In a playground there is a merry-go-round of mass 120 kg and radius 4 m. The radius of gyration is 3m. A child of mass 30 kg runs at a speed of 5 m/sec tangent to the rim of the merry-go-round when it is at rest and then jumps on it. Neglect friction and find the angular velocity of the merry-go-round and child

Options

A. 0.2 rad/sec
B. 0.1 rad/sec
C. 0.4 rad/sec
D. 0.8 rad/sec

Answer

To solve the problem, we will use the principle of conservation of angular momentum. The total angular momentum before the child jumps onto the merry-go-round must equal the total angular momentum after the child has jumped on.

### Step-by-Step Solution:

1. **Identify the given values:**
- Mass of the merry-go-round, \( M = 120 \, \text{kg} \)
- Radius of the merry-go-round, \( R = 4 \, \text{m} \)
- Radius of gyration of the merry-go-round, \( k = 3 \, \text{m} \)
- Mass of the child, \( m = 30 \, \text{kg} \)
- Speed of the child, \( v = 5 \, \text{m/s} \)

2. **Calculate the moment of inertia of the merry-go-round:**
The moment of inertia \( I \) of the merry-go-round can be calculated using the formula:
\[ I = M k^2 \]
Substituting the values:
\[ I = 120 \, \text{kg} \times (3 \, \text{m})^2 = 120 \times 9 = 1080 \, \text{kg} \cdot \text{m}^2 \]

3. **Calculate the angular momentum of the child before jumping on the merry-go-round:**
The angular momentum \( L \) of the child can be calculated using the formula:
\[ L = m \cdot v \cdot R \]
Substituting the values:
\[ L = 30 \, \text{kg} \times 5 \, \text{m/s} \times 4 \, \text{m} = 600 \, \text{kg} \cdot \text{m}^2/\text{s} \]

4. **Total initial angular momentum:**
Since the merry-go-round is initially at rest, its initial angular momentum is 0. Therefore, the total initial angular momentum is:
\[ L_{\text{initial}} = 600 \, \text{kg} \cdot \text{m}^2/\text{s} \]

5. **Calculate the total moment of inertia after the child jumps on the merry-go-round:**
The moment of inertia of the child when he is on the merry-go-round is given by:
\[ I_{\text{child}} = m \cdot R^2 \]
Substituting the values:
\[ I_{\text{child}} = 30 \, \text{kg} \times (4 \, \text{m})^2 = 30 \times 16 = 480 \, \text{kg} \cdot \text{m}^2 \]
Therefore, the total moment of inertia after the child jumps on is:
\[ I_{\text{total}} = I + I_{\text{child}} = 1080 \, \text{kg} \cdot \text{m}^2 + 480 \, \text{kg} \cdot \text{m}^2 = 1560 \, \text{kg} \cdot \text{m}^2 \]

6. **Apply conservation of angular momentum:**
The total angular momentum after the child jumps on is given by:
\[ L_{\text{final}} = I_{\text{total}} \cdot \omega \]
Setting the initial and final angular momentum equal:
\[ 600 = 1560 \cdot \omega \]
Solving for \( \omega \):
\[ \omega = \frac{600}{1560} = \frac{15}{39} = \frac{5}{13} \approx 0.3846 \, \text{rad/s} \]

### Final Answer:
The angular velocity of the merry-go-round and the child after the child jumps on is approximately \( 0.3846 \, \text{rad/s} \).