus, we have:
\[ 0.1 = 0.0075 \omega_f \]

### Step 5: Solve for the Final Angular Velocity
Rearranging the equation to find \( \omega_f \):
\[ \omega_f = \frac{0.1}{0.0075} = \frac{100}{7.5} \approx 13.33 \, \text{rad/s} \]

### Conclusion
The final angular velocity of the disc after adding the mass is approximately \( 13.33 \, \text{rad/s} \).