A rotating table completes one rotation is 10 sec. and its moment of ineratia is 100 kg-`m^2`. A person of 50 kg. mass stands at the centre of the rotating table. If the person moves 2m. from the centre, the angular velocity of the rotating table in rad/sec. will be:

Options

A. `(2pi)/(30)`
B. `(20pi)/(30)`
C. `(2pi)/(3)`
D. `2pi`

Answer

To solve the problem, we will use the principles of rotational motion, specifically the conservation of angular momentum. Here's a step-by-step solution:

### Step 1: Calculate the initial angular velocity
The table completes one rotation in 10 seconds. The angular velocity (ω_initial) can be calculated using the formula:

\[ \omega_{\text{initial}} = \frac{2\pi \text{ radians}}{T} \]

where \( T \) is the time period (10 seconds).

\[ \omega_{\text{initial}} = \frac{2\pi}{10} = \frac{\pi}{5} \text{ rad/sec} \]

### Step 2: Calculate the initial angular momentum
The initial angular momentum (L_initial) can be calculated using the formula:

\[ L_{\text{initial}} = I_{\text{initial}} \cdot \omega_{\text{initial}} \]

Given that the moment of inertia (I_initial) of the table is 100 kg·m²:

\[ L_{\text{initial}} = 100 \cdot \frac{\pi}{5} = 20\pi \text{ kg·m²/sec} \]

### Step 3: Calculate the change in moment of inertia when the person moves
When the person of mass 50 kg moves 2 meters from the center, we need to calculate the additional moment of inertia contributed by the person. The moment of inertia (I_person) of the person can be calculated using:

\[ I_{\text{person}} = m \cdot r^2 \]

where \( m \) is the mass of the person (50 kg) and \( r \) is the distance from the center (2 m):

\[ I_{\text{person}} = 50 \cdot (2^2) = 50 \cdot 4 = 200 \text{ kg·m²} \]

### Step 4: Calculate the final moment of inertia
Now, we can find the final moment of inertia (I_final) of the system:

\[ I_{\text{final}} = I_{\text{initial}} + I_{\text{person}} = 100 + 200 = 300 \text{ kg·m²} \]

### Step 5: Apply conservation of angular momentum
Since there is no external torque acting on the system, we can apply the conservation of angular momentum:

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

Thus,

\[ I_{\text{initial}} \cdot \omega_{\text{initial}} = I_{\text{final}} \cdot \omega_{\text{final}} \]

Substituting the known values:

\[ 100 \cdot \frac{\pi}{5} = 300 \cdot \omega_{\text{final}} \]

### Step 6: Solve for the final angular velocity
Now we can solve for \( \omega_{\text{final}} \):

\[ 20\pi = 300 \cdot \omega_{\text{final}} \]

\[ \omega_{\text{final}} = \frac{20\pi}{300} = \frac{\pi}{15} \text{ rad/sec} \]

### Final Answer
The angular velocity of the rotating table after the person moves 2 meters from the center is:

\[ \omega_{\text{final}} = \frac{\pi}{15} \text{ rad/sec} \]

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