A uniform disk of mass `300kg` is rotating freely about a vertical axis through its centre with constant angular velocity `omega .` A boy of mass `30kg` starts from the centre and moves along a radius to the edge of the disk. The angular velocity of the disk now is

Options

A. (a)`(omega_(0))/(6)`
B. (b)`(omega_(0))/(5)`
C. (c)`(4omega_(0))/(5)`
D. (d)`(5omega_(0))/(6)`

Answer

To solve the problem, we need to apply the principle of conservation of angular momentum. Here's a step-by-step breakdown of the solution:

### Step 1: Understand the System
We have a uniform disk with a mass of 300 kg rotating about a vertical axis with an initial angular velocity \( \omega_0 \). A boy with a mass of 30 kg starts from the center of the disk and moves to its edge.

### Step 2: Write the Expression for Angular Momentum
The angular momentum \( L \) of a rotating system is given by the product of its moment of inertia \( I \) and its angular velocity \( \omega \):
\[ L = I \cdot \omega \]

### Step 3: Calculate Initial Angular Momentum
Initially, the boy is at the center of the disk, so his contribution to the moment of inertia is zero. The moment of inertia of the disk is given by:
\[ I_{\text{disk}} = \frac{1}{2} M R^2 \]
where \( M = 300 \, \text{kg} \) and \( R \) is the radius of the disk.

Thus, the initial angular momentum \( L_i \) is:
\[ L_i = I_{\text{disk}} \cdot \omega_0 = \left(\frac{1}{2} \cdot 300 \cdot R^2\right) \cdot \omega_0 = 150 R^2 \omega_0 \]

### Step 4: Calculate Final Angular Momentum
When the boy moves to the edge of the disk, his contribution to the moment of inertia becomes significant. The moment of inertia of the boy when he is at the edge is:
\[ I_{\text{boy}} = m_{\text{boy}} \cdot R^2 = 30 \cdot R^2 \]
Thus, the total moment of inertia when the boy is at the edge is:
\[ I_f = I_{\text{disk}} + I_{\text{boy}} = \frac{1}{2} \cdot 300 \cdot R^2 + 30 \cdot R^2 = 150 R^2 + 30 R^2 = 180 R^2 \]
The final angular momentum \( L_f \) is then:
\[ L_f = I_f \cdot \omega_f = 180 R^2 \cdot \omega_f \]

### Step 5: Apply Conservation of Angular Momentum
According to the conservation of angular momentum:
\[ L_i = L_f \]
Substituting the expressions we found:
\[ 150 R^2 \omega_0 = 180 R^2 \omega_f \]

### Step 6: Solve for Final Angular Velocity
We can cancel \( R^2 \) from both sides (assuming \( R \neq 0 \)):
\[ 150 \omega_0 = 180 \omega_f \]
Now, solving for \( \omega_f \):
\[ \omega_f = \frac{150}{180} \omega_0 = \frac{5}{6} \omega_0 \]

### Conclusion
The final angular velocity of the disk when the boy reaches the edge is:
\[ \omega_f = \frac{5}{6} \omega_0 \]