A circular platform is mounted on a vertical frictionless axle. Its radius is `r = 2 m` and its moment of inertia`I = 200 kg m^2`. It is initially at rest. A `70 kg` man stands on the edge of the platform and begins to walk along the edge at speed `v_0 = 1 m s^-1` relative to the ground. The angular velocity of the platform is.

Options

A. `1.2 rad s^-1`
B. `0.4 rad s^-1`
C. `0.7 rad s^-1`
D. `2 rad s^-1`

Explanation

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

### Step 1: Understand the system
The circular platform has a moment of inertia \( I = 200 \, \text{kg m}^2 \) and a radius \( r = 2 \, \text{m} \). A man of mass \( m = 70 \, \text{kg} \) walks at a speed \( v_0 = 1 \, \text{m/s} \) relative to the ground. Initially, both the platform and the man are at rest.

**Hint:** Identify the key parameters of the system and their initial conditions.

### Step 2: Write down the conservation of angular momentum
Since there are no external torques acting on the system, the total angular momentum before the man starts walking must equal the total angular momentum after he starts walking.

**Hint:** Remember that angular momentum is conserved in the absence of external torques.

### Step 3: Calculate the initial angular momentum
Initially, both the platform and the man are at rest, so the initial angular momentum \( L_{\text{initial}} \) is:
\[ L_{\text{initial}} = 0 \]

**Hint:** Consider the state of the system before any motion occurs.

### Step 4: Calculate the final angular momentum
When the man starts walking, he has an angular momentum due to his motion. The angular momentum of the man can be calculated using:
\[ L_{\text{man}} = m \cdot v_0 \cdot r \]
Substituting the values:
\[ L_{\text{man}} = 70 \, \text{kg} \cdot 1 \, \text{m/s} \cdot 2 \, \text{m} = 140 \, \text{kg m}^2/\text{s} \]

The platform also has angular momentum given by:
\[ L_{\text{platform}} = I \cdot \omega \]
where \( \omega \) is the angular velocity of the platform.

**Hint:** Break down the contributions to angular momentum from both the man and the platform.

### Step 5: Set up the conservation equation
Setting the initial angular momentum equal to the final angular momentum gives:
\[ 0 = I \cdot \omega + m \cdot v_0 \cdot r \]
Substituting the known values:
\[ 0 = 200 \cdot \omega + 140 \]

**Hint:** This equation will allow you to solve for the unknown angular velocity \( \omega \).

### Step 6: Solve for angular velocity \( \omega \)
Rearranging the equation to solve for \( \omega \):
\[ 200 \cdot \omega = -140 \]
\[ \omega = -\frac{140}{200} = -0.7 \, \text{rad/s} \]

The negative sign indicates that the platform rotates in the opposite direction to the man's motion.

**Hint:** Pay attention to the sign of your result, as it indicates the direction of rotation.

### Final Answer
The angular velocity of the platform is:
\[ \omega = 0.7 \, \text{rad/s} \, \text{(in the opposite direction to the man's walking direction)} \]